\iota_S(M) \otimes^{\dagger} \iota_S(L) \arrow{u}{\tilde{\rho}^{\dagger}} \arrow[equal]{rr} && (\iota_S(M) \otimes \iota_S(L))[-d] \arrow{u}{\tilde{\rho}}
\end{tikzcd}
\end{equation}

Most of the rest of the section is devoted to the technical proof of this result. Before discussing the details, let us briefly explain what is the general strategy behind the proof and why implementing it correctly turns out to be quite subtle.
Consider the diagram

\begin{equation*}
\begin{tikzcd}
\Dcal^0(S) \times \Dcal^{0,loc}(S) \arrow{rr}{- \boxtimes -} \arrow{d}{\beta_S^0 \times \beta_S^{0,loc}} && \Dcal^0(S \times S) \arrow{rr}{\Delta_S^*[-d]} \arrow{d}{\beta_{S \times S}^0} && \Dcal(S) \arrow{d}{\Bti_S^*} \\
\Perv(S) \times \Loc_p(S) \arrow{rr}{- \boxtimes -} && \Perv(S \times S) \arrow{rr}{\Delta_S^*[-d]} && D^b(\Perv(S))
\end{tikzcd}
\end{equation*}

where both squares are commutative up to canonical natural isomorphism. By construction, the vertical and horizontal composites recover (the restrictions of) the shifted tensor product functors (<ref>) and (<ref>), respectively; in particular they take values in $\Dcal^0(S)$ and $\Perv(S)$, respectively.
If the functor

\begin{equation}\label{Delta^*-Perv}
\Delta_S^*[-d]: D^b(\Perv(S \times S)) \rightarrow D^b(\Perv(S))
\end{equation}

was $t$-exact for the obvious $t$-structures, then the corresponding functor $\Delta_S^*[-d]: \Dcal(S \times S) \rightarrow \Dcal(S)$ would send $\Dcal^0(S \times S)$ to $\Dcal^0(S)$. Hence one could refine the previous diagram to a diagram of the form

\begin{equation*}
\begin{tikzcd}
\Dcal^0(S) \times \Dcal^{0,loc}(S) \arrow{rr}{- \boxtimes -} \arrow{d}{\beta_S^0 \times \beta_S^{0,loc}} && \Dcal^0(S \times S) \arrow{rr}{\Delta_S^*[-d]} \arrow{d}{\beta_{S \times S}^0} && \Dcal^0(S) \arrow{d}{\beta^0_S} \\
\Perv(S) \times \Loc_p(S) \arrow{rr}{- \boxtimes -} && \Perv(S \times S) \arrow{rr}{\Delta_S^*[-d]} && \Perv(S)
\end{tikzcd}
\end{equation*}

and one could try to deduce that the functor (<ref>) is naturally isomorphic to the composite

\begin{equation*}
\Mcal^0(S) \times \Mcal^{0,loc}(S) \xrightarrow{- \boxtimes -} \Mcal^0(S \times S) \xrightarrow{\Delta_S^*[-d]} \Mcal^0(S)
\end{equation*}

by using the general functoriality of universal abelian factorizations from [55]. If this was possible, then the conclusion of <Ref> would follow trivially from the fact that lifting exact functors to universal abelian factorizations is compatible with composition.

Needless to say, the functor (<ref>) is not $t$-exact, hence the argument sketched here simply does not work. On the other hand, it is well-known that the composite functor

\begin{equation*}
\Delta_{S,*} \circ \Delta_S^*[-d]: D^b(\Perv(S \times S)) \rightarrow D^b(\Perv(S \times S))
\end{equation*}

can be modeled as an exact functor $C^b(\Perv(S \times S)) \rightarrow C^b(\Perv(S \times S))$ by means of a suitable Čech complex construction. For instance, in the simple case where $S$ is the affine line $\A^1_k$, one can use the total complex associated to the exact functor $C^b(\Perv(\A^2_k)) \rightarrow C^{[0,1]}(C^b(\Perv(\A^2_k)))$

\begin{equation}\label{cech:A1}
K^{\bullet} \rightsquigarrow [j_! j^* K^{\bullet} \rightarrow K^{\bullet}],
\end{equation}

where $j$ denotes the affine open immersion of the complement of the diagonal in $\A^2_k$. The key advantage is that the complex in (<ref>) lifts canonically to the level of perverse motives.
In general, the model for $\Delta_{S,*} \circ \Delta_S^*[-d]$ depends on the choice of a finite affine covering of the open complement of $\Delta_S(S)$ in $S \times S$, but the derived functor of the latter is essentially independent of such a choice. In fact, this idea is exploited in the proof of <cit.> to construct the inverse image functor $\Delta_S^*$ - or, more generally, inverse image functors along closed immersions - in the setting of perverse motives. We are going to prove <Ref> by a similar method.

§.§ Some Čech complex computations

For the reader's convenience, we start by recalling the Čech complex construction just mentioned in the setting of perverse sheaves and perverse motives.

Let $s: S \hookrightarrow Y$ be a closed immersion in $\Var_k$, and let $u: U \hookrightarrow Y$ denote the complementary open immersion.
Suppose that we are given a finite affine open covering $\Ucal = \left\{U_i \subset U \right\}_{i = 1}^{d}$ of $U$. For every subset $I \subset \left\{1,\dots,d\right\}$, set

\begin{equation*}
U_I := \bigcap_{i \in I} U_i
\end{equation*}

and let $j_I: U_I \hookrightarrow Y$ denote the corresponding open immersion (with the agreement that $U_{\emptyset} = Y$ and $j_{\emptyset} = \id_Y$). Note that each $j_I$ is affine, because $Y$ is separated over $k$ by definition. For every choice of subsets $I, I' \subset \left\{1,\dots,d\right\}$ satisfying $I' \subset I$, there exists a canonical natural transformation of functors $\Perv(Y) \rightarrow \Perv(Y)$

\begin{equation*}
\alpha^{\Ucal}_{I,I'}: j_{I,!} j_I^* K \rightarrow j_{I',!} j_{I'}^* K
\end{equation*}

induced by the co-unit along the inclusion of $U_I$ into $U_{I'}$. For each $r = 0,\dots,d$, set

\begin{equation*}
U^{(r)}_{\Ucal} := \coprod_{\substack{I \subset \left\{1,\dots,d \right\} \\ \# I = r}} U_I
\end{equation*}

and let $J_r: U^{(r)} \rightarrow Y$ denote the canonical morphism. We introduce the Čech complex functor

\begin{equation*}
\check{C}^{\bullet}_{Y,\Ucal}: \Perv(Y) \rightarrow C^{[-d,0]}(\Perv(Y)), \quad K \rightsquigarrow [J_{d,!} J_d^* K \to \dots \to J_{1,!} J_1^* K \to K],
\end{equation*}

where, for each $r = 1,\dots,d$, the arrow

\begin{equation*}
J_{r,!} J_r^* K = \bigoplus_{\substack{I \subset \left\{1,\dots,d\right\} \\ \# I = r}} j_{I,!} j_I^* K \rightarrow \bigoplus_{\substack{I' \subset \left\{1,\dots,d\right\} \\ \# I' = r-1}} j_{I',!} j_{I'}^* K = J_{r-1,!} J_{r-1}^* K
\end{equation*}

is induced by the usual alternating sum of the arrows $\alpha^{\Ucal}_{I,I'}$. We extend the Čech complex functor to an exact functor

\begin{equation*}
\check{C}_{Y,\Ucal}: C^b(\Perv(Y)) \xrightarrow{\check{C}_{Y,\Ucal}^{\bullet}} C^{[-d,0]}(C^b(\Perv(Y))) \xrightarrow{\Tot^{\bullet}} C^b(\Perv(Y))
\end{equation*}

and we consider its trivial derived functor

\begin{equation}\label{cech-D^bPcal}
\check{C}_{Y,\Ucal}: D^b(\Perv(Y)) \rightarrow D^b(\Perv(Y)).
\end{equation}

The argument used in the proof of <cit.> implies that the functor (<ref>) is canonically left adjoint to the inclusion $D^b_{S}(\Perv(Y)) \hookrightarrow D^b(\Perv(Y))$, hence canonically isomorphic to (the co-restriction of) the functor $s_* s^*: D^b(\Perv(Y)) \rightarrow D^b(\Perv(Y))$.

The construction carries over to the setting of perverse motives: one can define a Čech complex functor

\begin{equation*}
\check{C}^{\bullet}_{Y,\Ucal}: \Mcal^0(Y) \rightarrow C^{[-d,0]}(\Mcal^0(Y)), \quad M \rightsquigarrow [J_{d,!} J_d^* M \to \dots \to J_{1,!} J_1^* M \to M]
\end{equation*}

and use it to obtain a triangulated functor

\begin{equation*}
\check{C}_{Y,\Ucal}: D^b(\Mcal^0(Y)) \rightarrow D^b(\Mcal^0(Y))
\end{equation*}

which is canonically left adjoint to the inclusion $D^b_{S}(\Mcal^0(Y)) \hookrightarrow D^b(\Mcal^0(Y))$, hence canonically isomorphic to (the co-restriction of) the functor $s_* s^*: D^b(\Mcal^0(Y)) \rightarrow D^b(\Mcal^0(Y))$.
Note that we have a canonical natural isomorphism between exact functors $C^b(\Mcal^0(Y)) \rightarrow C^b(\Perv(Y))$

\begin{equation*}
\check{C}_{Y,\Ucal} (\iota_Y(M^{\bullet})) \xrightarrow{\sim} \iota_Y(\check{C}_{Y,\Ucal}(M^{\bullet}))
\end{equation*}

defined as the composite

\begin{equation*}
\begin{tikzcd}
\check{C}_{Y,\Ucal} (\iota_Y(M^{\bullet})) \arrow[equal]{d} && \iota_Y(\check{C}_{Y,\Ucal}(M^{\bullet})) \arrow[equal]{d} \\
\Tot^{\bullet} \check{C}^{\bullet}_{Y,\Ucal}(\iota_Y(M^{\bullet})) \arrow{r}{\sim} & \Tot^{\bullet} \iota_Y(\check{C}^{\bullet}_{Y,\Ucal}(M^{\bullet})) \arrow[equal]{r} & \iota_Y(\Tot^{\bullet} \check{C}^{\bullet}_{Y,\Ucal}(M^{\bullet}))
\end{tikzcd}
\end{equation*}

where the left-most horizontal arrow is induced by the family of natural isomorphisms of functors $C^b(\Mcal^0(Y)) \rightarrow C^b(\Perv(Y))$

\begin{equation*}
\check{C}^r_{Y,\Ucal}(\iota_Y(M^{\bullet})) := J_{r,!} J_r^* \iota_Y(M^{\bullet}) \xrightarrow{\sim} J_{r,!} \iota_{U_{\Ucal}^{(r)}}(J_r^* M^{\bullet}) \xrightarrow{\sim} \iota_Y (J_{r,!} J_r^* M^{\bullet}) =: \iota_Y(\check{C}^r_{Y,\Ucal}(M^{\bullet}))
\end{equation*}

as $r = 0,\dots,d$ varies. In the following, we consider the induced natural isomorphism of triangulated functors $D^b(\Mcal^0(Y)) \rightarrow D^b(\Perv(Y))$

\begin{equation*}
\check{C}_{Y,\Ucal} (\iota_Y(M^{\bullet})) \xrightarrow{\sim} \iota_Y(\check{C}_{Y,\Ucal}(M^{\bullet})).
\end{equation*}

We need to provide an analogous Čech complex construction on the level of triangulated étale motives. This is possible since the stable homotopic $2$-functor $\DA_{ct}^{\et}(\cdot,\Q)$ canonically underlies a stable homotopic algebraic derivator in the sense of <cit.>.

Keep the notation and assumptions of <Ref>.
For every choice of subsets $I, I' \subset \left\{1,\dots,d\right\}$ satisfying $I' \subset I$, consider the canonical natural transformation of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}(Y,\Q)$

\begin{equation*}
\alpha^{\Ucal}_{I,I'}: j_{I,!} j_I^* A \rightarrow j_{I',!} j_{I'}^* A
\end{equation*}

induced by the co-unit along the inclusion of $U_I$ into $U_{I'}$. By construction, the diagram of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow D^b_c(Y,\Q)$

\begin{equation*}
\begin{tikzcd}
j_{I,!} j_I^* \Bti_Y^*(A) \arrow{r}{\alpha^{\Ucal}_{I,I'}} \isoarrow{d} & j_{I',!} j_{I'}^* \Bti_Y^*(A) \isoarrow{d} \\
j_{I,!} \Bti_{U_I}^* (j_I^* A) \isoarrow{d} &  j_{I',!} \Bti_{U_{I'}}^* (j_{I'}^* A) \isoarrow{d} \\
\Bti_Y^* (j_{I,!} j_I^* A) \arrow{r}{\alpha^{\Ucal}_{I,I'}} & \Bti_Y^* (j_{I',!} j_{I'}^* A)
\end{tikzcd}
\end{equation*}

is commutative under Beilinson's equivalence $D^b(\Perv(Y)) \xrightarrow{\sim} D^b_c(Y,\Q)$.
Let $[-1,0]$ denote the partially ordered set $\left\{-1 < 0\right\}$; for every integer $n \geq 0$ we regard the product category $[-1,0]^n$ as a small diagram (with the convention that $[-1,0]^0$ is just the singleton category). We are particularly interested in the diagram $[-1,0]^d$: in this case, for every subset $I \subset \left\{0,\dots,d\right\}$, we let $\underline{I}$ denote the object $(\epsilon_1,\dots,\epsilon_d) \in [-1,0]^d$ whose $i$-th coordinate $\epsilon_i$ equals $-1$ if $i \in I$ and $0$ otherwise. Note that the assignment $I \rightsquigarrow \underline{I}$ defines an anti-isomorphism between the powerset of $\left\{1,\dots,d\right\}$ (regarded as a poset with the usual inclusion ordering) and the category $[-1,0]^d$.
Given two subsets $I, I' \subset \left\{1,\dots,d\right\}$ satisfying $I' \subset I$, we let $e_{I,I'}: \underline{I} \rightarrow \underline{I}'$ denote the corresponding morphism in $[-1,0]^d$.
Following the idea of <cit.>, we can construct a triangulated functor

\begin{equation*}
\check{C}^{\square}_{Y,\Ucal}: \DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}((Y,[-1,0]^{d}),\Q)
\end{equation*}

providing a coherent lifting of the $[-1,0]^d$-indexed diagram of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}(Y,\Q)$

\begin{equation*}
\underline{I}(A) := j_{I,!} j_I^* A, \qquad e_{I,I'}(A) := \alpha_{I,I'}^{\Ucal}(A): j_{I,!} j_I^* A \rightarrow j_{I',!} j_{I'}^* A.
\end{equation*}

Concretely, one can consider the two obvious morphisms of diagram of $k$-varieties

\begin{equation*}
\begin{tikzcd}
(\underline{I} \mapsto U_I)_{\underline{I} \in [-1,0]^d} \arrow{r}{\xi} \arrow{d}{\nu} & (\underline{I} \mapsto Y)_{\underline{I} \in [-1,0]^d} \\
\end{tikzcd}
\end{equation*}

and define the functor $\check{C}^{\square}_{Y,\Ucal}$ by the formula

\begin{equation*}
\check{C}^{\square}_{Y,\Ucal} := \xi_! \circ \nu^*.
\end{equation*}

Now, for every integer $n \geq 1$, let

\begin{equation*}
\Cof_n: \DA_{ct}^{\et}((Y,[-1,0]^n),\Q) \rightarrow \DA_{ct}^{\et}((Y,[-1,0]^{n-1}),\Q)
\end{equation*}

denote the co-fiber with respect to the last coordinate of $[-1,0]^n = [-1,0]^{n-1} \times [-1,0]$. We let

\begin{equation*}
\Tot^{\square} := \Cof_1 \circ \dots \circ \Cof_d: \DA_{ct}^{\et}((Y,[-1,0]^d),\Q) \rightarrow \DA_{ct}^{\et}(Y,\Q)
\end{equation*}

denote the associated derived total complex functor, and we set

\begin{equation*}
\check{C}_{Y,\Ucal} := \Tot^{\square} \circ \check{C}^{\square}_{Y,\Ucal}: \DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}(Y,\Q).
\end{equation*}

Similarly, we can define a functor

\begin{equation*}
\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}: \DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}((Y,[-1,0]^{d}),\Q)
\end{equation*}

providing a coherent lifting of the $[-1,0]^d$-indexed diagram of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}(Y,\Q)$

\begin{equation*}
\underline{I}(A) :=
\begin{cases}
J_{d,!} J_d^* A, & I = \left\{1,\dots,d\right\} \\
0, & I \neq \left\{1,\dots,d\right\}.
\end{cases}
\end{equation*}

Note that there is an obvious natural transformation of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow \DA_{ct}^{\et}((Y,[-1,0]^d),\Q)$

\begin{equation}\label{nat:cech-sigmacech}
\check{C}^{\square}_{Y,\Ucal}(A) \rightarrow \sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(A).
\end{equation}

One can repeat the entire construction replacing $\DA_{ct}^{\et}(\cdot,\Q)$ with any stable homotopic algebraic derivator in the sense of <cit.>. As a consequence of <cit.>, the stable homotopic $2$-functor $D^b_c(\cdot,\Q)$ canonically underlies a stable homotopic algebraic derivator, and the Betti realization

\begin{equation*}
\Bti^*: \DA_{ct}^{\et}(\cdot,\Q) \rightarrow D^b_c(\cdot,\Q)
\end{equation*}

can be canonically extended over diagrams of $k$-varieties. Thus we can define a functor

\begin{equation*}
\check{C}^{\square}_{Y,\Ucal}: D^b_c(Y,\Q) \rightarrow D^b_c((Y,[-1,0]^{d}),\Q)
\end{equation*}

providing a coherent lifting of the $[-1,0]^d$-indexed diagram of functors $D^b_c(Y,\Q) \rightarrow D^b_c(Y,\Q)$

\begin{equation*}
\underline{I}(K^{\bullet}) := j_{I,!} j_I^* K^{\bullet}, \qquad e_{I,I'}(A) := \alpha_{I,I'}^{\Ucal}(K^{\bullet}): j_{I,!} j_I^* K^{\bullet} \rightarrow j_{I',!} j_{I'}^* K^{\bullet},
\end{equation*}

as well as a functor

\begin{equation*}
\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}: D^b_c(Y,\Q) \rightarrow D^b_c((Y,[-1,0]^{d}),\Q)
\end{equation*}

providing a coherent lifting of the $[-1,0]^d$-indexed diagram of functors $D^b_c(Y,\Q) \rightarrow D^b_c(Y,\Q)$

\begin{equation*}
\underline{I}(K^{\bullet}) :=
\begin{cases}
J_{d,!} J_d^* K^{\bullet}, & I = \left\{1,\dots,d\right\} \\
0, & I \neq \left\{1,\dots,d\right\}.
\end{cases}
\end{equation*}

As in the previous case, they are related by a canonical natural transformation of functors $D^b_c(Y,\Q) \rightarrow D^b_c((Y,[-1,0]^d),\Q)$

\begin{equation*}
\check{C}^{\square}_{Y,\Ucal}(K^{\bullet}) \rightarrow \sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(K^{\bullet}).
\end{equation*}

Note that, moreover, there exist canonical natural isomorphisms of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow D^b_c((Y,[-1,0]^d),\Q)$

\begin{equation*}
\check{C}^{\square}_{Y,\Ucal} \Bti_Y^*(A) \xrightarrow{\sim} \Bti_{(Y,[-1,0]^d)}^*(\check{C}^{\square}_{Y,\Ucal}(A))
\end{equation*}

\begin{equation*}
\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal} \Bti_Y^*(A) \xrightarrow{\sim} \Bti_{(Y,[-1,0]^d)}^*(\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(A))
\end{equation*}

rendering the diagram of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow D^b_c((Y,[-1,0]^d),\Q)$

\begin{equation*}
\begin{tikzcd}
\check{C}^{\square}_{Y,\Ucal}(\Bti_Y^*(A)) \arrow{r}{\sim} \arrow{d} & \Bti_{(Y,[-1,0]^d)}^* (\check{C}^{\square}_{Y,\Ucal}(A)) \arrow{d} \\
\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(\Bti_Y^*(A)) \arrow{r}{\sim} & \Bti_{(Y,[-1,0]^d)}^* (\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(A))
\end{tikzcd}
\end{equation*}

commutative. We can also define a total complex functor

\begin{equation*}
\Tot^{\square} := \Cof_1 \circ \dots \circ \Cof_d: D^b_c((Y,[-1,0]^d),\Q) \rightarrow D^b_c(Y,\Q).
\end{equation*}

By construction, the composite

\begin{equation*}
\check{C}_{Y,\Ucal} := \Tot^{\square} \circ \check{C}^{\square}_{Y,\Ucal}: D^b_c(Y,\Q) \rightarrow D^b_c(Y,\Q)
\end{equation*}

recovers the functor (<ref>) under Beilinson's equivalence $D^b(\Perv(Y)) \xrightarrow{\sim} D^b_c(Y,\Q)$. Therefore we obtain a canonical natural isomorphism of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow D^b_c(Y,\Q)$

\begin{equation*}
\check{C}_{Y,\Ucal} \Bti_Y(A) \xrightarrow{\sim} \Bti_Y^* (\check{C}_{Y,\Ucal}(A))
\end{equation*}

by taking the composite

\begin{equation*}
\begin{tikzcd}
\check{C}_{Y,\Ucal} \Bti_Y(A) \arrow[equal]{d} && \Bti_Y^* (\check{C}_{Y,\Ucal}(A)). \arrow[equal]{d} \\
\Tot^{\square} \check{C}^{\square}_{Y,\Ucal} (\Bti_Y^*(A)) \arrow{r}{\sim} & \Tot^{\square} \Bti_{Y,[-1,0]^d}^* (\check{C}^{\square}_{Y,\Ucal}(A)) \arrow[equal]{r} & \Bti_Y^*(\Tot^{\square} \check{C}^{\square}_{Y,\Ucal}(A))
\end{tikzcd}
\end{equation*}

By construction, the diagram of functors $\DA_{ct}^{\et}(Y,\Q) \rightarrow D^b_c(Y,\Q)$

\begin{equation}\label{dia:cech-s_*s^*-Bti}
\begin{tikzcd}
\check{C}_{Y,\Ucal} \Bti_Y(A) \arrow{rr}{\sim} \arrow[equal]{d} && \Bti_Y^* (\check{C}_{Y,\Ucal}(A)) \arrow[equal]{d} \\
s_* s^* \Bti_Y^*(A) \arrow{r}{\theta} & s_* \Bti_S^* (s^* A) & \Bti_Y^*(s_* s^* A) \arrow{l}{\bar{\theta}}
\end{tikzcd}
\end{equation}

is commutative.

We need to compare the total Čech complex functor on the level of perverse motives with the analogous functor on triangulated motives. This turns out to be quite subtle, since the two functors are constructed in different ways. In order to proceed, it is convenient to introduce yet a further subdiagram of $\Dcal^0(S)$:

Given a closed immersion $s: S \hookrightarrow Y$ in $\Sm_k$, we introduce the full additive subcategory

\begin{equation*}
\Dcal^{0,s}(Y) := \left\{B \in \Dcal^0(Y) \; | \; s^{\dagger} B \in \Dcal^0(S) \right\} \subset \Dcal^0(Y).
\end{equation*}

We can state the promised result about Čech complex functors as follows:

Let $s: S \hookrightarrow Y$ be a closed immersion of constant codimension $d$ in $\Sm_k$. Suppose that the open complement $U$ of $S$ in $Y$ admits a finite affine open covering $\Ucal$ with exactly $d$ elements. Then there exist a canonical natural isomorphism of functors $\Dcal^{0,s}(Y) \rightarrow \Mcal^0(Y)$

\begin{equation}\label{cech-Dcal-Mp}
\pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \xrightarrow{\sim} \check{C}_{Y,\Ucal}[-d] (\pi_Y(B))
\end{equation}

such that the diagram of functors $\Dcal^{0,s}(Y) \rightarrow \Perv(Y)$

\begin{equation}\label{dia:cech-lem}
\begin{tikzcd}
\Bti_Y^* (\check{C}_{Y,\Ucal}[-d] B) \arrow[equal]{d} & \check{C}_{Y,\Ucal}[-d] \Bti_Y^*(B) \arrow[equal]{r} \arrow{l}{\sim} & \check{C}_{Y,\Ucal}[-d] \iota_Y \pi_Y(B) \isoarrow{d} \\
\iota_Y \pi_Y (\check{C}_{Y,\Ucal}[-d] B) \arrow{rr}{\sim} && \iota_Y (\check{C}_{Y,\Ucal}[-d] \pi_Y(B))
\end{tikzcd}
\end{equation}

is commutative.

First of all, let us check that the composite functor

\begin{equation*}
\check{C}_{Y,\Ucal}[-d] \circ \pi_Y: \Dcal^{0,s}(Y) \rightarrow \Mcal^0(Y) \rightarrow D^b(\Mcal^0(Y))
\end{equation*}

really takes values in $\Mcal^0(Y)$ - or, equivalently, that its composition with the conservative $t$-exact functor $\iota_Y: D^b(\Mcal^0(Y)) \rightarrow D^b(\Perv(Y))$ takes values in $\Perv(Y)$. To this end, for every object $B \in \Dcal^{0,s}(Y)$ we compute

\begin{align*}
\iota_Y (\check{C}_{Y,\Ucal}[-d](\pi_Y(B))) = & \check{C}_{Y,\Ucal}[-d] (\iota_Y \pi_Y(B)) \\
= & \check{C}_{Y,\Ucal}[-d](\beta_Y(B)) \\
= & s_* s^*[-d] \beta_Y(B) \\
= & s_* s^*[-d] \Bti_Y^*(B) && \textup{as $B \in \Dcal^0(Y)$}\\
= & s_* \Bti_S^*(s^*[-d] B) \\
= & s_* \beta_S(s^*[-d] B)  && \textup{as $B \in \Dcal^{0,s}(Y)$}.
\end{align*}

The latter object belongs indeed to $\Perv(Y)$ in view of the definition of $\beta_S$ and by the $t$-exactness of $s_*: D^b_c(S,\Q) \rightarrow D^b_c(Y,\Q)$.

In order to define the sought-after natural isomorphism (<ref>), we start by constructing a natural transformation between exact functors $\Dcal^0(Y) \rightarrow C^{[-d,0]}(\Mcal^0(Y))$

\begin{equation*}
\pi_Y(\check{C}_{Y,\Ucal}[-d](B))[d] \rightarrow \check{C}^{\bullet}_{Y,\Ucal}(\pi_Y(B)).
\end{equation*}

With the notation of <Ref> and <Ref>, this amounts to giving a natural transformation of functors $\Dcal^0(Y) \rightarrow \Mcal^0(Y)$

\begin{equation}\label{nat:cech-tr-complex-d}
\pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \rightarrow J_{d,!} J_d^* \pi_Y(B)
\end{equation}

such that the composite

\begin{equation}\label{nat:cech-tr-composite-d}
\pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \rightarrow J_{d,!} J_d^* \pi_Y(B) \rightarrow J_{d-1,!} J_{d-1}^* \pi_Y(B) = \bigoplus_{\substack{I \subset \left\{1,\dots,d\right\} \\ \# I = d-1}} j_{I,!} j_I^* \pi_Y(B)
\end{equation}

vanishes. We define (<ref>) by taking the composite

\begin{equation*}
\begin{tikzcd}
\pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \arrow[equal]{d} && J_{d,!} J_d^* \pi_Y(B), \arrow[equal]{d} \\
\pi_Y(\Tot^{\square}[-d] \check{C}^{\square}_{Y,\Ucal}(B)) \arrow{r} & \pi_Y(\Tot^{\square}[-d] \sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(B)) \arrow[equal]{r} & \pi_Y(J_{d,!} J_d^* B)
\end{tikzcd}
\end{equation*}

where the first horizontal arrow is induced by (<ref>).
The vanishing of (<ref>) follows from the existence of a commutative diagram of the form

\begin{equation*}
\begin{tikzcd}
\pi_Y(\check{C}_{Y,\Ucal}[-d] B) \arrow{r} \arrow{dr} & \pi_Y(J_{d,!} J_d^* B) \arrow{r} \arrow[equal]{d} & \bigoplus_{\substack{I \subset \left\{1,\dots,d\right\} \\ \# I = d-1}} \pi_Y(j_{I,!} j_I^* B) \arrow[equal]{d} \\
& J_{d,!} J_d^* \pi_Y(B) \arrow{r} & \bigoplus_{\substack{I \subset \left\{1,\dots,d\right\} \\ \# I = d-1}} j_{I,!} j_I^* \pi_Y(B)
\end{tikzcd}
\end{equation*}

where the upper horizontal composite vanishes by construction.
We deduce a natural transformation of functors $\Dcal^0(Y) \rightarrow D^{[0,d]}(\Mcal^0(Y))$

\begin{equation}\label{cech-D^0Y}
\begin{tikzcd}
\pi_Y(\check{C}_{Y,\Ucal}[-d] B) \arrow[equal]{d} & \check{C}_{Y,\Ucal}[-d](\pi_Y(B)). \arrow[equal]{d} \\
\Tot^{\bullet} \pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \arrow{r} & \Tot^{\bullet} \check{C}^{\bullet}_{Y,\Ucal}[-d](\pi_Y(B))
\end{tikzcd}
\end{equation}

In an analogous way, we obtain a natural transformation of functors $\Dcal^0(Y) \rightarrow D^{[-d,0]}(\Perv(Y))$

\begin{equation}\label{cech-D^0PervY}
\beta_Y(\check{C}_{Y,\Ucal}[-d] (B)) \rightarrow \check{C}_{Y,\Ucal}[-d] (\beta_Y(B))
\end{equation}

making the diagram of functors $\Dcal^0(Y) \rightarrow D^{[0,d]}(\Perv(Y))$

\begin{equation}\label{dia:cech-iota-pi}
\begin{tikzcd}[font=\small]
\beta_Y(\check{C}_{Y,\Ucal}[-d](B)) \arrow[equal]{dd} \arrow{r} & \check{C}_{Y,\Ucal}[-d] (\beta_Y(B)) \arrow[equal]{d} \\
& \check{C}_{Y,\Ucal}[-d] (\iota_Y \pi_Y(B)) \isoarrow{d} \\
\iota_Y \pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \arrow{r} & \iota_Y(\check{C}_{Y,\Ucal}[-d](\pi_Y(B)))
\end{tikzcd}
\end{equation}

commutative. We claim that the value of (<ref>) on a given object $B \in \Dcal^0(Y)$ is an isomorphism if and only if $B \in \Dcal^{0,s}(Y)$. Using the commutativity of the diagram (<ref>) and the conservativity of $\iota_Y: D^{[0,d]}(\Mcal^0(Y)) \rightarrow D^{[0,d]}(\Perv(Y))$, we see that this is equivalent to showing that the value of (<ref>) on a given object $B \in \Dcal^0(Y)$ is an isomorphism if and only if $B \in \Dcal^{0,s}(Y)$. To this end, taking the image under ${^p H^0}$ of the commutative diagram (<ref>), the arrow (<ref>) can be identified with the canonical arrow in $\Perv(Y)$

\begin{equation}\label{arrow-cech-s_*s^*}
{^p H^0} \Bti_Y^*(s_* s^{\dagger} B) = {^p H^0} s_* \Bti_S^*(s^{\dagger} B) = s_* {^p H^0} \Bti_S^* (s^{\dagger} B) \rightarrow s_* \Bti_S^* (s^{\dagger} B).
\end{equation}

Here, the latter arrow in the composite can be also written as

\begin{equation*}
s_* {^p H^0} s^{\dagger} \Bti_Y^*(B) \rightarrow s_* s^{\dagger} \Bti_Y^*(B)
\end{equation*}

and is induced by the left-$t$-exactness of the functor $s^{\dagger} = s^*[-d]: D^b_c(Y,\Q) \rightarrow D^b_c(S,\Q)$ (which follows precisely from the fact that $U$ admits an affine open covering with $d$ elements). Using the conservativity of $s_*: D^b_c(S,\Q) \rightarrow D^b_c(Y,\Q)$, we see that the arrow (<ref>) is invertible if and only if $ \Bti_S^*(s^{\dagger} B) \in \Perv(S)$.

We define the natural isomorphism (<ref>) by taking the restriction of (<ref>) to $\Dcal^{0,s}(Y)$. By construction, the diagram (<ref>) coincides with the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
\Bti_Y^* (\check{C}_{Y,\Ucal}[-d] B) \arrow[equal]{dr} \arrow[equal]{dd} &&& \check{C}_{Y,\Ucal}[-d] \Bti_Y^*(B) \arrow[equal]{d} \arrow{lll}{\sim} \\
& \beta_Y (\check{C}_{Y,\Ucal}[-d] B) \arrow{r}{\sim} \arrow[equal]{dl} & \check{C}_{Y,\Ucal}[-d] \beta_Y(B) \arrow[equal]{ur} \arrow[equal]{r} & \check{C}_{Y,\Ucal}[-d] \iota_Y \pi_Y(B) \isoarrow{d} \\
\iota_Y \pi_Y (\check{C}_{Y,\Ucal}[-d] B) \arrow{rrr}{\sim} &&& \iota_Y (\check{C}_{Y,\Ucal}[-d] \pi_Y(B))
\end{tikzcd}
\end{equation*}

where the two triangles are commutative by definition and the bottom piece is already known to be commutative. Hence, in order to conclude the proof it suffices to show that the top piece is commutative as well. Unwinding the various definitions, the commutativity of the upper piece follows from the commutativity of the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
\Bti_Y^*(\Tot^{\square}[-d] \check{C}^{\square}_{Y,\Ucal}(B)) \arrow[equal]{r} \arrow{d} & \Tot^{\square}[-d] \Bti_{(Y,[-1,0]^d)}^* (\check{C}^{\square}_{Y,\Ucal}(B)) \arrow{d} & \Tot^{\square}[-d] \check{C}^{\square}_{Y,\Ucal}(\Bti_Y^*(B)) \arrow{l}{\sim} \arrow{d} \\
\Bti_Y^* (\Tot^{\square}[-d] \sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(B)) \arrow[equal]{d} \arrow[equal]{r} & \Tot^{\square}[-d] \Bti_{(Y,[-1,0]^d)}^* (\sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(B)) & \Tot^{\square}[-d] \sigma_{\leq (-1,\dots,-1)} \check{C}^{\square}_{Y,\Ucal}(\Bti_Y^*(B)) \arrow[equal]{d} \arrow{l}{\sim} \\
\Bti_Y^* (J_{d,!} J_d^* B) \arrow[equal]{d} && J_{d,!} J_d^* \Bti_Y^*(B) \arrow{ll}{\sim} \arrow[equal]{d} \\
\beta_Y (J_{d,!} J_d^* B) && J_{d,!} J_d^* \beta_Y(B) \arrow{ll}{\sim}
\end{tikzcd}
\end{equation*}

which in turn follows from the commutativity of the four inner pieces, which holds by construction. This concludes the proof.

Let $s: S \hookrightarrow Y$ be a closed immersion of constant codimension $d$ in $\Sm_k$. Then there exists a natural isomorphism of functors $\Dcal^{0,s}(Y) \rightarrow \Mcal^0(S)$

\begin{equation}\label{formula:zeta_S}
\zeta_s: \pi_S(s^{\dagger} B) \xrightarrow{\sim} s^{\dagger} \pi_Y(B)
\end{equation}

such that the diagram of functors $\Dcal^{0,s}(Y) \rightarrow \Perv(S)$

\begin{equation}\label{dia:zeta_S}
\begin{tikzcd}
\Bti_S^* (s^{\dagger} B) \arrow[equal]{dd} & s^{\dagger} \Bti_Y^*(B) \arrow{l}{\theta} \arrow[equal]{d} \\
& s^{\dagger} \iota_Y \pi_Y(B) \arrow{d}{\tilde{\theta}} \\
\iota_S \pi_S (s^{\dagger} B) \arrow{r}{\zeta_s} & \iota_S(s^{\dagger} \pi_V(B))
\end{tikzcd}
\end{equation}

is commutative.

Note that, for every closed immersion $s: S \hookrightarrow Y$ as in the statement, the faithfulness and conservativity of the functor $\iota_S: \Mcal^0(S) \hookrightarrow \Perv(S)$ implies that a natural transformation as in (<ref>) rendering the diagram (<ref>) commutative is necessarily uniquely determined and invertible, provided it exists.

We start with a preliminary observation. Let $s: S \hookrightarrow Y$ be as in the statement; let $v: V \hookrightarrow Y$ be an open immersion, and form the Cartesian square in $\Sm_k$

\begin{equation*}
\begin{tikzcd}
U \arrow{r}{s_U} \arrow{d}{u} & V \arrow{d}{v} \\
S \arrow{r}{s} & Y.
\end{tikzcd}
\end{equation*}

Suppose for a moment that the thesis is known to hold both for $s$ and for $s_U$. We claim that, in this situation, the diagram of functors $\Dcal^{0,s}(Y) \rightarrow \Mcal^0(U)$

\begin{equation*}
\begin{tikzcd}[font=\small]
u^* \pi_S(s^{\dagger} B) \arrow{r}{\zeta_s} & u^* s^{\dagger} \pi_Y(B) \\
\pi_U(u^* s^{\dagger} B) \arrow[equal]{u} \arrow[equal]{d} & s^{\dagger} v^* \pi_Y(B) \arrow[equal]{u} \arrow[equal]{d} \\
\pi_U(s_U^{\dagger} v^* B) \arrow{r}{\zeta_{s_U}} & s_U^{\dagger} \pi_V(v^* B)
\end{tikzcd}
\end{equation*}

is commutative. In order to show this, it suffices to check the commutativity of its image under the faithful functor $\iota_U: \Mcal^0(U) \hookrightarrow \Perv(U)$

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_U (u^* \pi_S(s^{\dagger} B)) \arrow{r}{\zeta_s} & \iota_U (u^* s^{\dagger} \pi_Y(B)) \\
\iota_U \pi_U(u^* s^{\dagger} B) \arrow[equal]{u} \arrow[equal]{d} & \iota_U (s_U^{\dagger} v^* \pi_Y(B)) \arrow[equal]{u} \arrow[equal]{d} \\
\iota_U \pi_U(s_U^{\dagger} v^* B) \arrow{r}{\zeta_{s_U}} & \iota_U (s_U^{\dagger} \pi_V(v^* B))
\end{tikzcd}
\end{equation*}

that we can decompose as

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_U (u^* \pi_S(s^{\dagger} B)) \arrow{rrrr}{\zeta_s} &&&& \iota_U (u^* s^{\dagger} \pi_Y(B)) \\
& u^* \iota_S \pi_S (s^{\dagger} B) \arrow{ul}{\tilde{\theta}} \arrow{rr}{\zeta_s} && u^* \iota_S(s^{\dagger} \pi_Y(B)) \arrow{ur}{\tilde{\theta}} \\
& u^* \Bti_S^* (s^{\dagger} B) \arrow[equal]{u} \arrow{d}{\theta} & u^* s^{\dagger} \Bti_Y^*(B) \arrow[equal]{d} \arrow{l}{\theta} \arrow[equal]{r} & u^* s^{\dagger} \iota_Y \pi_Y(B) \arrow{u}{\tilde{\theta}} \\
\iota_U \pi_U (u^* s^{\dagger} B) \arrow[equal]{uuu} \arrow[equal]{ddd} \arrow[equal]{r} & \Bti_U^* (u^* s^{\dagger} B) \arrow[equal]{d} & s_U^{\dagger} v^* \Bti_Y^*(B) \arrow{d}{\theta} \arrow[equal]{r} & s_U^{\dagger} v^* \iota_V \pi_V(B) \arrow[equal]{u} \arrow{d}{\tilde{\theta}} & \iota_U(s_U^{\dagger} v^* \pi_Y(B)) \arrow[equal]{uuu} \arrow[equal]{ddd} \\
& \Bti_U^* (s_U^{\dagger} v^* B) & s_U^{\dagger} \Bti_V^*(v^* B) \arrow{l}{\theta} \arrow[equal]{dr} & s_U^{\dagger} \iota_V(v^* \pi_Y(B)) \arrow[equal]{d} \arrow{ur}{\tilde{\theta}} \\
&&& s_U^{\dagger} \iota_V \pi_V(v^* B) \arrow{dr}{\tilde{\theta}} \\
\iota_U \pi_U(s_U^{\dagger} v^* B) \arrow{rrrr}{\zeta_{s_U}} \arrow[equal]{uur} &&&& \iota_U (s_U^{\dagger} \pi_V(v^* B))
\end{tikzcd}
\end{equation*}

Here, the two five-term pieces involving $\zeta_s$ and $\zeta_{s_U}$ are commutative by assumption, the six-term piece in the upper-right part as well as the two five-term pieces in the upper-left part and in the central part are commutative by <cit.>, the central six-term piece is commutative by construction, and the remaining three pieces are commutative by naturality.

As a consequence of this observation we deduce that, in order to show the existence of the natural isomorphism $\zeta_s$ along a given closed immersion $s: S \hookrightarrow Y$, it suffices to show that, for some finite open covering $\left\{v_i: V_i \hookrightarrow Y\right\}_{i \in I}$, the isomorphisms $\zeta_{s_i}$ along all restrictions $s_i: s^{-1}(V_i) \hookrightarrow V_i$ exist: this follows from the fact that both $\Mcal^0(\cdot)$ and $\Perv(\cdot)$ are stacks for the Zariski topology. Therefore, since both $Y$ and $S$ are smooth by assumption (hence, in particular, the closed immersion $s$ is regular of codimension $d$), we may reduce to the case where $Y$ is affine and connected and the open complement $U$ of $S$ in $Y$ admits a finite affine open covering $\Ucal$ with exactly $d$ elements: for example, for each point of $S$ one can take the distinguished open subsets associated to the elements of some regular sequence of length $d$ defining $S$ inside $Y$ locally around the chosen point.

In this situation, fix one such open covering $\Ucal$; we define the sought-after natural isomorphism $\zeta_s$ by declaring that its image under the fully faithful functor $s_*: \Mcal^0(S) \hookrightarrow \Mcal^0(Y)$ is the composite

\begin{equation*}
\begin{tikzcd}
s_* \pi_S(s^{\dagger} B) \arrow[equal]{d} && s_* s^{\dagger} \pi_Y(B), \arrow[equal]{d} \\
\pi_Y(s_* s^{\dagger} B) \arrow[equal]{r} & \pi_Y(\check{C}_{Y,\Ucal}[-d](B)) \arrow{r}{\sim} & \check{C}_{Y,\Ucal}[-d](\pi_Y(B))
\end{tikzcd}
\end{equation*}

where the bottom-left horizontal arrow is given by <Ref>. In order to check the commutativity of the diagram (<ref>), it suffices to check the commutativity of its image under the fully faithful functor $s_*: \Mcal^0(S) \hookrightarrow \Mcal^0(Y)$

\begin{equation*}
\begin{tikzcd}
s_* \Bti_S^* (s^{\dagger} B) \arrow[equal]{dd} & s_* s^{\dagger} \Bti_Y^*(B) \arrow{l}{\theta} \arrow[equal]{d} \\
& s_* s^{\dagger} \iota_Y \pi_Y(B) \arrow{d}{\tilde{\theta}} \\
s_* \iota_S \pi_S (s^{\dagger} B) \arrow{r}{\zeta_s} & s_* \iota_S (s^{\dagger} \pi_Y(B)),
\end{tikzcd}
\end{equation*}

which in turn is equivalent to showing the commutativity of the outer rectangle of the composite diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
s_* \Bti_S^* (s^{\dagger} B) \arrow[equal]{dd} & s_* s^{\dagger} \Bti_Y^*(B) \arrow{l}{\theta} \arrow[equal]{d} \\
& s_* s^{\dagger} \iota_Y \pi_Y(B) \arrow{d}{\tilde{\theta}} \\
s_* \iota_S \pi_S (s^{\dagger} B) \arrow{r}{\zeta_s} & s_* \iota_S (s^{\dagger} \pi_Y(B)) \\
\iota_Y(s_* \pi_S (s^{\dagger} B)) \arrow{r}{\zeta_s} \arrow{u}{\bar{\tilde{\theta}}} & \iota_Y (s_* s^{\dagger} \pi_Y(B)) \arrow{u}{\bar{\tilde{\theta}}}.
\end{tikzcd}
\end{equation*}

Expanding the definition of the bottom horizontal arrow in the latter, we obtain the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
& s_* \Bti_S^* (s^{\dagger} B) \arrow[equal]{dl} & s_* s^{\dagger} \Bti_Y^*(B) \arrow{l}{\theta} \arrow[equal]{dr} \arrow[equal]{d} \\
s_* \iota_S \pi_S (s^{\dagger} B) && \check{C}_{Y,\Ucal}[-d] \Bti_Y^*(B) \arrow[equal]{d} \arrow{ddl}{\sim} & s_* s^{\dagger} \iota_Y (\pi_Y(B)) \arrow{d}{\tilde{\theta}} \\
\iota_Y(s_* \pi_S(s^{\dagger} B)) \arrow{u}{\bar{\tilde{\theta}}} \arrow[equal]{d} & \Bti_Y^* (s_* s^{\dagger} B) \arrow[equal]{dl} \arrow[equal]{d} \arrow{uu}{\bar{\theta}} & \check{C}_{Y,\Ucal}[-d] \iota_Y \pi_Y(B) \isoarrow{dd} \arrow[equal]{ur} & s_* \iota_S(s^{\dagger} \pi_Y(B)) \\
\iota_Y \pi_Y (s_* s^{\dagger} B) \arrow[equal]{dr} & \Bti_Y^* (\check{C}_{Y,\Ucal}[-d] B) \arrow[equal]{d} && \iota_Y(s_* s^{\dagger} \pi_Y(B)) \arrow{u}{\bar{\tilde{\theta}}} \\
& \iota_Y \pi_Y (\check{C}_{Y,\Ucal}[-d] B) \arrow{r}{\sim} & \iota_Y(\check{C}_{Y,\Ucal}[-d] \pi_Y(B)) \arrow[equal]{ur}
\end{tikzcd}
\end{equation*}

Here, the top-left piece is commutative by <cit.>, the top-central piece and the bottom-right piece are commutative by construction, the bottom-central piece is commutative by <Ref>, while the other two pieces are commutative by naturality.

§.§ Proof of the main result and applications

We can finally compare the shifted tensor product of <Ref> with the one obtained from the external tensor product, as announced at the beginning of this section:

We want to apply the abstract criterion of <cit.>. To this end, we have to exhibit the existence of a natural isomorphism of functors $\Dcal^0(S) \times \Dcal^{0,loc}(S) \rightarrow \Mcal^0(S)$

\begin{equation}\label{iso:prop-dist-otimes}
\chi_S: (\pi_S(A) \otimes \pi_S(L))[-d] \xrightarrow{\sim} \pi_S((A \otimes L)[-d])
\end{equation}

rendering the diagram of functors $\Dcal^0(S) \times \Dcal^{0,loc}(S) \rightarrow \Perv(S)$

\begin{equation}\label{dia:prop-dist-otimes}
\begin{tikzcd}[font=\small]
(\iota_S \pi_S(A) \otimes \iota_S \pi_S(L))[-d] \arrow{dd}{\tilde{\rho}} \arrow[equal]{r} & (\Bti_S^*(A) \otimes \Bti_S^*(L))[-d] \arrow{r}{\rho} & \Bti_S^*(A \otimes L)[-d] \\
&& \Bti_S^*((A \otimes L)[-d]) \arrow[equal]{u} \arrow[equal]{d} \\
\iota_S(\pi_S(A) \otimes \pi_S(L))[-d] \arrow{rr}{\chi_S} && \iota_S \pi_S((A \otimes L)[-d])
\end{tikzcd}
\end{equation}

commutative. We define the natural isomorphism (<ref>) as the composite

\begin{equation*}
\begin{tikzcd}
(\pi_S(A) \otimes \pi_S(L))[-d] \arrow[equal]{d} && \pi((A \otimes L)[-d]), \arrow[equal]{d} \\
\Delta_S^{\dagger}(\pi_S(A) \boxtimes \pi_S(L)) \arrow[equal]{r} & \Delta_S^{\dagger} \pi_{S \times S}(A \boxtimes L) & \pi_S(\Delta_S^{\dagger}(A \boxtimes L)) \arrow{l}{\zeta_{\Delta_S}}
\end{tikzcd}
\end{equation*}

where the isomorphism

\begin{equation}\label{zeta:AL}
\zeta_{\Delta_S}: \pi_S(\Delta_S^{\dagger}(A \boxtimes L)) \rightarrow \Delta_S^{\dagger} \pi_{S \times S}(A \boxtimes L)
\end{equation}

is provided by specializing the natural transformation (<ref>) to the object $A \boxtimes L$. The fact that (<ref>) is actually invertible follows from <Ref>, since the object $A \boxtimes L$ belongs to $\Dcal^{0,\Delta_S}(S \times S)$ by construction: this is just a rephrasing of the discussion in <Ref>.
In order to check that the diagram (<ref>) is commutative, it suffices to express all the arrows in terms of the external tensor product: in this way, we obtain the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
\Delta_S^*[-d] (\iota_S \pi_S(A) \boxtimes \iota_S \pi_S(L)) \arrow[equal]{r} \arrow{d}{\tilde{\rho}} & \Delta_S^*[-d] (\Bti_S^*(A) \boxtimes \Bti_S^*(L)) \arrow{r}{\rho} & \Delta_S^*[-d] \Bti_{S \times S}^* (A \boxtimes L) \arrow{d}{\tilde{\theta}} \arrow[equal]{dl} \\
\Delta_S^*[-d] \iota_{S \times S}(\pi_S(A) \boxtimes \pi_S(L)) \arrow{d}{\tilde{\theta}} \arrow[equal]{r} & \Delta_S^*[-d] \iota_{S \times S} \pi_{S \times S}(A \boxtimes L) \arrow{d}{\tilde{\theta}} & \Bti_S^*(\Delta_S^*(A \boxtimes L)[-d]) \arrow[equal]{d} \\
\iota_S(\Delta_S^*(\pi_S(A) \boxtimes \pi_S(L)))[-d] \arrow[equal]{d} \arrow[equal]{r} & \iota_S(\Delta_S^* \pi_{S \times S}(A \boxtimes L))[-d] \arrow[equal]{d} & \Bti_S^* (\Delta_S^*[-d](A \boxtimes L)) \arrow[equal]{d} \\
\iota_S(\Delta_S^*[-d] (\pi_S(A) \boxtimes \pi_S(L))) \arrow[equal]{r} & \iota_S(\Delta_S^*[-d] \pi_{S \times S}(A \boxtimes L)) & \iota_S \pi_S (\Delta_S^*[-d](A \boxtimes L)) \arrow{l}{\zeta_{\Delta_S}}
\end{tikzcd}
\end{equation*}

where the right-most piece is commutative by <Ref>, the top piece is commutative by <cit.>, while the other pieces are commutative by naturality. This concludes the proof.

Of course, the result of <Ref> admits a symmetric version where the roles of the two variables are exchanged: there exists a canonical isomorphism of functors $\Mcal^{0,loc}(S) \times \Mcal^0(S) \rightarrow \Mcal^0(S)$

\begin{equation*}
L \otimes^{\dagger} M \xrightarrow{\sim} (L \otimes M)[-d]
\end{equation*}

such that the diagram of functors $\Mcal^{0,loc}(S) \times \Mcal^0(S) \rightarrow \Perv(S)$

\begin{equation*}
\begin{tikzcd}
\iota_S(L \otimes^{\dagger} M) \arrow{r}{\sim} & \iota_S((L \otimes M)[-d]) \arrow[equal]{r} & \iota_S(L \otimes M)[-d] \\
\iota_S(L) \otimes^{\dagger} \iota_S(M) \arrow{u}{\tilde{\rho}^{\dagger}} \arrow[equal]{rr} && (\iota_S(L) \otimes \iota_S(M))[-d] \arrow{u}{\rho}
\end{tikzcd}
\end{equation*}

is commutative.

Recall that, for both stable homotopic $2$-functors $\DA_{ct}^{\et}(\cdot,\Q)$ and $D^b_c(\cdot,\Q)$ are endowed with Verdier duality functors defined, over any given $S \in \Var_k$, by the formula

\begin{equation*}
\Dbb_S(A) := \Homcal_S(A, a_S^! \unit_{\Spec(k)}),
\end{equation*}

where $a_S: S \rightarrow \Spec(k)$ denotes the structural projection. Since, as $S$ varies in $\Var_k$, the Betti realization functors $\Bti_S^*: \DA_{ct}^{\et}(S,\Q) \rightarrow D^b(S,\Q)$ commute with the six operations on the two sides, they also commute with Verdier duality in a canonical way. The fact that $\Dbb_S$ defines a duality means that, in both cases, the canonical arrow

\begin{equation*}
A \rightarrow \Dbb_S(\Dbb_S(A))
\end{equation*}

is invertible. Using this, one can deduce a canonical identification

\begin{equation}\label{formula:homcal-Verdier}
\Homcal_S(A,B) \xrightarrow{\sim} \Dbb_S(A \otimes \Dbb_S(B))
\end{equation}

in both cases; again, these are compatible with one another under the Betti realization.
Since, for every $S \in \Var_k$, the equivalence

\begin{equation*}
\Dbb_S: D^b_c(S,\Q) \xrightarrow{\sim} D^b_c(S,\Q)^{op}
\end{equation*}

is $t$-exact for the perverse $t$-structures, it restricts to an equivalence

\begin{equation*}
\Dbb_S: \Perv(S) \xrightarrow{\sim} \Perv(S)^{op}
\end{equation*}

and so, by <cit.>, it canonically lifts to an equivalence

\begin{equation*}
\Dbb_S: \Mcal(S) \xrightarrow{\sim} \Mcal(S)^{op}
\end{equation*}

that one might still call Verdier duality. Note that these functors are obtained for free without even the need to know that internal homomorphisms of perverse motives actually exist! In fact, one might try to use the above formula (<ref>) in the setting of perverse motives as a way of defining internal homomorphisms there. In this way one would obtain, over every $S \in \Var_k$, a bi-triangulated functor

\begin{equation}\label{formula:homcal-VerdierM}
\Homcal_S(-,-): D^b(\Mcal(S))^{op} \times D^b(\Mcal(S)) \rightarrow D^b(\Mcal(S))
\end{equation}

endowed with a canonical natural isomorphism of functors $D^b(\Mcal(S))^{op} \times D^b(\Mcal(S)) \rightarrow D^b(\Perv(S))$

\begin{equation*}
\iota_S \Homcal_S(M_1^{\bullet},M_2^{\bullet}) \xrightarrow{\sim} \Homcal_S(\iota_S(M_1^{\bullet}),\iota_S(M_2^{\bullet})).
\end{equation*}

However, with this approach it would remain unclear whether the functor (<ref>) defines a right adjoint to the internal tensor product as expected: the issue is that, for a general object $M^{\bullet} \in D^b(\Mcal(S))$, none of the two functors

\begin{equation*}
- \otimes M^{\bullet}, \Homcal_S(M^{\bullet},-): D^b(\Mcal(S)) \rightarrow D^b(\Mcal(S))
\end{equation*}

is exact - and, even worse, none of them is obtained directly from the lifting principles of universal abelian factorizations. However, the result of <Ref> puts us in the desired situation at least for some choices of $M^{\bullet}$. This is what we want to prove now.

In order to state the final result of this section, recall that, for any additive category $\Dcal$, we let $\Rbf(\Dcal)$ denote the category of finitely presented abelian presheaves on $\Dcal$ (i.e. cokernels of representable presheaves) and we let $\A(\Dcal)$ denote Freyd's abelian hull of $\Dcal$. By construction, every object of $\A(\Dcal)$ can be written as the kernel of an arrow in $\Rdf(\Dcal)$. Moreover, for every additive functor $\beta: \Dcal \rightarrow \Acal$ into an abelian category $\Acal$, the corresponding universal abelian factorization $\A(\beta)$ is defined as a Serre quotient of $\A(\Dcal)$, and so the image of $\Rbf(\Dcal)$ in $\A(\beta)$ generates the whole of $\A(\beta)$ under kernels.

Let $S \in \Sm_k$ be connected of dimension $d$, and let $\Dbb_S$ denote the motivic Verdier duality functor defined in <cit.>. Let $L \in \Mcal^{0,loc}(S)$, and suppose that $L$ lies in the image of $\mathbf{R}(\Dcal^{0,loc}(S))$ under the quotient functor $\pi_S: \A(\Dcal^{0,loc}(S)) \rightarrow \Mcal^{0,loc}(S)$. Then there is a canonical adjunction

\begin{equation*}
- \otimes L[-d] : \Mcal^0(S) \rightleftarrows \Mcal^0(S): \Dbb_S(L[-d] \otimes \Dbb_S(-))
\end{equation*}

compatible with the corresponding adjunction on the underlying categories of perverse sheaves in the sense of <cit.>.

As a consequence of <Ref> and <Ref>, it suffices to construct, for every $L$ as in the statement, an adjunction

\begin{equation}\label{adj:L-otimesdiamond}
- \otimes^{\dagger} L: \Mcal^0(S) \leftrightarrows \Mcal^0(S): \Dbb_S(L \otimes^{\dagger} \Dbb_S(-))
\end{equation}

compatible with the corresponding adjunction on the underlying categories of perverse sheaves in the sense of <cit.>. As a consequence of the faithfulness of the functor $\iota_S: \Mcal^0(S) \hookrightarrow \Perv(S)$, we see that such an adjunction (or, more precisely, its associated unit and co-unit transformations) is uniquely determined, provided it exists. We construct the adjoint pair in the statement in two steps of increasing generality:

(Step 1) Suppose first that $L = \pi_S(A)$ for some $A \in \Dcal^{0,loc}(S)$. In this case, we obtain the sought-after adjunction by applying <cit.> to the diagrams

\begin{equation*}
\begin{tikzcd}[font=\small]
\Dcal^0(S) \arrow{rr}{- \otimes^{\dagger} A} \arrow{d}{\beta^0_S} && \Dcal^0(S) \arrow{d}{\beta^0_S} \\
\Perv(S) \arrow{rr}{- \otimes^{\dagger} \beta_S^0(A)} && \Perv(S)
\end{tikzcd}
\end{equation*}

\begin{equation*}
\begin{tikzcd}[font=\small]
\Dcal^0(S) \arrow{rr}{\Dbb_S} \arrow{d}{\beta^0_S} && \Dcal^0(S) \arrow{rr}{A \otimes^{\dagger} -} \arrow{d}{\beta^0_S} && \Dcal^0(S) \arrow{rr}{\Dbb_S} \arrow{d}{\beta^0_S} && \Dcal^0(S) \arrow{d}{\beta^0_S} \\
\Perv(S) \arrow{rr}{\Dbb_S} && \Perv(S) \arrow{rr}{\beta_S^0(A) \otimes^{\dagger} -} && \Perv(S) \arrow{rr}{\Dbb_S} && \Perv(S).
\end{tikzcd}
\end{equation*}

Note that, as a consequence of <cit.>, the natural bijections

\begin{equation*}
\Hom_{\Mcal^0(S)}(M \otimes^{\dagger} \pi_S(A), N) = \Hom_{\Mcal^0(S)}(M, \Dbb_S(\pi_S(A) \otimes^{\dagger} \Dbb_S(N)))
\end{equation*}

are functorial with respect to $A \in \Dcal^{0,loc}(S)$.

(Step 2) Let $L \in \mathbf{R}(\Dcal^{0,loc}(S))$. Choose a presentation of the form $L = \coker\left\{\pi_S(A) \rightarrow \pi_S(B)\right\}$ for some morphism $A \rightarrow B$ in $\Dcal^{0,loc}(S)$. Then we obtain a natural bijection

\begin{equation*}
\Hom_{\Mcal^0(S)}(M \otimes^{\dagger} L, N) = \Hom_{\Mcal^0(S)}(M, \Dbb_S(L \otimes^{\dagger} \Dbb_S(N)))
\end{equation*}

by combining the two chains of bijections

\begin{equation*}
\begin{aligned}
& \Hom_{\Mcal^0(S)}(M \otimes^{\dagger} L, N) \\
= & \Hom_{\Mcal^0(S)}(M \otimes^{\dagger} \coker\left\{\pi_S(A) \rightarrow \pi_S(B) \right\},N) \\
= & \Hom_{\Mcal^0(S)}(\coker\left\{M \otimes^{\dagger} \pi_S(A) \rightarrow M \otimes^{\dagger} \pi_S(B) \right\},N) \\
= & \ker\left\{\Hom_{\Mcal^0(S)}(M \otimes^{\dagger} \pi_S(B),N) \rightarrow \Hom_{\Mcal^0(S)}(M \otimes^{\dagger} \pi_S(A),N) \right\}
\end{aligned}
\end{equation*}

\begin{equation*}
\begin{aligned}
& \Hom_{\Mcal^0(S)}(M, \Dbb_S(L \otimes^{\dagger} \Dbb_S(N))) \\
= & \Hom_{\Mcal^0(S)}(M,\Dbb_S(\coker\left\{\pi_S(A) \rightarrow \pi_S(B) \right\} \otimes^{\dagger} \Dbb_S(N))) \\
= & \Hom_{\Mcal^0(S)}(M,\Dbb_S(\coker\left\{\pi_S(A) \otimes^{\dagger} \Dbb_S(N) \rightarrow \pi_S(B) \otimes^{\dagger} \Dbb_S(N) \right\})) \\
= & \Hom_{\Mcal^0(S)}(M,\ker\left\{\Dbb_S(\pi_S(B) \otimes^{\dagger} \Dbb_S(N)) \rightarrow \Dbb_S(\pi_S(A) \otimes^{\dagger} \Dbb_S(N)) \right\}) \\
= & \ker\left\{\Hom_{\Mcal^0(S)}(M,\Dbb_S(\pi_S(B) \otimes^{\dagger}\Dbb_S(N))) \rightarrow \Hom_{\Mcal^0(S)}(M,\Dbb_S(\pi_S(A) \otimes^{\dagger} \Dbb_S(N)))\right\}
\end{aligned}
\end{equation*}

via the previous case. It is clear from the construction that the adjunction obtained in this way is compatible with the corresponding adjunction on perverse sheaves.

In <Ref> we will explain how to prove that the same result holds in full generality, by combining our own results with certain related recent advances in the theory of Nori motives. We stress, however, that the results just mentioned rely crucially on the construction of internal homomorphisms obtained in <Ref>, where, in turn, the essential ingredient is provided precisely by the tiny-looking <Ref>.

§ CONSTRUCTION OF THE UNIT CONSTRAINT

By <Ref> we have a canonical monoidal structure on perverse motives endowed with mutually compatible associativity and commutativity constraints. In order to complete the proof of <Ref>, we need to construct a unit constraint compatible with the rest of the structure. This is achieved in the present section building on the auxiliary results of <Ref>. The extension techniques developed in [54] allow us to complete our construction by working with smooth varieties only: this will often prove convenient in order to avoid technical issues related to singular varieties. Over smooth varieties, we will often make use of shifted inverse image functors and shifted tensor products as defined in <Ref> and <Ref>, respectively.

As usual, we write perverse Nori motives with the symbol "$\Mcal^0$" rather than as "$\Mcal$" in all those passages that depend closely on the construction of the tensor product, but not in the statement of the main result:

The internal tensor structure on $D^b(\Mcal(\cdot))$ constructed in  <Ref> carries a canonical internal unit constraint in the sense of <cit.>, with respect to which the morphism of $\Var_k$-fibered categories $\iota: D^b(\Mcal(\cdot)) \rightarrow D^b(\Perv(\cdot))$ is canonically unitary in the sense of <cit.>.

In the first place, we need to construct the unit section for perverse motives.

We adopt the following notation:

* We use the symbol $\unit = (\unit_S)_{S \in \Var_k}$ to denote the canonical unit sections of the monoidal $\Var_k$-fibered categories $\DA_{ct}^{\et}(\cdot,\Q)$ and $D^b_c(\cdot,\Q)$.

* We let $w: \Bti^*(\unit) \xrightarrow{\sim} \unit$ denote the canonical isomorphism of unit sections defined in [7].

* We use the symbol $u$ to denote the internal unit constraints of $\DA_{ct}^{\et}(\cdot,\Q)$ and $D^b_c(\cdot,\Q)$.

§.§ Construction of the unit section

In the first place, we need to define the unit section of the $\Var_k$-fibered category $D^b(\Mcal(\cdot))$.

The $\Var_k$-fibered category $D^b(\Mcal(\cdot))$ possesses a canonical unit section $\unit^{\Mcal} = (\unit^{\Mcal}_S)_{S \in \Var_k}$, and the morphism of $\Var_k$-fibered categories $\iota: D^b(\Mcal(\cdot)) \rightarrow D^b(\Perv(\cdot))$ induces a canonical isomorphism of unit sections

\begin{equation*}
\tilde{w}: \iota(\unit^{\Mcal}) \xrightarrow{\sim} \unit.
\end{equation*}

The basic technical issue is that, if a variety $S \in \Var_k$ is not smooth, the unit sheaf $\unit_S \in D^b_c(S,\Q)$ is not a perverse sheaf (not even up to decomposing $S$ into connected components and shifting appropriately); in fact, a priori it is not obvious at all that the unit sheaf belongs to the essential image of the triangulated functor $\iota: D^b(\Mcal(S)) \rightarrow D^b(\Perv(S))$. However, the structure of stable homotopic $2$-functor on $D^b(\Mcal(\cdot))$ allows one to easily overcome this issue by working directly over smooth varieties only: this idea can be made rigorous via the embedding method of <cit.>, which we apply to the setting of fibered categories over $\Var_k$ with respect to the full subcategory $\Sm_k$ of smooth varieties.

The core of the proof of <Ref> is constructing the isomorphisms witnessing the compatibility of the various motivic unit objects with respect to inverse image functors. Using a factorization method analogous to that of [53], we easily reduce to the cases of smooth morphisms and closed immersions, which we treat separately. While the case of smooth morphisms is quite straightforward, the case of closed immersions is slightly tricky and requires some preliminary discussion.
Our approach consists in reducing to the case of closed immersions of codimension $1$, which can be treated directly by means of suitable localization triangles; it is plausible that the Čech complex constructions described in <Ref> could lead to an alternative approach, but we have not tried to follow this direction.

Our preliminary results can be summarized as follows:

Regard $D^b(\Mcal(\cdot))$ and $D^b(\Perv(\cdot))$ as triangulated $\Sm_k$-fibered categories with inverse image functors $f^{\dagger}$ as in <Ref>.

Let $K$ be a section of the $\Sm_k$-fibered category $D^b(\Perv(\cdot))$ such that, for every connected $S \in \Sm_k$, the object $K_S \in D^b(\Perv(S))$ lies in degree $0$. Suppose that we are given, for every $S \in \Sm_k$, and object $K^{\Mcal}_S \in \Mcal(S)$ together with an isomorphism in $D^b(\Perv(S))$

\begin{equation}\label{w_S:lem}
w_S: \iota_S(K_S^{\Mcal}) \xrightarrow{\sim} K_S.
\end{equation}

Then the following statements hold:

* Fix a morphism $f: T \rightarrow S$ in $\Sm_k$. Then there exists at most one morphism in $D^b(\Mcal(T))$

\begin{equation}\label{KM*}
K^{\Mcal,\dagger}_f: f^{\dagger} K^{\Mcal}_S \rightarrow K^{\Mcal}_T
\end{equation}

making the diagram in $\Perv(T)$

\begin{equation*}
\begin{tikzcd}
f^{\dagger} \iota_S(K^{\Mcal}_S) \arrow{r}{\tilde{\theta}_f} \arrow{d}{\tilde{w}_S} & \iota_T(f^{\dagger} K^{\Mcal}_S) \arrow{r}{K_f^{\Mcal,\dagger}} & \iota_T(K^{\Mcal}_T) \arrow{d}{\tilde{w}_T} \\
f^{\dagger} K_S \arrow{rr}{K_f^{\dagger}} && K_T
\end{tikzcd}
\end{equation*}

commute; if this is the case, it is automatically an isomorphism.

* Suppose that, for every morphism $f: T \rightarrow S$ in $\Sm_k$, the natural isomorphism (<ref>) exists. Then the natural isomorphisms (<ref>) turn the collection of objects $\left\{K^{\Mcal}_S\right\}_{S \in \Sm_k}$ into a section $K$ of the $\Sm_k$-fibered category $D^b(\Mcal(\cdot))$, and the morphisms (<ref>) define an isomorphism between sections of $D^b(\Perv(\cdot))$

\begin{equation}\label{w:lem}
w: \iota(K^{\Mcal}) \xrightarrow{\sim} K.
\end{equation}

* Suppose that the isomorphism (<ref>) is known to exist whenever $f$ is either a smooth morphism or a closed immersion of codimension $1$ between smooth connected varieties. Then the natural isomorphism (<ref>) exists for every morphism $f: T \rightarrow S$ in $\Sm_k$.

* The uniqueness of $K^{\Mcal,\dagger}_f$ follows from the fact that $\iota_T: \Mcal(T) \rightarrow \Perv(T)$ is faithful, while its invertibility follows from the fact that $\iota_T$ is also conservative (being exact).

* We only need to check that the isomorphisms (<ref>) are compatible with composition: given two composable morphisms $f: T \rightarrow S$ and $g: S \rightarrow V$ in $\Sm_k$, we have to show that the diagram in $\Mcal(T)$

\begin{equation*}
\begin{tikzcd}
f^{\dagger} g^{\dagger} K^{\Mcal}_V \arrow[equal]{d} \arrow{r}{K_g^{\Mcal,\dagger}} & f^{\dagger} K^{\Mcal}_S \arrow{d}{K_f^{\Mcal,\dagger}} \\
(gf)^{\dagger} K^{\Mcal}_V \arrow{r}{K_{gf}^{\Mcal,\dagger}} & K^{\Mcal}_T
\end{tikzcd}
\end{equation*}

is commutative or, equivalently, that its image under the faithful functor $\iota_T: \Mcal(T) \rightarrow \Perv(T)$

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_T(f^{\dagger} g^{\dagger} K^{\Mcal}_V) \arrow[equal]{d} \arrow{r}{K_g^{\Mcal,\dagger}} & \iota_T(f^{\dagger} K^{\Mcal}_S) \arrow{d}{K_f^{\Mcal,\dagger}} \\
\iota_T((gf)^{\dagger} K^{\Mcal}_V) \arrow{r}{K_{gf}^{\Mcal,\dagger}} & \iota_T(K^{\Mcal}_T).
\end{tikzcd}
\end{equation*}

is commutative. But the latter coincides with the outer rectangle of the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
\iota_T(f^{\dagger} g^{\dagger} K_V^{\Mcal}) \arrow{rrrr}{K_g^{\Mcal,\dagger}} \arrow[equal]{dddd} &&&& \iota_T(f^{\dagger} K_S^{\Mcal}) \arrow{dddd}{K_f^{\Mcal,\dagger}} \\
& f^{\dagger} \iota_S(g^{\dagger} K_V^{\Mcal}) \arrow{ul}{\tilde{\theta}_f} \arrow{rr}{K_g^{\Mcal,\dagger}} && f^{\dagger} \iota_S(K^{\Mcal}_S) \arrow{ur}{\tilde{\theta}_f} \arrow{d}{w_S} \\
& f^{\dagger} g^{\dagger} \iota_V(K^{\Mcal}_V) \arrow{u}{\tilde{\theta}_g} \arrow{r}{\tilde{w}_V} \arrow[equal]{d} & f^{\dagger} g^{\dagger} K_V \arrow{r}{K_g^{\dagger}} \arrow[equal]{d} & f^{\dagger} K_S \arrow{d}{K_f^{\dagger}} \\
& (gf)^{\dagger} \iota_V(K^{\Mcal}_V) \arrow{r}{\tilde{w}_V} \arrow{dl}{\tilde{\theta}_{gf}} & (gf)^{\dagger} K_V \arrow{r}{K_{gf}^{\dagger}} & K_T \\
\iota_T((gf)^{\dagger} K^{\Mcal}_V) \arrow{rrrr}{K_{gf}^{\Mcal,\dagger}} &&&& \iota_Q(K^{\Mcal}_T) \arrow{ul}{w_T}
\end{tikzcd}
\end{equation*}

where the three five-term pieces are commutative by construction while all the other inner pieces are commutative by naturality.

* As one can easily deduce from the proof of the previous point, for every two given composable morphisms $f: T \rightarrow S$ and $g: S \rightarrow V$ in $\Sm_k$ such that the isomorphisms $K^{\Mcal,\dagger}_f$ and $K^{\Mcal,\dagger}_g$ are already known to exist, the natural isomorphism $K^{\Mcal,\dagger}_{gf}$ exists as well and equals the composite

\begin{equation*}
(gf)^{\dagger} K^{\Mcal}_V = f^{\dagger} g^{\dagger} K^{\Mcal}_V \xrightarrow{K^{\Mcal,\dagger}_g} f^{\dagger} K^{\Mcal}_S \xrightarrow{K^{\Mcal,\dagger}_f} K^{\Mcal}_T.
\end{equation*}

This implies that it suffices to know that the isomorphism $K^{\Mcal,\dagger}_f$ exists whenever $f$ is a smooth morphism or a closed immersion in $\Sm_k$. Even better, once one knows that $K^{\Mcal,\dagger}_p$ exists for every smooth morphism $p$ (in particular, for every open immersion), it suffices to show that the isomorphism $K^{\Mcal,\dagger}_z$ exists for every closed immersion $z$ between smooth connected varieties.

In order to conclude, we now claim that, in order to know the existence of $K^{\Mcal,\dagger}_z$ for every closed immersion between smooth varieties $z: Z \hookrightarrow S$, it suffices to know that it exists whenever $z$ is a closed immersion of pure codimension $1$. To see this, recall that, given a closed immersion $z: Z \hookrightarrow S$ in $\Sm_k$, it is always possible to find a finite open covering $\left\{S_i \subset S \right\}_{i \in I}$ such that each closed immersion $z_i: z^{-1}(S_i) \hookrightarrow S_i$ can be written as a composite of closed immersions of pure codimension $1$. In this situation, for each index $i \in I$ form the Cartesian square

\begin{equation*}
\begin{tikzcd}
Z_i \arrow{r}{z_i} \arrow{d}{u_{Z,i}} & S_i \arrow{d}{u_i} \\
Z \arrow{r}{z} & S.
\end{tikzcd}
\end{equation*}

Arguing as in the proof of the previous point one easily sees that, if the four isomorphism $K^{\Mcal,\dagger}$, $K^{\Mcal,\dagger}$, $K^{\Mcal,\dagger}$, $K^{\Mcal,\dagger}$ all exist, then they make the diagram in $\Mcal(Z_i)$

\begin{equation*}
\begin{tikzcd}
z_i^{\dagger} u_i^{\dagger} K_S \arrow{rr}{K^{\Mcal,\dagger}_{u_i}} \arrow[equal]{dd} && z_i^{\dagger} K_{S_i} \arrow{d}{K^{\Mcal,\dagger}_{z_i}} \\
&& K_{Z_i} \\
u_{Z,i}^{\dagger} z^{\dagger} K_S \arrow{rr}{K^{\Mcal,\dagger}_z} && u_{Z,i}^{\dagger} K_Z \arrow{u}{K^{\Mcal,\dagger}_{u_{Z,i}}}
\end{tikzcd}
\end{equation*}

commute. Conversely, since the $\Ucal_S$-fibered categories $\Mcal(\cdot)$ and $\Perv(\cdot)$ are stacks for the Zariski topology, the existence of the isomorphisms $K^{\Mcal,\dagger}_{z_i}$, $K^{\Mcal,\dagger}_{u_i}$ and $K^{\Mcal,\dagger}_{u_{Z,i}}$ for all $i \in I$ implies the existence of $K^{\Mcal,\dagger}_z$. This proves the claim and concludes the proof.

By <cit.>, it suffices to construct the motivic unit section $\unit^{\Mcal}$ and the isomorphism $w: \iota(\unit^{\Mcal}) \xrightarrow{\sim} \unit$ over $\Sm_k$. To this end, we apply the criterion of <Ref> taking $K$ to be the shifted unit section of the $\Sm_k$-fibered category $D^b(\Perv(\cdot))$: for every connected $S \in \Sm_k$ we set $K_S := \unit_S[\dim(S)]$, and we extend the definition to general $S \in \Sm_k$ in the obvious way. In a similar manner, we define the shifted motivic unit objects: for every connected $S \in \Sm_k$ we set $K^{\Mcal}_S := \pi_S(\unit_S[\dim(S)])$, and we extend the definition to general $S \in \Sm_k$ in the obvious way. Note that we have a canonical isomorphism in $\Perv(S)$

\begin{equation}\label{w_S}
\tilde{w}_S: \iota_S(K^{\Mcal}_S) \xrightarrow{\sim} K_S,
\end{equation}

which is defined as the composite

\begin{equation*}
\iota_S(K^{\Mcal}_S) := \iota_S \pi_S(\unit_S[\dim(S)]) = \Bti_S^*(\unit_S[\dim(S)]) = \unit_S[\dim(S)] =: K_S
\end{equation*}

in the case when $S$ is connected and by the obvious extension in the general case. Finally, we define the unshifted motivic unit objects by setting $\unit^{\Mcal,\dagger}_S := K^{\Mcal}_S[-\dim(S)]$ when $S \in \Sm_k$ is connected, and extending the definition to general $S \in \Sm_k$ in the obvious way. As above, we have a canonical isomorphism in $D^b(\Perv(S))$

\begin{equation}\label{w_S:unit}
\iota_S(\unit^{\Mcal}_S) \xrightarrow{\sim} \unit_S,
\end{equation}

defined as the composite

\begin{equation*}
\iota_S(\unit^{\Mcal}_S) = \iota_S(K^{\Mcal}_S[-\dim(S)]) = \iota_S(K^{\Mcal}_S)[-\dim(S)] \xrightarrow{\sim} K_S[-\dim(S)] = \unit_S
\end{equation*}

when $S \in \Sm_k$ is connected and by the obvious extension in the general case.  Clearly, turning the collection $(\unit^{\Mcal}_S)_{S \in \Sm_k}$ into a section of $D^b(\Mcal(\cdot))$ with respect to the usual inverse image functors $f^*$ in such a way that the isomorphisms (<ref>) define an isomorphism of sections is the same as turning the collection $(K^{\Mcal}_S)_{S \in \Sm_k}$ into a section of $D^b(\Mcal(\cdot))$ with respect to the shifted inverse image functors $f^{\dagger}$ in such a way that the isomorphisms (<ref>) define an isomorphism of sections.

By <Ref>(3), in order to achieve the latter task, it suffices to construct the isomorphisms $K^{\Mcal,\dagger}_f$ in the cases where $f$ is either a smooth morphism or a closed immersion of codimension $1$. We treat these two cases separately as follows:

* Let $p: P \rightarrow S$ be a smooth morphism in $\Sm_k$. We define the isomorphism in $\Mcal(P)$

\begin{equation*}
K^{\Mcal,\dagger}_p: p^{\dagger} K^{\Mcal}_S \xrightarrow{\sim} K^{\Mcal}_P
\end{equation*}

by taking the composite

\begin{equation*}
p^{\dagger} K^{\Mcal}_S := p^{\dagger} \pi_S(\unit_S[\dim(S)]) = \pi_P(p^{\dagger} \unit_S[\dim(S)]) \xrightarrow{\unit_p^{\dagger}} \pi_P(\unit_P[\dim(P)]) =: K^{\Mcal}_P.
\end{equation*}

The associated diagram in $\Perv(P)$

\begin{equation*}
\begin{tikzcd}[font=\small]
p^{\dagger} \iota_S(K^{\Mcal}_S) \arrow{r}{\tilde{\theta}_p} \arrow{d}{\tilde{w}_S} & \iota_P(p^{\dagger} K^{\Mcal}_S) \arrow{r}{K_p^{\Mcal,\dagger}} & \iota_P(K^{\Mcal}_P) \arrow{d}{\tilde{w}_P} \\
p^{\dagger} K_S \arrow{rr}{K_p^{\dagger}} && K_P
\end{tikzcd}
\end{equation*}

which can be more explicitly written in the form

\begin{equation*}
\begin{tikzcd}[font=\small]
p^{\dagger} \Bti_S^*(\unit_S[\dim(S)]) \arrow{r}{\theta_p} \arrow{d}{w_S} & \Bti_P^*(p^{\dagger} \unit_S[\dim(S)]) \arrow{r}{\unit_p^{\dagger}} & \Bti_P^*(\unit_P[\dim(P)]) \arrow{d}{w_P} \\
p^{\dagger} \unit_S[\dim(S)] \arrow{rr}{\unit_p^{\dagger}} && \unit_P[\dim(P)]
\end{tikzcd}
\end{equation*}

is commutative since $w: \Bti^*(\unit) \xrightarrow{\sim} \unit$ is a morphism of sections.

* Let $z: Z \hookrightarrow S$ be a closed immersion of codimension $1$ in $\Sm_k$. In order to define the isomorphism in $\Mcal(Z)$

\begin{equation*}
\unit_z^{\Mcal,*}: z^{\dagger} K_S^{\Mcal} \xrightarrow{\sim} K^{\Mcal}_Z
\end{equation*}

we describe the morphism in $D^b(\Mcal(S))$

\begin{equation}\label{unit_z-adj}
K^{\Mcal}_S \rightarrow z_* K^{\Mcal}_Z[1]
\end{equation}

corresponding to it under adjunction. To this end, let $u: U \hookrightarrow S$ denote the open immersion complementary to $z$. Consider the distinguished triangle in $\DA_{ct}^{\et}(S,\Q)$

\begin{equation}\label{loc-rot-DA}
z_* \unit_Z[\dim(Z)] \rightarrow u_! u^* \unit_S[\dim(S)] \rightarrow \unit_S[\dim(S)] \xrightarrow{+}
\end{equation}

obtained from the usual localization triangle

\begin{equation*}
u_! u^* \unit_S[\dim(S)] \rightarrow \unit_S[\dim(S)] \rightarrow z_* z^* \unit_S[\dim(S)] \xrightarrow{+}
\end{equation*}

after inserting the isomorphism $\unit_z^*: z^* \unit_S \xrightarrow{\sim} \unit_Z$ and rotating backwards. Note that all the terms in (<ref>) belong to $\Dcal^0(S)$ by construction. Hence, applying the functor $\pi_S: \Dcal^0(S) \rightarrow \Mcal(S)$ and using the canonical isomorphisms $\beta_S \circ z_* = z_* \circ \beta_Z$ and $\beta_S \circ u_! u^* = u_! \circ \beta_U \circ u^* = u_! u^* \circ \beta_S$, we obtain a short exact sequence in $\Mcal(S)$

\begin{equation*}
0 \rightarrow z_* K^{\Mcal}_Z \rightarrow u_! u^* K^{\Mcal}_S \rightarrow K^{\Mcal}_S \rightarrow 0.
\end{equation*}

We define (<ref>) to be the second arrow of the distinguished triangle in $D^b(\Mcal^0(S))$

\begin{equation*}
u_! u^* K^{\Mcal}_S \rightarrow K^{\Mcal}_S \rightarrow z_* K^{\Mcal}_Z[1] \xrightarrow{+}
\end{equation*}

obtained by taking the triangle associated to this short exact sequence and rotating forwards.
By construction, showing the commutativity of the associated diagram in $\Perv(Z)$

\begin{equation*}
\begin{tikzcd}[font=\small]
z^{\dagger} \iota_S(K^{\Mcal}_S) \arrow{d}{\tilde{w}_S} \arrow{r}{\tilde{\theta}_z} & \iota_Z(z^{\dagger} K^{\Mcal}_S) \arrow{r}{K^{\Mcal,\dagger}_z} & \iota_Z(K^{\Mcal}_Z) \arrow{d}{\tilde{w}_Z} \\
z^{\dagger} K_S \arrow{rr}{K_z^{\dagger}} && K_Z
\end{tikzcd}
\end{equation*}

is equivalent to showing the commutativity of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_S(K^{\Mcal}_S) \arrow{dd}{\tilde{w}_S} \arrow{r} & \iota_S(z_* K^{\Mcal}_Z)[1] \arrow{d}{\bar{\tilde{\theta}}_z} \\
& z_* \iota_Z(K^{\Mcal}_Z)[1] \arrow{d}{\tilde{w}_Z} \\
K_S \arrow{r} & z_* K_Z[1]
\end{tikzcd}
\end{equation*}

Since the latter can be written equivalently in the form

\begin{equation*}
\begin{tikzcd}[font=\small]
\Bti_S^*(\unit_S[\dim(S)]) \arrow{d}{\tilde{w}_S} \arrow{r} & z_* \Bti_Z(\unit_Z[\dim(Z)])[1] \arrow{d}{w_z} \\
\unit_S[\dim(S)] \arrow{r} & (z_* \unit_Z[\dim(Z)])[1]
\end{tikzcd}
\end{equation*}

its commutativity is in the end equivalent to that of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
z^* \Bti_S^*(\unit_S) \arrow{d}{w_S} \arrow{r}{\theta_z} & \Bti_Z^* (z^* \unit_S) \arrow{r}{\unit_z^*} & \Bti^*_Z(\unit_Z) \arrow{d}{w_z} \\
z^* \unit_S \arrow{rr}{\unit_z^*} && \unit_Z
\end{tikzcd}
\end{equation*}

which indeed holds since $w: \Bti^*(\unit) \xrightarrow{\sim} \unit$ is a morphism of sections.

This concludes the proof.

§.§ Construction of the unit constraint

In order to prove <Ref>, we need to show that the $\Var_k$-fibered category $D^b(\Mcal(\cdot))$ admits an internal unit constraint with respect to the motivic unit section constructed in <Ref>, compatibly with the usual unit constraint on $D^b(\Perv(\cdot))$ under the morphism of $\Var_k$-fibered categories $\iota: D^b(\Mcal(\cdot)) \rightarrow D^b(\Perv(\cdot))$.

As done in the previous subsection, it is useful to invoke the results of <cit.> to reduce the construction to smooth varieties; this time, the main reason is that we want to apply <Ref>, which was formulated and proved for smooth varieties only.

The core of the proof of <Ref> is checking the compatibility of the action of the unit objects with respect to inverse image functors. Using the factorization ideas of [53], we easily reduce to the cases of smooth morphisms and closed immersions, which we treat separately. Again, the case of smooth morphisms does not create any additional difficulties, while the case of closed immersion is a bit more subtle and requires some preliminary work. What we have to do is to rephrase the compatibility conditions considered in a form which only involves $t$-exact functors by replacing inverse images under closed immersions by the corresponding direct images.

For sake of brevity, we formulate our preliminary results only for the right unit action; of course, the same conclusions apply to the left unit action as well.

Regard $D^b(\Mcal(\cdot))$ as a $\Sm_k$-fibered category, and consider the motivic uni section $\unit^{\Mcal} = (\unit^{\Mcal}_S)_{S \in \Sm_k}$ constructed in <Ref>. Suppose that we are given, for every $S \in \Sm_k$, a natural isomorphism of functors $D^b(\Mcal(S)) \rightarrow D^b(\Mcal(S))$

\begin{equation*}
u_r = u_{r,S}: M^{\bullet} \otimes \unit^{\Mcal}_S \xrightarrow{\sim} M^{\bullet}.
\end{equation*}

Then the following statements hold:

* For every morphism $f: T \rightarrow S$ in $\Sm_k$, let $D_f$ denote the diagram of functors $D^b(\Mcal(S)) \rightarrow D^b(\Mcal(T))$

\begin{equation*}
\begin{tikzcd}
f^* M^{\bullet} \otimes f^* \unit^{\Mcal}_S \arrow{r}{m} \arrow{d}{\unit_f^*} & f^* (M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{d}{u_r} \\
f^* M^{\bullet} \otimes \unit^{\Mcal}_T \arrow{r}{u_r} & f^* M^{\bullet}.
\end{tikzcd}
\end{equation*}

Then, in order for $D_f$ to be commutative for every morphism $f$ in $\Sm_k$, it suffices that it be commutative for every smooth morphism and for every closed immersion.

* Let $z: Z \hookrightarrow S$ be a closed immersion in $\Sm_k$, and let $D'_z$ denote the diagram $D'_z$ of functors $D^b(\Mcal(Z)) \rightarrow D^b(\Mcal(S))$

\begin{equation*}
\begin{tikzcd}
z_* N^{\bullet} \otimes z_* z^* \unit^{\Mcal}_S \arrow{d}{\unit_z^*} && z_* N^{\bullet} \otimes \unit^{\Mcal}_S \arrow{ll}{\eta} \arrow{d}{u_r} \\
z_* N^{\bullet} \otimes z_* \unit^{\Mcal}_Z \arrow{r}{\bar{m}} & z_* (N^{\bullet} \otimes \unit^{\Mcal}_Z) \arrow{r}{u_r} & z_* N^{\bullet}.
\end{tikzcd}
\end{equation*}

Then $D_z$ is commutative if and only if $D'_z$ is commutative.

Both assertions follow by a standard application of the factorization method of [53]. Since unit constraints are not discussed in [53], for the reader's convenience we include the argument:

* Given a morphism $f: T \rightarrow S$ in $\Sm_k$, choose a factorization of it in $\Var_k$ of the form

\begin{equation*}
f: T \xrightarrow{t} P \xrightarrow{p} S
\end{equation*}

with $t$ a closed immersion and $p$ a smooth morphism; note that $P$ is smooth since so is $S$ as in the statement, so this factorization exists in $\Sm_k$. Then, by construction, the diagram $D_f$ coincides with the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
f^* M^{\bullet} \otimes f^* \unit^{\Mcal}_S \arrow{rrrr}{m} \arrow{dddd}{\unit^{\Mcal,*}} \arrow[equal]{dr} &&&& f^* (M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow[equal]{dl} \arrow{dddd}{u_r} \\
& t^* p^* M^{\bullet} \otimes t^* p^* \unit^{\Mcal}_S \arrow{r}{m} \arrow{d}{\unit^{\Mcal,*}} & t^* (p^* M^{\bullet} \otimes p^* \unit^{\Mcal}_S) \arrow{r}{m} \arrow{d}{\unit^{\Mcal,*}} & t^* p^* (M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{dd}{u_r} \\
& t^* p^* M^{\bullet} \otimes t^* \unit_P \arrow{r}{m} \arrow{d}{\unit^{\Mcal,*}} & t^* (p^* M^{\bullet} \otimes \unit_P) \arrow{dr}{u_r} \\
& t^* p^* M^{\bullet} \otimes \unit_T \arrow{rr}{u_r} && t^* p^* M^{\bullet} \\
f^* M^{\bullet} \otimes \unit^{\Mcal}_T \arrow[equal]{ur} \arrow{rrrr}{m} &&&& f^* M^{\bullet} \arrow[equal]{ul}
\end{tikzcd}
\end{equation*}

where the two central trapezoids are essentially the diagrams $D_p$ and $D_z$. Since all the remaining pieces of the diagram are already known to be commutative, the thesis follows.

* Let $D''_z$ denote the diagram of functors $D^b(\Mcal(Z)) \rightarrow D^b(\Mcal(Z))$

\begin{equation*}
\begin{tikzcd}
z^* z_* N^{\bullet} \otimes z^* \unit^{\Mcal}_S \arrow{r}{m} \arrow{d}{\unit_z^{\Mcal,*}} & z^* (z_* N^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{d}{u_r} \\
z^* z_* N^{\bullet} \otimes \unit^{\Mcal}_Z \arrow{r}{u_r} & z^* z_* N^{\bullet}.
\end{tikzcd}
\end{equation*}

We claim that the commutativity of $D_z$ and of $D'_z$ are both equivalent to the commutativity of $D''_z$. We prove each equivalence directly as follows:

* It is clear that the commutativity of $D_z$ implies that of $D''_z$. Conversely, $D_z$ can be written as the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
z^* M^{\bullet} \otimes z^* \unit^{\Mcal}_S \arrow{rrr}{m} \arrow{dr}{\eta} \arrow{ddd}{\unit^*} &&& z^* (M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{ddd}{u_r} \arrow{dl}{\eta} \\
& z^* z_* z^* M^{\bullet} \otimes z^* \unit^{\Mcal}_S \arrow{r}{m} \arrow{d}{\unit^*} & z^* (z_* z^* M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{d}{u_r} \\
& z^* z_* z^* M^{\bullet} \otimes \unit^{\Mcal}_Z  \arrow{r}{u_r} & z^* z_* z^* M^{\bullet} \\
z^* M^{\bullet} \otimes \unit^{\Mcal}_Z \arrow{rrr}{u_r} \arrow{ur}{\eta} &&& z^* M^{\bullet} \arrow{ul}{\eta}
\end{tikzcd}
\end{equation*}

where the central square is essentially $D''_z$ and the lateral pieces are already known to be commutative. This shows that $D_z$ is commutative if and only if so is $D''_z$.

* The diagram $D'_z$ can be written as the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
z_* N^{\bullet} \otimes z_* z^* \unit^{\Mcal}_S \arrow{dr}{\eta} \arrow{dd}{\unit^{\Mcal,*}} &&&& z_* N^{\bullet} \otimes \unit^{\Mcal}_S \arrow{llll}{\eta} \arrow{dd}{u_r} \arrow{dll}{\eta} \\
& z_* z^* (z_* N^{\bullet} \otimes z_* z^* \unit^{\Mcal}_S) \arrow{dd}{\unit^{\Mcal,*}} & z_* z^* (z_* N^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{l}{\eta} \arrow{dr}{u_r} \\
z_* N^{\bullet} \otimes z_* \unit^{\Mcal}_Z \arrow{dr}{\eta} \arrow{dd}{\eta} && z_*(z^* z_* N^{\bullet} \otimes z^* \unit^{\bullet}_S) \arrow{u}{m} \arrow{dr}{\unit^{\Mcal,*}} & z_* z^* z_* N^{\bullet} \arrow{r}{\epsilon} & z_* B \\
& z_* z^* (z_* N^{\bullet} \otimes z_* \unit^{\Mcal}_Z) & z_* (z^* z_* N^{\bullet} \otimes z^* z_* z^* \unit^{\Mcal}_S) \arrow{d}{\unit^{\Mcal,*}} \arrow{u}{\epsilon} \arrow{uul}{m} & z_* (z^* z_* N^{\bullet} \otimes \unit^{\bullet}_Z) \arrow{u}{u_r} \arrow{dr}{\epsilon} \\
z_* z^* (z_* N^{\bullet} \otimes z_* \unit^{\bullet}_Z) && z_*(z^* z_* N^{\bullet} \otimes z^* z_* \unit^{\bullet}_Z) \arrow{ll}{m} \arrow{rr}{\epsilon} \arrow{ul}{m} \arrow{ur}{\epsilon} && z_* (N^{\bullet} \otimes \unit^{\bullet}_Z) \arrow{uu}{u_r}
\end{tikzcd}
\end{equation*}

where all arrows are invertible by construction and all pieces except possibly the square

\begin{equation*}
\begin{tikzcd}[font=\small]
z_*(z^* z_* N^{\bullet} \otimes z^* \unit^{\Mcal}_S) \arrow{r}{m} \arrow{d}{\unit_z^{\Mcal,*}} & z_* z^* (z_* N^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{d}{u_r} \\
z_*(z^* z_* N^{\bullet} \otimes \unit^{\Mcal}_Z) \arrow{r}{u_r} & z_* z^* z_* N^{\bullet}
\end{tikzcd}
\end{equation*}

are commutative by naturality. Hence $D'_z$ is commutative if and only if the above square is commutative; since the functor $z_*: D^b(\Mcal(Z)) \rightarrow D^b(\Mcal(S))$ is fully faithful, this is the case if and only if $D''_z$ is commutative.

This proves the claim and concludes the proof.

In order to be able to apply <Ref>(2), we need to use the following variant of <Ref>:

Let $z: Z \hookrightarrow S$ be a closed immersion in $\Sm_k$. Then the bi-exact functor

\begin{equation*}
z_*(-) \otimes z_*(-[-\dim(Z)]) = (z_*(-) \otimes z_*(-))[-\dim(Z)]: \Mcal^0(Z) \times \Mcal^{0,loc}(Z) \rightarrow \Mcal^0(S)
\end{equation*}

is canonically isomorphic to the functor obtained by applying <cit.> to the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
\Dcal^0(Z) \times \Dcal^{0,loc}(Z) \arrow{rrrr}{(z_*(-) \otimes z_*(-))[-\dim(Z)]} \arrow{d}{\beta^0_Z \times \beta^{0,loc}_Z} &&&& \Dcal^0(S) \arrow{d}{\beta^0_S} \\
\Perv(Z) \times \Perv(Z) \arrow{rrrr}{(z_*(-) \otimes z_*(-))[-\dim(Z)]} &&&& \Perv(S).
\end{tikzcd}
\end{equation*}

As the reader can check, the argument used for the proof of <Ref> goes through modulo the obvious changes; we omit the details.

By <cit.>, it suffices to construct the motivic unit constraint and check the unitarity of $\iota: D^b(\Mcal(\cdot)) \rightarrow D^b(\Perv(\cdot))$ (via the isomorphism $w: \iota(\unit^{\Mcal}) \xrightarrow{\sim} \unit$ constructed in <Ref>) over $\Sm_k$. To this end, we apply the criterion of <Ref>.

In the first place, fix $S \in \Sm_k$ and consider the two diagrams

\begin{equation*}
\begin{tikzcd}[font=\small]
\Dcal^0(S) \arrow{rr}{- \otimes^{\dagger} \unit_S[\dim(S)]} \arrow{d}{\beta_S^0} && \Dcal^0(S) \arrow{d}{\beta_S^0} \\
\Perv(S) \arrow{rr}{- \otimes^{\dagger} \unit_S[\dim(S)]} && \Perv(S) \\
& \Downarrow \\
\Dcal^0(S) \arrow{rr}{\id} \arrow{d}{\beta_S^0} && \Dcal^0(S) \arrow{d}{\beta_S^0} \\
\Perv(S) \arrow{rr}{\id} && \Perv(S)
\end{tikzcd}
\qquad \qquad \qquad
\begin{tikzcd}[font=\small]
\Dcal^0(S) \arrow{rr}{\unit_S[\dim(S)] \otimes^{\dagger} -} \arrow{d}{\beta_S^0} && \Dcal^0(S) \arrow{d}{\beta_S^0} \\
\Perv(S) \arrow{rr}{\unit_S[\dim(S)] \otimes^{\dagger} -} && \Perv(S) \\
& \Downarrow \\
\Dcal^0(S) \arrow{rr}{\id} \arrow{d}{\beta_S^0} && \Dcal^0(S) \arrow{d}{\beta_S^0} \\
\Perv(S) \arrow{rr}{\id} && \Perv(S).
\end{tikzcd}
\end{equation*}

The shifted right (resp. left) unit isomorphisms of functors $\Dcal^0(S) \rightarrow \Dcal^0(S)$

\begin{equation*}
u_r = u_{r,S}: A \otimes^{\dagger} \unit_S[\dim(S)] \xrightarrow{\sim} A, \qquad u_l = u_{l,S}: \unit_S[\dim(S)] \otimes^{\dagger} A \xrightarrow{\sim} A
\end{equation*}

and the shifted right (resp. left) unit isomorphism of functors $\Perv(S) \rightarrow \Perv(S)$

\begin{equation*}
u_r = u_{r,S}: K \otimes^{\dagger} \unit_S[\dim(S)] \xrightarrow{\sim} K, \qquad u_l = u_{l,S}: \unit_S[\dim(S)] \otimes^{\dagger} K \xrightarrow{\sim} K
\end{equation*}

satisfy the compatibility condition of <cit.>, because the same holds for the corresponding natural isomorphisms on $\DA_{ct}^{\et}(\cdot,\Q)$ and on $D^b_c(\cdot,\Q)$ - this is just a way of rephrasing the unitarity of $\Bti^*: \DA_{ct}^{\et}(\cdot,\Q) \rightarrow D^b_c(\cdot,\Q)$. Moreover, in both cases the shifted left and right unit isomorphisms coincide up to a sign (up to a sign $(-1)^{\dim(S)}$) when evaluated on the shifted object $\unit_S[\dim(S)]$.
Therefore, applying <cit.>, we obtain two canonical natural isomorphisms between functors $\Mcal(S) \rightarrow \Mcal^0(S)$

\begin{equation}\label{utilde:Mcal}
\tilde{u}_r = \tilde{u}_{r,S}: M \otimes^{\dagger} \unit_S^{\Mcal}[\dim(S)] \xrightarrow{\sim} M, \qquad \tilde{u}_l = \tilde{u}_{l,S}: \unit_S^{\Mcal}[\dim(S)] \otimes^{\dagger} M \xrightarrow{\sim} M
\end{equation}

compatible with the corresponding shifted unit isomorphisms on $\Perv(\cdot)$; they coincide (up to a sign $(-1)^{\dim(S)}$) when evaluated on the shifted object $\unit_S^{\Mcal}[\dim(S)]$. Applying <Ref>, unshifting the unit objects and the tensor product functors, and applying <Ref>, we obtain two unit natural isomorphisms of functors $D^b(\Mcal^0(S)) \rightarrow D^b(\Mcal^0(S))$

\begin{equation}\label{unit-Mp}
\tilde{u}_r = \tilde{u}_{r,S}: M^{\bullet} \otimes \unit_S^{\Mcal} \xrightarrow{\sim} M^{\bullet}, \qquad \tilde{u}_l = \tilde{u}_{l,S}: \unit_S^{\Mcal} \otimes M^{\bullet} \xrightarrow{\sim} M^{\bullet}
\end{equation}

again compatible with the corresponding unit isomorphisms on $D^b(\Perv(\cdot))$;
they coincide (with no sign) when evaluated on the unshifted object $\unit_S^{\Mcal}$, thereby satisfying condition ($u$ITS-1) of <cit.>.

It remains to show that the natural isomorphisms (<ref>) satisfy condition ($u$ITS-2) of <cit.>. For sake of simplicity, we only write the proof for the right unit action; of course, the same argument will apply to the left unit action as well. What we have to show in the case of the right unit action is that, for every morphism $f: T \rightarrow S$ in $\Sm_k$, the diagram $D_f$ associated to $f$ as in the statement of <Ref> is commutative. To this end, applying <Ref>(1) we reduce to considering the cases where $f$ is either a smooth morphism or a closed immersion; applying <Ref>(2), we can also rephrase the case of a closed immersion $z$ in terms of the commutativity of the alternative diagram $D'_z$ introduced there. Here are the details:

* Let $p: P \rightarrow S$ be a smooth morphism in $\Sm_k$. The diagram of functors $D^b(\Mcal^0(S)) \rightarrow D^b(\Mcal^0(P))$

\begin{equation*}
\begin{tikzcd}[font=\small]
p^* M^{\bullet} \otimes p^* \unit^{\Mcal}_S \arrow{r}{\tilde{m}} \arrow{d}{\unit_p^{\Mcal,*}} & p^*(M^{\bullet} \otimes \unit^{\Mcal}_S) \arrow{d}{\tilde{u}_r} \\
p^* M^{\bullet} \otimes \unit^{\Mcal}_P \arrow{r}{\tilde{u}_r} & p^* M^{\bullet}
\end{tikzcd}
\end{equation*}

is commutative if and only if so is the diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
p^{\dagger} M^{\bullet} \otimes^{\dagger} p^{\dagger} K^{\Mcal}_S \arrow{r}{\tilde{m}} \arrow{d}{\unit_p^{\Mcal,*}} & p^{\dagger}(M^{\bullet} \otimes^{\dagger} K^{\Mcal}_S) \arrow{d}{\tilde{u}_r} \\
p^{\dagger} M^{\bullet} \otimes K^{\Mcal}_P \arrow{r}{\tilde{u}_r} & p^{\dagger} M^{\bullet}.
\end{tikzcd}
\end{equation*}

Since all the functors involved in the latter diagram are both dg-enhanced and $t$-exact for the obvious $t$-structures, by <Ref>(1) they coincide with the trivial derived functors of the corresponding exact functors $\Mcal^0(S) \rightarrow \Mcal^0(P)$. Hence (see <Ref>) the above diagram is commutative if and only if the corresponding diagram of functors $\Mcal^0(S) \rightarrow \Mcal^0(P)$

\begin{equation*}
\begin{tikzcd}[font=\small]
p^{\dagger} M \otimes p^{\dagger} \unit^{\Mcal}_S \arrow{r}{\tilde{m}} \arrow{d}{\unit_p^{\Mcal,*}} & p^{\dagger}(M \otimes \unit^{\Mcal}_S) \arrow{d}{\tilde{u}_r} \\
p^{\dagger} M \otimes \unit^{\Mcal}_P \arrow{r}{\tilde{u}_r} & p^{\dagger} M
\end{tikzcd}
\end{equation*}

is commutative. Since all the functors and natural transformations involved in the latter diagram come from the lifting principles of universal abelian factorizations, it suffices to show that the corresponding diagram of functors in $\DA_{ct}^{\et}(\cdot,\Q)$ is commutative, which follows from the validity of axiom ($u$ITS-2) there.

* Let $z: Z \hookrightarrow S$ be a closed immersion in $\Sm_k$. By <Ref>(2), it suffices to show that the diagram of triangulated functors $D^b(\Mcal^0(Z)) \rightarrow D^b(\Mcal^0(S))$

\begin{equation*}
\begin{tikzcd}[font=\small]
z_* M^{\bullet} \otimes z_* z^* \unit^{\Mcal}_S \arrow{d}{\unit_z^{\Mcal,*}} && z_* M^{\bullet} \otimes \unit^{\Mcal}_S \arrow{ll}{\eta} \arrow{d}{\tilde{u}_r} \\
z_* M^{\bullet} \otimes z_* \unit^{\Mcal}_Z \arrow{r}{\bar{\tilde{m}}} & z_* (M^{\bullet} \otimes \unit^{\Mcal}_Z) \arrow{r}{\tilde{u}_r} & z_* M^{\bullet}
\end{tikzcd}
\end{equation*}

is commutative. Since all the functors involved in this diagram are dg-enhanced and $t$-exact for the obvious $t$-structures, by Proposition <ref>(1) they coincide with the trivial derived functors of the corresponding exact functors $\Mcal^0(Z) \rightarrow \Mcal^0(S)$. Hence, applying Proposition <ref>(2), we see that the above diagram is commutative if and only if the corresponding diagram of functors $\Mcal^0(Z) \rightarrow \Mcal^0(S)$

\begin{equation*}
\begin{tikzcd}[font=\small]
z_* M \otimes z_* z^* \unit^{\Mcal}_S \arrow{d}{\unit_z^{\Mcal,*}} && z_* M \otimes \unit^{\Mcal}_S \arrow{ll}{\eta} \arrow{d}{\tilde{u}_r} \\
z_* M \otimes z_* \unit^{\Mcal}_Z \arrow{r}{\bar{\tilde{m}}} & z_* (M \otimes \unit^{\Mcal}_Z) \arrow{r}{\tilde{u}_r} & z_* M
\end{tikzcd}
\end{equation*}

is commutative. As a consequence of <Ref> and <Ref>, all the functors and natural transformations appearing in the latter diagram come from the lifting principles of universal abelian factorizations. Thus it suffices to show that the corresponding diagram of functors in $\DA_{ct}^{\et}(\cdot,\Q)$ is commutative, which follows from the validity of axiom ($u$ITS-2) there.

This concludes the proof.

§.§ Compatibility with the associativity and commutativity constraints

In order to make the picture complete, it remains to study the compatibility between the unit constraint of <Ref> and the associativity and commutativity constraints constructed in <Ref> and <Ref>. Since the unit constraint has been defined in the language of internal tensor structures, whereas the associativity and commutativity constraints have been defined in the language of external tensor structures, we need to decide whether to check the two compatibility conditions in the internal or in the external setting. Working in the external setting turns out to be easier, since everything can be more directly related to the lifting principles of universal abelian factorizations. In order to explain why this is the case, we start with the following observation:

For every $S_1, S_2 \in \Sm_k$, consider the diagram

\begin{equation*}
\begin{tikzcd}
\Dcal^0(S_1) \arrow{rr}{- \boxtimes K_{S_2}} \arrow{d}{\beta_{S_1}^0} && \Dcal^0(S_1 \times S_2) \arrow{d}{\beta_{S_1 \times S_2}^0} \\
\Perv(S_1) \arrow{rr}{- \boxtimes K_{S_2}} && \Perv(S_1 \times S_2) \\
& \Downarrow \\
\Dcal^0(S_1) \arrow{rr}{pr^{12,\dagger}_1} \arrow{d}{\beta_{S_1}^0} && \Dcal^0(S_1 \times S_2) \arrow{d}{\beta_{S_1 \times S_2}^0} \\
\Perv(S_1) \arrow{rr}{pr^{12,\dagger}_1} && \Perv(S_1 \times S_2).
\end{tikzcd}
\end{equation*}

Therefore, applying <cit.>, we obtain a canonical natural isomorphism between functors $\Mcal^0(S_1) \rightarrow \Mcal^0(S_1 \times S_2)$

\begin{equation*}
M_1 \boxtimes K^{\Mcal}_{S_2} \xrightarrow{\sim} pr^{12,\dagger}_1 M_1.
\end{equation*}

As a consequence of <Ref> we see that, for every $S \in \Sm_k$, the motivic right unit isomorphism in (<ref>) coincides with the composite isomorphism

\begin{equation*}
M_1 \otimes^{\dagger} K^{\Mcal}_S = M_1 \otimes \unit^{\Mcal}_S := \Delta_S^* (M_1 \boxtimes \unit^{\Mcal}_S) = \Delta_S^{\dagger} (M_1 \boxtimes K^{\Mcal}_S) \xrightarrow{\sim} \Delta_S^{\dagger} pr^{12,\dagger}_1 M_1 = M_1.
\end{equation*}

Of course, the analogue result holds for the motivic left unit isomorphism.

We can now easily deduce the sought-after compatibility properties as follows:

The external unit constraint obtained from <Ref> and the external associativity constraint of <Ref> are compatible in the sense of <cit.>.

We have to show that the associativity and unit constraints satisfy condition ($au$ETS) from <cit.>. For sake of simplicity, we only check the commutativity of the third family of diagrams mentioned there; the commutativity of the other two families follows by a similar argument. Using the analogue of <cit.> for external tensor structures, we see that it suffices to consider smooth varieties only. In the end, what we have to show is that, for every $S_1, S_2, S_3 \in \Sm_k$, the diagram of functors $D^b(\Mcal^0(S_1)) \times D^b(\Mcal^0(S_2)) \rightarrow D^b(\Mcal^0(S_1 \times S_2 \times S_3))$

\begin{equation*}
\begin{tikzcd}
(M_1^{\bullet} \boxtimes M_2^{\bullet}) \boxtimes \unit^{\Mcal}_{S_3} \arrow{r}{a} \arrow{d}{u_r} & M_1^{\bullet} \boxtimes (M_2^{\bullet} \boxtimes \unit^{\Mcal}_{S_3}) \arrow{d}{u_r} \\
pr^{123,*}_{23} (M_1^{\bullet} \boxtimes M_2^{\bullet}) & M_1^{\bullet} \boxtimes pr^{12,*}_{2} M_2^{\bullet} \arrow{l}{m}
\end{tikzcd}
\end{equation*}

is commutative. After shifting the unit objects, using <Ref>, we reduce to showing that the diagram of functors $\Mcal^0(S_1) \times \Mcal^0(S_2) \rightarrow \Mcal^0(S_1 \times S_2 \times S_3)$

\begin{equation*}
\begin{tikzcd}
(M_1 \boxtimes M_2) \boxtimes K^{\Mcal}_{S_3} \arrow{r}{a} \arrow{d}{u_r} & M_1 \boxtimes (M_2 \boxtimes K^{\Mcal}_{S_3}) \arrow{d}{u_r} \\
pr^{123,\dagger}_{23} (M_1 \boxtimes M_2) & M_1 \boxtimes pr^{12,\dagger}_{2} M_2 \arrow{l}{m}
\end{tikzcd}
\end{equation*}

is commutative. Since all the functors and natural transformations appearing in the latter diagram come from the lifting principles of universal abelian factorizations, it suffices to check that the corresponding diagram on $\DA_{ct}^{\et}(\cdot,\Q)$ is commutative, which (after unshifting the unit objects) follows from the validity of axiom ($au$ETS) there.

The internal unit constraint of <Ref> and the internal commutativity constraint obtained from <Ref> are compatible in the sense of <cit.>.

We have to show that the commutativity and unit constraints satisfy condition ($cu$ETS) from <cit.>. Using the analogue of <cit.> for external tensor structure, we see that it suffices to consider smooth varieties only. In the end, what we have to show is that, for every $S_1, S_2 \in \Sm_k$, the diagram of functors $D^b(\Mcal(S_1)) \rightarrow D^b(\Mcal(S_1 \times S_2))$

\begin{equation*}
\begin{tikzcd}
M_1^{\bullet} \boxtimes \unit^{\Mcal}_{S_2} \arrow{r}{c} \arrow{d}{u_r} & \tau^* (\unit^{\Mcal}_{S_2} \boxtimes M_1^{\bullet}) \arrow{d}{u_l} \\
pr^{12,*}_1 M_1^{\bullet} \arrow[equal]{r} & \tau^* pr^{21,*}_1 M_1^{\bullet}
\end{tikzcd}
\end{equation*}

is commutative. After shifting the unit object, using <Ref>, we reduce to showing that the diagram of functors $\Mcal^0(S_1) \rightarrow \Mcal^0(S_1 \times S_2)$

\begin{equation*}
\begin{tikzcd}
M_1 \boxtimes K^{\Mcal}_{S_2} \arrow{r}{c} \arrow{d}{u_r} & \tau^* (K^{\Mcal}_{S_2} \boxtimes M_1) \arrow{d}{u_l} \\
pr^{12,\dagger}_1 M_1 \arrow[equal]{r} & \tau^* pr^{21,\dagger}_1 M_1
\end{tikzcd}
\end{equation*}

is commutative. Since all the functors and natural transformations appearing in the latter diagram come from the lifting principles of universal abelian factorizations, it suffices to check that the corresponding diagram on $\DA_{ct}^{\et}(\cdot,\Q)$ is commutative, which (after unshifting the unit object) follows from the validity of axiom ($cu$ETS) there.

Combine the results of <Ref>, <Ref>, <Ref> and <Ref> obtained previously with the results of <Ref>, <Ref> and <Ref> obtained in the present section.

§ INTERNAL HOMOMORPHISMS

In this section, we complete the construction of the six functor formalism on perverse motives by showing the existence of internal homomorphisms and their compatibility with those on the underlying categories of perverse sheaves. As usual, we state the main result in the language of stable homotopic $2$-functors:

The monoidal stable homotopic $2$-functor $D^b(\Mcal(\cdot))$ is closed in the sense of <cit.>.
Moreover, for every $S \in \Var_k$, the canonical natural transformation between functors $D^b(\Mcal(S))^{op} \times D^b(\Mcal(S)) \rightarrow D^b(\Perv(S))$

\begin{equation*}
\iota_S\Homcal_S(M_1^{\bullet},M_2^{\bullet}) \rightarrow \Homcal_S(\iota_S(M_1^{\bullet}),\iota_S(M_2^{\bullet}))
\end{equation*}

is invertible.

The strategy of the proof consists of two main steps: on the one hand, an abstract construction of internal homomorphisms as an ind-adjoint of the internal tensor product; on the other side, the adjunction for distinguished motivic local systems obtained in <Ref>. As usual, following the conventions of <Ref>, we write the categories of perverse motives as "$\Mcal^0$" in all constructions and proofs specifically involving the tensor product, and simply as "$\Mcal$" otherwise.

The abstract construction of internal homomorphism is inspired by Arapura's construction of motivic direct images in <cit.>, based on a suitable adjoint functor theorem for derived categories of Grothendieck abelian categories from [31]. Compared to Arapura's case, our construction requires a longer preparation in order to extend the internal tensor product to the level of derived ind-categories.

§.§ Recollections on ind-categories

We start by reviewing some facts about ind-categories of (Noetherian) abelian categories and their derived categories. Recall from <cit.> that, for every small category $\Ccal$, the ind-category $\Ind \Ccal$ is the full subcategory of $\Fun(\Ccal^{op},\Sets)$ consisting of all those presheaves which are filtered colimits of representables; the Yoneda embedding $\Ccal \hookrightarrow \Fun(\Ccal^{op},\Sets)$ can be refined to a fully faithful functor $\Ccal \hookrightarrow \Ind \Ccal$.

Let $\Acal$ be an essentially small abelian category. Then:

* The category $\Ind \Acal$ is a Grothendieck abelian category, and the canonical embedding $\Acal \hookrightarrow \Ind \Acal$ is exact.

* If $\Acal$ is Noetherian, then its essential image in $\Ind \Acal$ consists of the subcategory of Noetherian objects inside $\Ind \Acal$.

* The canonical triangulated functor $D^b(\Acal) \rightarrow D(\Ind \Acal)$ is fully faithful and induces an equivalence

\begin{equation*}
D^b(\Acal) \xrightarrow{\sim} D^b_{\Acal}(\Ind \Acal) := \left\{A^{\bullet} \in D(\Ind \Acal) \; | \; H^n(A^{\bullet}) = 0 \; \forall |n| \gg 0, \; H^n(A^{\bullet}) \in \Acal \; \forall n \in \Z \right\}.
\end{equation*}

The first statement is proved in <cit.>; the second one is an easy consequence of <cit.>; the third one follows from <cit.>.

We need to extend certain additive functors of abelian categories to the level of ind-categories and to their derived categories. The following result describes this in detail:

Let $F: \Acal_1 \rightarrow \Acal_2$ be a functor between abelian categories. Then:

* Up to canonical natural isomorphism, there exists a unique functor

\begin{equation*}
\Ind(F): \Ind \Acal_1 \rightarrow \Ind \Acal_2
\end{equation*}

extending $F$ and commuting with arbitrary small filtered colimits.

* If $F$ is additive, then so it $\Ind(F)$.

* If $F$ is faithful (resp. full) then so is $\Ind(F)$.

* If $F$ is left-exact (resp. right-exact), then so is $\Ind(F)$.

Moreover, the assignment $F \mapsto \Ind(F)$ is canonically compatible with composition of functors.

The first statement is proved in <cit.> and does not require the assumption that $\Acal_1$ and $\Acal_2$ are abelian; the second statement follows directly from the construction of $\Ind(F)$; the third statement is <cit.>, and the fourth statement is <cit.>. Finally, the compatibility with composition follows directly from the construction.

Similarly, we need to extend certain natural transformations to the level of ind-categories. This can be done as follows:

Let $\Acal_1$ and $\Acal_2$ be two abelian categories. Then:

* Given two functors $F,G: \Acal_1 \rightarrow \Acal_2$, and a natural transformation $\alpha: F \rightarrow G$, the following hold:

(i) There exists a unique natural transformation of functors $\Ind \Acal_1 \rightarrow \Ind \Acal_2$

\begin{equation*}
\Ind(\alpha): \Ind(F) \rightarrow \Ind(G)
\end{equation*}

such that, for every object $A = \varinjlim_{i \in I} A_i$ of $\Ind \Acal_1$, the diagram in $\Ind \Acal_2$

\begin{equation}\label{dia:indalpha}
\begin{tikzcd}
\varinjlim_{i \in I} F(A_i) \arrow{rr}{\varinjlim_{i \in I} \alpha(A_i)} \isoarrow{d} && \varinjlim_{i \in I} G(A_i) \isoarrow{d} \\
\Ind(F)(\varinjlim_{i \in I} A_i) \arrow{rr}{\Ind(\alpha)(A)} && \Ind(G)(\varinjlim_{i \in I} A_i)
\end{tikzcd}
\end{equation}

is commutative.

(ii) The natural transformation $\Ind(\alpha)$ is invertible if and only if $\alpha$ is invertible.

* Given three functors $F,G,H: \Acal_1 \rightarrow \Acal_2$, for every choice of natural transformations $\alpha: F \rightarrow G$ and $\beta: G \rightarrow H$, we have the equality

\begin{equation*}
\Ind(\beta \circ \alpha) = \Ind(\beta) \circ \Ind(\alpha).
\end{equation*}

In the first statement, (i) follows from the fact that the commutativity of the diagram (<ref>) uniquely determines $\Ind(\alpha)$, while (ii) is a consequence of the second statement of the proposition; the latter follows from the explicit definition of $\Ind(\alpha)$ and $\Ind(\beta)$ via commutative diagrams of the form (<ref>).

The main case of interest for us concerns the unit and co-unit transformations associated to an adjoint pair, in which case we can apply the following criterion:

Let $\Acal_1$ and $\Acal_2$ be abelian categories, and let

\begin{equation*}
F: \Acal_1 \leftrightarrows \Acal_2 :G
\end{equation*}

be an adjoint pair of functors. Then the functors $\Ind(F)$ and $\Ind(G)$ fit into an adjunction

\begin{equation*}
\Ind(F): \Ind \Acal_1 \leftrightarrows \Ind \Acal_2 :\Ind(G)
\end{equation*}

compatible with the original one.

Indeed, one can use <Ref> to extend the unit and co-unit transformations to the level of ind-categories, as well as to check that the extended natural transformations still satisfy the triangular identities of adjunctions.

We also need to extend some short exact sequences of exact functors to the level of ind-categories. To this end, we note the following result:

Let $\Acal_1$ and $\Acal_2$ be essentially small abelian categories. Then, for every short exact sequence of functors $\Acal_1 \rightarrow \Acal_2$

\begin{equation*}
0 \rightarrow F' \rightarrow F \rightarrow F'' \rightarrow 0,
\end{equation*}

the induced sequence of functors $\Ind \Acal_1 \rightarrow \Ind \Acal_2$

\begin{equation*}
0 \rightarrow \Ind(F') \rightarrow \Ind(F) \rightarrow \Ind(F'') \rightarrow 0
\end{equation*}

is exact as well.

Since $\Ind \Acal_2$ is a Grothendieck abelian category by <Ref>(1), the thesis follows from the fact that filtered colimits of exact sequences in Grothendieck abelian categories are exact by definition.

§.§ Indization of perverse motives

In order to construct motivic internal homomorphisms, we first need to extend part of the functoriality of the stable homotopic $2$-functor $D^b(\Mcal(\cdot))$ to the assignment $S \mapsto D(\Ind \Mcal(S))$.

Specifically, we are interested in direct and inverse images under open and closed immersions. By <Ref>, for every closed immersion $i: Z \hookrightarrow S$, the fully faithful exact functor $i_*: \Mcal(Z) \rightarrow \Mcal(S)$ extends to a fully faithful exact functor

\begin{equation*}
i_*: \Ind \Mcal(Z) \rightarrow \Ind \Mcal(S).
\end{equation*}

Similarly, for every open immersion $j: U \hookrightarrow S$, the exact functor $j^*: \Mcal(S) \rightarrow \Mcal(U)$ extends to an exact functor

\begin{equation*}
j^*: \Ind \Mcal(S) \rightarrow \Ind \Mcal(U).
\end{equation*}

Moreover, if $j$ is affine, the fully faithful exact functors $j_!, j_*: \Mcal(U) \rightarrow \Mcal(S)$ extend to fully faithful exact functors

\begin{equation*}
j_!, j_*: \Ind \Mcal(U) \rightarrow \Ind \Mcal(S)
\end{equation*}

and, using <Ref>, we obtain canonical adjunction

\begin{equation*}
j_!: \Ind \Mcal(U) \leftrightarrows \Ind \Mcal(S): j^*, \qquad \qquad j^*: \Ind \Mcal(S) \leftrightarrows \Ind \Mcal(U): j_*.
\end{equation*}

The result of <cit.> can be generalized to the setting of ind-categories as follows:

For every closed immersion $i: Z \hookrightarrow S$, the triangulated functor $i_*: D(\Ind \Mcal(Z)) \rightarrow D(\Ind \Mcal(S))$ admits a canonical left adjoint

\begin{equation*}
i^*: D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(Z))
\end{equation*}

sending $D^b(\Mcal(S))$ to $D^b(\Mcal(Z))$.

Following the argument for the proof of <cit.>, we start by establishing an auxiliary result:

Given a closed immersion $i: Z \hookrightarrow S$ with complementary open immersion $j: U \hookrightarrow S$, the triangulated functor $i_*: D(\Ind \Mcal(Z)) \rightarrow D(\Ind \Mcal(S))$
is fully faithful and induces an equivalence

\begin{equation*}
D(\Ind \Mcal(Z)) \xrightarrow{\sim} D_Z(\Ind \Mcal(S)) := \ker \left\{j^*: D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(U)) \right\}.
\end{equation*}

As in the proof of <cit.>, one reduces formally to the case where $S \in \Aff_k$ and $Z = V(f)$ for some $f \in \Ocal(S)$. In this case, consider the exact functor

\begin{equation*}
\Phi_f: \Ind \Mcal(S) \rightarrow \Ind \Mcal(Z).
\end{equation*}

We claim that the restriction to $D_Z(\Ind \Mcal(S))$ of its trivial derived functor

\begin{equation*}
\Phi_f: D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(Z))
\end{equation*}

provides a quasi-inverse to $i_*$.
The natural isomorphism of functors $D(\Ind \Mcal(Z)) \rightarrow D(\Ind \Mcal(Z))$

\begin{equation*}
\Phi_f(i_* M^{\bullet}) = M^{\bullet}
\end{equation*}

is induced by the natural isomorphism (<ref>) after passing to the ind-completions and deriving. The natural isomorphisms of functors $D_Z(\Ind \Mcal(S)) \rightarrow D_Z(\Ind \Mcal(S))$

\begin{equation*}
i_* \Phi_f(M^{\bullet}) = M^{\bullet}
\end{equation*}

is obtained from the distinguished triangles of functors $D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(S))$

\begin{equation*}
i_* \Psi_f(j^* M^{\bullet}) \rightarrow \Omega_f(M^{\bullet}) \rightarrow M^{\bullet} \xrightarrow{+}, \qquad j_! j^* M^{\bullet} \rightarrow \Omega_f(M^{\bullet}) \rightarrow i_* \Phi_f(M^{\bullet}) \xrightarrow{+}
\end{equation*}

induced from the short exact sequences of exact functors $\Ind \Mcal(S) \rightarrow \Ind \Mcal(S)$
extending (<ref>) and (<ref>), taking into account <Ref> above.

We have to show that the functor $i_*: D(\Ind \Mcal(Z)) \rightarrow D(\Ind \Mcal(S))$ admits a left adjoint sending $D^b(\Mcal(S))$ to $D^b(\Mcal(Z))$. As a consequence of <Ref>, this is equivalent to showing that the inclusion functor $D_Z(\Ind \Mcal(S)) \subset D(\Ind \Mcal(S))$ admits a left adjoint sending $D^b(\Mcal(S))$ to $D^b_Z(\Mcal(S))$. The latter can be constructed using Čech complexes in the same way as in the proof of <cit.>; see also <Ref>.

* For every pair of composable closed immersions $z': Z' \hookrightarrow Z$ and $z: Z \hookrightarrow S$ in $\Var_k$, the natural isomorphism of exact functors $\Mcal(Z')) \rightarrow \Mcal(Z)$

\begin{equation*}
\overline{\conn}_{z',z}: (z z')_* M \xrightarrow{\sim} z_* z'_* M
\end{equation*}

induces an analogous isomorphism of the corresponding exact functors $\Ind \Mcal(Z') \rightarrow \Ind \Mcal(S)$ via <Ref>, whence an analogous natural isomorphism of their trivial derived functors $D(\Ind \Mcal(Z')) \rightarrow D(\Ind \Mcal(S))$. By adjunction, we obtain a canonical isomorphism of functors $D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(Z'))$

\begin{equation*}
\conn_{z',z}: (z z')^* M^{\bullet} \xrightarrow{\sim} {z'}^* z^* M^{\bullet}.
\end{equation*}

* Consider a Cartesian square in $\Var_k$

\begin{equation*}
\begin{tikzcd}
V_Z \arrow{r}{z_V} \arrow{d}{i_Z} & V \arrow{d}{i} \\
Z \arrow{r}{z} & S
\end{tikzcd}
\end{equation*}

where both $z$ and $i$ are closed immersions. Then the natural transformation of functors $D(\Ind \Mcal(V)) \rightarrow D(\Ind \Mcal(Z))$

\begin{equation*}
z^* i_* M^{\bullet} \xrightarrow{\eta} z^* i_* z_{V,*} z_V^* M^{\bullet} = z^* z_* i_{Z,*} z_V^* M^{\bullet} \xrightarrow{\epsilon} i_{Z,*} z_V^* M^{\bullet}
\end{equation*}

is invertible. This is equivalent to the invertibility of the unit arrow, which can be shown via the associated localization triangle; the existence of the latter follows directly from the construction of <cit.>.

* Consider a Cartesian square in $\Var_k$

\begin{equation*}
\begin{tikzcd}
V_Z \arrow{r}{z_V} \arrow{d}{v_Z} & V \arrow{d}{v} \\
Z \arrow{r}{z} & S
\end{tikzcd}
\end{equation*}

where $z$ is a closed immersion and $v$ is an affine open immersion. Since the natural transformation of exact functors $\Mcal(Z) \rightarrow \Mcal(V)$

\begin{equation*}
v^* z_* M \xrightarrow{\eta} z_{V,*} z_V^* v^* z_* M = z_{V,*} v_Z^* z^* z_* M \xrightarrow{\epsilon} z_{V,*} v_Z^* M
\end{equation*}

is invertible by proper base-change, by <Ref> the same is true for the induced natural transformation of exact functors $\Ind \Mcal(Z) \rightarrow \Ind \Mcal(V)$, and hence also for the natural transformation of their trivial derived functors $D(\Ind \Mcal(Z)) \rightarrow D(\Ind \Mcal(V))$. We deduce that the corresponding natural transformation between left adjoints $D(\Ind \Mcal(V)) \rightarrow D(\Ind \Mcal(Z))$

\begin{equation*}
v_{Z,!} z_V^* M^{\bullet} \xrightarrow{\eta} v_{Z,!} z_V^* v^* v_! M^{\bullet} = v_{Z,!} v_Z^* z^* v_! M^{\bullet} \xrightarrow{\epsilon} z^* v_! M^{\bullet}
\end{equation*}

is invertible as well. Moreover, it is clear from the construction that the natural transformation of functors $D(\Ind \Mcal(S)) \rightarrow D(\Ind \Mcal(V_Z))$

\begin{equation*}
z_V^* v^* M^{\bullet} \xrightarrow{\eta} z_V^* v^* z_* z^* M^{\bullet} \xrightarrow{\sim} z_V^* z_{V,*} v_Z^* z^* M^{\bullet} \xrightarrow{\epsilon} v_Z^* z^* M^{\bullet}
\end{equation*}

is invertible.

* In fact, following <cit.>, one could turn the assignment $S \mapsto D(\Ind \Mcal(S))$ into a $\Var_k$-fibered category. We omit the details.

§.§ Abstract construction of internal homomorphisms

After the above preliminaries, we are ready to describe our abstrac construction of internal homomorphisms on the derived ind-categories of perverse motives. Here are the details:

Fix $S \in \Var_k$. For every object $M^{\bullet} \in C^b(\Mcal^0(S))$, we construct a canonical triangulated functor

\begin{equation}\label{ext-otimesK}
- \otimes M^{\bullet}: D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))
\end{equation}

extending the functor $- \otimes M^{\bullet}: D^b(\Mcal^0(S)) \rightarrow D^b(\Mcal^0(S))$.
Since the latter is, by definition, the composite

\begin{equation*}
D^b(\Mcal^0(S)) \xrightarrow{- \boxtimes M^{\bullet}} D^b(\Mcal^0(S \times S)) \xrightarrow{\Delta_S^*} D^b(\Mcal^0(S)),
\end{equation*}

the task reduces to constructing the two single extensions

\begin{equation}\label{ext-boxK}
- \boxtimes M^{\bullet}: D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S \times S))
\end{equation}

\begin{equation}\label{ext-DeltaS}
\Delta_S^*: D(\Ind \Mcal^0(S \times S)) \rightarrow D(\Ind \Mcal^0(S))
\end{equation}

and, since the second extension has already been defined in <Ref>, it remains to construct the first one. To this end, choose $a,b \in \Z$ such that $M^{\bullet} \in C^{[a,b]}(\Mcal^0(S))$. Recall from <cit.> that we have a canonical identification

\begin{equation*}
\Ind C^{[a,b]}(\Mcal^0(S \times S)) = C^{[a,b]}(\Ind \Mcal^0(S \times S)).
\end{equation*}

Consider the exact functor

\begin{equation*}
- \boxtimes M^{\bullet}: \Ind \Mcal^0(S) \rightarrow \Ind C^{[a,b]}(\Mcal^0(S \times S)) = C^{[a,b]}(\Ind \Mcal^0(S \times S))
\end{equation*}

extending the functor $- \boxtimes M^{\bullet}: \Mcal^0(S) \rightarrow C^{[a,b]}(\Mcal^0(S \times S))$. By construction, the induced functor

\begin{equation*}
- \boxtimes M^{\bullet}: D(\Ind \Mcal^0(S)) \rightarrow D(C^{[a,b]}(\Ind \Mcal^0(S \times S))) \xrightarrow{\Tot^{\bullet}} D(\Ind \Mcal^0(S \times S))
\end{equation*}

does not depend on the choice of the integers $a,b$ and defines the sought-after extension (<ref>).
Using the independence from $a,b$ of the construction for each single $M^{\bullet}$, one can easily see that the construction of the functor (<ref>) is covariantly functorial with respect to $M^{\bullet} \in D^b(\Mcal^0(S))$.

Note that both functors (<ref>) and (<ref>) commute with small direct sums: for the first one, this follows from the construction (as the functor $\Tot^{\bullet}$ commutes with small direct sums); for the second one, this follows from the fact that it is a left adjoint. Therefore the functor (<ref>) commutes with small direct sums as well.
Since, by <cit.> the category $D(\Ind \Mcal^0(S))$ satisfies Brown representability, we deduce that, for every fixed $B^{\bullet} \in D(\Ind \Mcal^0(S))$, the functor

\begin{equation*}
D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S)), \quad A^{\bullet} \rightsquigarrow \Hom_{D(\Ind \Mcal^0(S))}(A^{\bullet} \otimes M^{\bullet},B^{\bullet})
\end{equation*}

is representable by an object $\Homcal_S(M^{\bullet},B^{\bullet}) \in D(\Ind \Mcal^0(S))$. By construction, the assignment $B^{\bullet} \rightsquigarrow \Homcal_S(M^{\bullet},B^{\bullet})$ is functorial in $B^{\bullet} \in D(\Ind \Mcal^0(S))$, and so defines a functor

\begin{equation}\label{HomcalK}
\Homcal_S(M^{\bullet},-): D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))
\end{equation}

which is a triangulated right adjoint to (<ref>). Since, as already noted above, the construction of (<ref>) is covariantly functorial in $M^{\bullet} \in D^b(\Mcal^0(S))$, we deduce that the construction of (<ref>) is contravariantly functorial in $M^{\bullet} \in D^b(\Mcal^0(S))$.

Note that it is not clear a priori whether (<ref>) also sends $D^b(\Mcal^0(S))$ to itself; in fact, showing this is one of the main points in the proof of <Ref>.

In order to understand the abstract internal homomorphisms functors just defined better, we now study their compatibility with open and closed immersions.

For every open immersion $j: U \hookrightarrow S$, we have a canonical natural isomorphism of functors $D(\Ind \Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(U))$

\begin{equation*}
\tilde{m} = \tilde{m}_j: j^* A^{\bullet} \otimes j^* M^{\bullet} \xrightarrow{\sim} j^*(A^{\bullet} \otimes M^{\bullet})
\end{equation*}

extending the given isomorphism of functors $D^b(\Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D^b(\Mcal^0(S))$. It is obtained as the composite

\begin{equation*}
j^* A^{\bullet} \otimes j^* M^{\bullet} := \Delta_U^*(j^* A^{\bullet} \boxtimes j^* M^{\bullet}) \xrightarrow{\tilde{m}} \Delta_U^* (j \times j)^* (A^{\bullet} \boxtimes M^{\bullet}) = j^* \Delta_U^* (A^{\bullet} \boxtimes M^{\bullet}) =: j^*(A^{\bullet} \otimes M^{\bullet})
\end{equation*}

where the first arrow is obtained from the natural isomorphism of bi-exact functors $\Mcal^0(S) \times \Mcal^0(S) \rightarrow \Mcal^0(U \times U)$

\begin{equation*}
\tilde{m}_{j,j}: j^* A^{\bullet} \boxtimes j^* M^{\bullet} \xrightarrow{\sim} (j \times j)^*(A^{\bullet} \boxtimes M^{\bullet})
\end{equation*}

after passing to the ind-completion in the first variable and then to the derived categories, while the second arrow is described in <Ref>(3).

Similarly, for every closed immersion $i: Z \hookrightarrow S$, we have a canonical natural isomorphism of functors $D(\Ind \Mcal^0(Z)) \times D^b(\Mcal^0(Z)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation*}
\bar{\tilde{m}} = \bar{\tilde{m}}_i: i_* A^{\bullet} \otimes i_* M^{\bullet} \xrightarrow{\sim} i_* (A^{\bullet} \otimes M^{\bullet})
\end{equation*}

extending the given isomorphism of functors $D^b(\Mcal^0(Z)) \times D^b(\Mcal^0(Z)) \rightarrow D^b(\Mcal^0(S))$. It is obtained as the composite

\begin{equation*}
i_* A^{\bullet} \otimes i_* M^{\bullet} = \Delta_S^*(i_* A^{\bullet} \boxtimes i_* M^{\bullet}) \xrightarrow{\bar{\tilde{m}}} \Delta_S^* (i \times i)_* (A^{\bullet} \boxtimes M^{\bullet}) \xrightarrow{\sim} i_* \Delta_Z^*(A^{\bullet} \boxtimes M^{\bullet}) =: i_* (A^{\bullet} \otimes M^{\bullet}).
\end{equation*}

where the first arrow is obtained from the natural isomorphism of bi-exact functors $\Mcal^0(Z) \times \Mcal^0(Z) \rightarrow \Mcal^0(S \times S)$

\begin{equation*}
\bar{\tilde{m}}_{i,i}: i_* A \boxtimes i_* M \xrightarrow{\sim} (i \times i)_* (A \boxtimes M)
\end{equation*}

after passing to the ind-completion in the first variable and then to the derived categories, while the second arrow is described in <Ref>(2).

Let $j: U \hookrightarrow S$ be an affine open immersion, and let $i: Z \hookrightarrow S$ denote the complementary closed immersion. Then the natural transformations of functors $D(\Ind \Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation}\label{aux-proj-DInd}
j_! (j^* A^{\bullet} \otimes j^* M^{\bullet}) \xrightarrow{\eta} j_!(j^* A \otimes j^* j_! j^* M) \xrightarrow{\tilde{m}} j_! j^*(A \otimes j_! j^* M) \xrightarrow{\epsilon} A^{\bullet} \otimes j_! j^* M^{\bullet}
\end{equation}

\begin{equation}\label{aux-sup-DInd}
A^{\bullet} \otimes i_* i^* M^{\bullet} \xrightarrow{\eta} i_* i^* A^{\bullet} \otimes i_* i^* M^{\bullet} \xrightarrow{\bar{\tilde{m}}} i_* (i^* A^{\bullet} \otimes i^* M^{\bullet})
\end{equation}

are invertible.

Showing that the natural transformation (<ref>) is invertible is equivalent to showing the invertibility of the co-unit transformation

\begin{equation*}
\epsilon: j_! j^*(A^{\bullet} \otimes j_! j^* M^{\bullet}) \rightarrow A^{\bullet} \otimes j_! j^* M^{\bullet}.
\end{equation*}

The latter can be written as the composite

\begin{equation*}
\begin{tikzcd}[font=\small]
j_! j^*(A^{\bullet} \otimes j_! j^* M^{\bullet}) \arrow[equal]{d} & A^{\bullet} \otimes j_! j^* M^{\bullet} \arrow[equal]{d} \\
j_! j^* \Delta_S^*(A^{\bullet} \boxtimes j_! j_* M^{\bullet}) \arrow[equal]{d} & \Delta_S^* (A^{\bullet} \boxtimes j_! j^* M^{\bullet}) \\
j_! \Delta_U^* (j \times j)^* (A^{\bullet} \boxtimes j_! j^* M^{\bullet}) \arrow{r}{\sim} & \Delta_S^* (j \times j)_! (j \times j)^* (A^{\bullet} \boxtimes j_! j^* M^{\bullet}) \arrow{u}{\epsilon}
\end{tikzcd}
\end{equation*}

where the central arrow is invertible by <Ref>(3). Thus it suffices to show the invertibility of the co-unit

\begin{equation*}
\epsilon: (j \times j)_! (j \times j)^* (A^{\bullet} \boxtimes j_! j^* M^{\bullet}) \rightarrow A^{\bullet} \boxtimes j_! j^* M^{\bullet},
\end{equation*}

which follows from the invertibility of the analogous natural transformation of functors $\Mcal^0(S) \times \Mcal^0(S) \rightarrow \Mcal^0(S \times S)$.
The same argument allows us to conclude that the co-unit transformation of functors $D(\Ind \Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation*}
\epsilon: j_! j^* A^{\bullet} \otimes j_! j^* M^{\bullet} \rightarrow j_! j^* A^{\bullet} \otimes M^{\bullet}
\end{equation*}

is invertible. By the localization triangle in the second variable, this implies the vanishing of the functor $D(\Ind \Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation*}
j_! j^* A^{\bullet} \otimes i_* i^* M^{\bullet} = 0.
\end{equation*}

By the localization triangle in the first variable, this in turn implies that the unit transformation of functors $D(\Ind \Mcal^0(S)) \times D^b(\Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation*}
\eta: A^{\bullet} \otimes i_* i^* M^{\bullet} \rightarrow i_* i^* A^{\bullet} \otimes i_* i^* M^{\bullet}
\end{equation*}

is invertible, which is equivalent to the invertibility of the natural transformation (<ref>).

Keep the notation and assumptions of <Ref>. Then there exist canonical natural isomorphisms of functors $D^b(\Mcal^0(S))^{op} \times D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$

\begin{equation*}
\Homcal_S(j_! j^* M^{\bullet}, B^{\bullet}) \xrightarrow{\sim} j_* \Homcal_U (j^* M^{\bullet}, j^* B^{\bullet}), \quad \Homcal_S(i_* i^* M^{\bullet},B^{\bullet}) \xrightarrow{\sim} i_* \Homcal_Z(i^* M^{\bullet}, i^! B^{\bullet}).
\end{equation*}

For every fixed $M^{\bullet} \in D^b(\Mcal^0(S))$, the natural isomorphisms of functors $D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))$ in the statement are obtained by adjunction from those of <Ref>. Since the latter natural isomorphisms are functorial in $M^{\bullet} \in D^b(\Mcal^0(S))$, so are those in the statement.

§.§ Proof of the main result

The strategy for the proof of <Ref> is based on two abstract results that we now describe. For sake of clarity we formulate them in the setting of perverse motives and perverse sheaves, but they might be useful in more general settings as well.

In the first place, we need a criterion ensuring that the abstract internal homomorphisms just constructed respect the categories $D^b(\Mcal(S))$.

Suppose that, for every $S \in \Var_k$ and every $M \in \Mcal^0(S)$ there exists a dense affine open immersion $j: U \hookrightarrow S$ such that the functor

\begin{equation*}
\Homcal_U(j^* M, -): D(\Ind \Mcal^0(U)) \rightarrow D(\Ind \Mcal^0(U))
\end{equation*}

sends $D^b(\Mcal^0(U))$ to itself. Then, for every $S \in \Var_k$, the functor

\begin{equation}\label{homcal-propstat}
\Homcal_S(-,-): D^b(\Mcal^0(S))^{op} \times D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))
\end{equation}

sends $D^b(\Mcal^0(S))^{op} \times D^b(\Mcal^0(S))$ to $D^b(\Mcal^0(S))$.

We have to show that, for every $S \in \Var_k$ and every $M^{\bullet} \in D^b(\Mcal^0(S))$, the functor

\begin{equation*}
\Homcal_S(M^{\bullet},-): D(\Ind \Mcal^0(S)) \rightarrow D(\Ind \Mcal^0(S))
\end{equation*}

sends $D^b(\Mcal^0(S))$ to itself. We argue by Noetherian induction on $S$. If $\dim(S) = 0$, the result holds by hypothesis if $M^{\bullet}$ is concentrated in degree $0$; the general case follows immediately by dévissage, since $D^b(\Mcal^0(S))$ is a triangulated subcategory of $D(\Ind \Mcal^0(S))$ and the functor (<ref>) is triangulated in the first variable.

For the inductive step, suppose that $\dim(S) > 0$ and that the result is known to hold for all proper closed subvarieties of $S$. As in the previous paragraph, we reduce immediately to the case where the complex $M^{\bullet}$ is concentrated in degree $0$, so that we may identify it with an object $M \in \Mcal^0(S)$. In this case, choose a dense affine open immersion $j: U \hookrightarrow S$ satisfying the hypothesis in the statement; let $i: Z \hookrightarrow S$ denote the complementary closed immersion. For every $B^{\bullet} \in D^b(\Mcal^0(S))$, applying the functor $\Homcal_S(-,B^{\bullet})$ to the localization triangle

\begin{equation*}
j_! j^* M \rightarrow M \rightarrow i_* i^* M \xrightarrow{+}
\end{equation*}

we obtain the distinguished triangle in $D(\Ind \Mcal^0(S))$

\begin{equation*}
\Homcal_S(i_* i^* M,B^{\bullet}) \rightarrow \Homcal_S(M,B^{\bullet}) \rightarrow \Homcal_S(j_! j^* M,B^{\bullet}) \xrightarrow{+}.
\end{equation*}

Since $D^b(\Mcal^0(S))$ is a triangulated subcategory of $D(\Ind \Mcal^0(S))$, it suffices to show that the two lateral terms in the latter triangle belong to $D^b(\Mcal^0(S))$ in order to conclude the same for the term in the middle. By <Ref> we have canonical isomorphisms

\begin{equation*}
\Homcal_S(i_* i^* M,B^{\bullet}) = i_* \Homcal_Z(i^* M, i^! B^{\bullet}), \qquad \Homcal_S(j_! j^* M, B^{\bullet}) = j_* \Homcal_U (j^* M, j^* B^{\bullet}).
\end{equation*}

Since we know that $i^* M, i^! B^{\bullet} \in D^b(\Mcal^0(S))$, applying the inductive hypothesis we deduce that $\Homcal_Z(i^* M,i^! B^{\bullet}) \in D^b(\Mcal^0(Z))$, and therefore that $i_* \Homcal_Z(i^* M,i^! B^{\bullet}) \in D^b(\Mcal^0(S))$. Similarly, since we know that $j^* B^{\bullet} \in D^b(\Mcal^0(U))$, applying the hypothesis on $j^* M$ we deduce that $\Homcal_U(j^* M,j^* B^{\bullet}) \in D^b(\Mcal^0(U))$. This concludes the proof.

In the second place, we need a criterion ensuring that the monoidal morphism $\iota: D^b(\Mcal^0(\cdot)) \rightarrow D^b(\Perv(\cdot))$ is compatible with internal homomorphisms.

Suppose that the hypotheses (and hence the conclusions) of <Ref> are satisfied. In addition, suppose that, for every $S \in \Var_k$ and every $M \in \Mcal^0(S)$, there exists a dense affine open immersion $j: U \hookrightarrow S$ such that the canonical natural transformation of functors $D^b(\Mcal^0(U)) \rightarrow D^b(\Perv(U))$

\begin{equation*}
\iota_U \Homcal_U(j^* M,B^{\bullet}) \rightarrow \Homcal_S(\iota_U(j^* M),\iota_U(B^{\bullet}))
\end{equation*}

is invertible. Then, for every $S \in \Var_k$, the canonical natural transformation of functors $D^b(\Mcal^0(S)) \rightarrow D^b(\Perv(S))$

\begin{equation*}
\iota_S \Homcal_S(M^{\bullet},B^{\bullet}) \rightarrow \Homcal_S(\iota_S(M^{\bullet}),\iota_S(B^{\bullet}))
\end{equation*}

is invertible.

As a preparation for the proof of <Ref>, we need to describe the relations among the various natural isomorphisms witnessing the compatibility of internal homomorphisms with open and closed immersions. For simplicity, we state these auxiliary results in the setting of (closed) monoidal stable homotopic $2$-functors of <cit.>.

Let $R: \Hbb_1 \rightarrow \Hbb_2$ be a monoidal morphism of monoidal stable homotopic $2$-functors. Then the following assertions hold:

* For every closed immersion $i: Z \hookrightarrow S$, and every $M \in \Hbb_1(Z)$, the diagram of functors $\Hbb_1(S) \rightarrow \Hbb_2(S)$

\begin{equation*}
\begin{tikzcd}
R_S(i_*(i^* A \otimes M)) \isoarrow{dd} & i_* R_Z(i^* A \otimes M) \arrow{l}{\bar{\theta}^{-1}} & i_*(R_Z(i^* A) \otimes R_Z(M)) \arrow{l}{\rho} \\
&& i_*(i^* R_S(A) \otimes R_Z(M)) \arrow{u}{\theta} \isoarrow{d} \\
R_S(A \otimes i_* M) & R_S(A) \otimes R_S(i_* M) \arrow{l}{\rho}  & R_S(A) \otimes i_* R_Z(M) \arrow{l}{\bar{\theta}^{-1}}
\end{tikzcd}
\end{equation*}

is commutative.

* For every open immersion $j: U \hookrightarrow S$, and every $M \in \Hbb_1(U)$, the diagram of functors $\Hbb_1(S) \rightarrow \Hbb_2(S)$

\begin{equation*}
\begin{tikzcd}
R_S(j_!(j^* A \otimes M)) \isoarrow{dd} & j_! R_U(j^* A \otimes M) \arrow{l}{\bar{\theta}} & j_!(R_U(j^* A) \otimes R_U(M)) \arrow{l}{\rho} \\
&& j_!(j^* R_S(A) \otimes R_U(M)) \arrow{u}{\theta} \isoarrow{d} \\
R_S(A \otimes j_! M) & R_S(A) \otimes R_S(j_! M) \arrow{l}{\rho} & R_S(A) \otimes j_! R_U(M) \arrow{l}{\bar{\theta}}
\end{tikzcd}
\end{equation*}

is commutative.

We only prove the second assertion; the first one can be proved in a similar way.
Developing the definitions of the two vertical isomorphisms in the second diagram, we obtain the more explicit diagram

\begin{equation*}
\begin{tikzcd}[font=\small]
R_S(j_!(j^* A \otimes M)) \arrow{d}{\eta} & j_! R_U(j^* A \otimes M) \arrow{l}{\bar{\theta}} & j_!(R_U(j^* A) \otimes R_U(M)) \arrow{l}{\rho} \\
R_S(j_!(j^* A \otimes j^* j_! M)) \arrow{dd}{m} && j_!(j^* R_S(A) \otimes \iota_U(M)) \arrow{u}{\theta} \arrow{d}{\eta} \\
&& j_!(j^* R_S(A) \otimes j^* j_! R_U(M)) \arrow{d}{m} \\
R_S(j_! j^*(A \otimes j_! M)) \arrow{d}{\epsilon} && j_! j^*(R_S(A) \otimes j_! R_U(M)) \arrow{d}{\epsilon} \\
R_S(A \otimes j_! M) & R_S(A) \otimes R_S(j_! M) \arrow{l}{\rho} & R_S(A) \otimes j_! R_U(M). \arrow{l}{\bar{\theta}}
\end{tikzcd}
\end{equation*}

By inserting, on the top side of the latter, the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
R_S(j_!(j^* A \otimes M)) \arrow{d}{\eta} & j_! R_U(j^* A \otimes M) \arrow{l}{\theta} \arrow{d}{\eta} & j_!(R_U(j^* A) \otimes R_U(M)) \arrow{l}{m} \arrow{d}{\eta} \\
R_S(j_!(j^* A \otimes j^* j_! M)) & j_! R_U(j^* A \otimes j^* j_! M) \arrow{l}{\theta} & j_!(R_U(j^* A) \otimes R_U(j^* j_! M)) \arrow{l}{m} & j_!(j^* R_S(A) \otimes R_U(M)) \arrow{ul}{\theta} \arrow{d}{\eta} \\
&&& R_S(A) \otimes R_U(j^* j_! M)) \arrow{ul}{\theta}
\end{tikzcd}
\end{equation*}

and, on its bottom side, the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
j_! R_U(j^*(A \otimes j_! M)) \arrow{d}{\bar{\theta}} \\
R_S(j_! j^*(A \otimes j_! M)) \arrow{dr}{\epsilon} & j_! j^* (R_S(A \otimes j_! M)) \arrow{d}{\epsilon} \arrow{ul}{\theta} & j_! j^* (R_S(A) \otimes R_S(j_! M)) \arrow{l}{m} \arrow{d}{\epsilon} & j_! j^*(R_S(A) \otimes j_! R_U(M)) \arrow{d}{\epsilon} \arrow{l}{\bar{\theta}} \\
& R_S(A \otimes j_! M) & R_S(A) \otimes R_S(j_! M) \arrow{l}{m} & j_! (j^* R_S(A) \otimes j_! R_U(M))\arrow{l}{\bar{\theta}}
\end{tikzcd}
\end{equation*}

we are reduced to considering the outer part of the diagram

\begin{equation*}
\begin{tikzcd}[font=\tiny]
R_S(j_!(j^* A \otimes j^* j_! M)) \arrow{dd}{m} & j_! R_U(j^* A \otimes j^* j_! M) \arrow{l}{\theta} \arrow{dd}{m} & j_!(R_U(j^* A) \otimes R_U(j^* j_! M)) \arrow{l}{m} \\
&& j_!(j^* R_S(A) \otimes R_U(j^* j_! M))) \arrow{u}{\theta} \arrow{r}{\eta} & j_!(j^* R_S(A) \otimes R_U(M)) \arrow{d}{\eta} \\
R_S(j_! j^*(A \otimes j_! M)) & j_! R_U(j^*(A \otimes j_! M)) \arrow{l}{\theta} & j_!(j^* R_S(A) \otimes j^* R_S(j_! M)) \arrow{d}{m} \arrow{u}{\theta} & j_!(j^* R_S(A) \otimes j^* j_! R_U(M)) \arrow{d}{m} \arrow{l}{\theta} \\
& j_! j^* (R_S(A \otimes j_! M)) \arrow{u}{\theta} & j_! j^* (R_S(A) \otimes R_S(j_! M)) \arrow{l}{m}  & j_! j^*(R_S(A) \otimes j_! R_U(M)) \arrow{l}{\theta}
\end{tikzcd}
\end{equation*}

where the two lateral pieces are commutative by naturality while the central rectangle is commutative by definition.

Let $R: \Hbb_1 \rightarrow \Hbb_2$ be a monoidal morphism of monoidal stable homotopic $2$-functors. Then the following assertions hold:

* For every closed immersion $i: Z \hookrightarrow S$, the diagram of functors $\Hbb_1(S) \rightarrow \Hbb_2(S)$

\begin{equation*}
\begin{tikzcd}
R_S i_* \Homcal_Z(M, i^! B) \arrow{r}{\bar{\theta}} & i_* R_Z \Homcal_Z(M, i^! B) \arrow{r}{\rho} & i_* \Homcal_Z(R_Z(M), R_Z(i^! B)) \arrow{d}{\theta^!} \\
&& i_* \Homcal_Z(R_Z(M), i^! R_S(B)) \\
R_S \Homcal_S(i_* M, B) \isoarrow{uu} \arrow{r}{\rho} & \Homcal_S(R_S(i_* M), R_S(B)) \arrow{r}{\bar{\theta}} & \Homcal_S(i_* R_Z(M), R_S(B)) \isoarrow{u}
\end{tikzcd}
\end{equation*}

is commutative.

* For every open immersion $j: U \hookrightarrow S$, and every $K \in \Hbb_1(U)$ the diagram of functors $\Hbb_1(S) \rightarrow \Hbb_2(S)$

\begin{equation*}
\begin{tikzcd}
R_S j_* \Homcal_U(M, j^! B) \arrow{r}{\bar{\theta}} & j_* R_U \Homcal_U(M, j^! B) \arrow{r}{\rho} & j_* \Homcal_U(R_U(M), R_U(j^! B)) \arrow{d}{\theta^!} \\
&& j_* \Homcal_U(R_U(M), j^! R_S(B)) \\
R_S \Homcal_S(j_! M, B) \isoarrow{uu} \arrow{r}{\rho} & \Homcal_S(R_S(j_! M), R_S(B)) \arrow{r}{\bar{\theta}} & \Homcal_S(j_! R_U(M), R_S(B)) \isoarrow{u}
\end{tikzcd}
\end{equation*}

is commutative.

By adapting the argument in the proof of <cit.>, one sees that the two statements are formal consequences of the corresponding statements of <Ref>. We omit the details.

We have to show that, for every $S \in \Var_k$ and every $M^{\bullet} \in D^b(\Mcal^0(S))$, the natural transformation of functors $D^b(\Mcal^0(S)) \rightarrow D^b(\Perv(S))$

\begin{equation*}
\iota_S \Homcal_S(M^{\bullet},B^{\bullet}) \rightarrow \Homcal_S(\iota_S(M^{\bullet}),\iota_S(B^{\bullet}))
\end{equation*}

is invertible. We argue by Noetherian induction on $S$. If $\dim(S) = 0$, the result holds by hypothesis if $M^{\bullet}$ is concentrated in degree $0$; the general case follows immediately by dévissage.

For the inductive step, assume that $\dim(S) > 0$ and that the result is known to hold for all proper closed subvarieties of $S$. As in the previous paragraph, we may reduce immediately to the case where the complex $M^{\bullet}$ is concentrated in degree $0$, so that we may identify it with an object $M \in \Mcal^0(S)$. In this case, choose a Zariski-dense affine open immersion $j: U \hookrightarrow S$ satisfying the hypothesis in the statement; let $i: Z \hookrightarrow S$ denote the complementary closed immersion. For every $B^{\bullet} \in D^b(\Mcal^0(S))$, we obtain the morphism of distinguished triangles in $D^b(\Mcal^0(S))$

\begin{equation}\label{dia_tr_homcal}
\begin{tikzcd}
\iota_S \Homcal_S(i_* i^* M,B^{\bullet}) \arrow{r} \arrow{d} & \iota_S\Homcal_S(M,B^{\bullet}) \arrow{r} \arrow{d} & \iota_S \Homcal_S(j_! j^* M,B^{\bullet}) \arrow{r}{+1} \arrow{d} & {}\\
\Homcal_S(\iota_S(i_* i^* M), \iota_S(B^{\bullet})) \arrow{r} & \Homcal_S(\iota_S(M), \iota_S(B^{\bullet})) \arrow{r} & \Homcal_S^{\Mcal}(\iota_S(j_! j^* M), \iota_S(B^{\bullet})) \arrow{r}{+1} & {}.
\end{tikzcd}
\end{equation}

In order to show that the central vertical arrow is an isomorphism, it suffices to show that so are the two lateral vertical arrows. But the two arrows in question are the two marked arrows in the diagrams in $D^b(\Perv(S))$

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_S i_* \Homcal_Z(i^* M, i^! B^{\bullet}) \arrow{r}{\sim} & i_* \iota_Z \Homcal_Z(i^* M, i^! B^{\bullet}) \arrow{r}{\sim} & i_* \Homcal_Z(\iota_Z(i^* M), \iota_Z(i^! B^{\bullet})) \isoarrow{d} \\
&& i_* \Homcal_Z(\iota_Z(i^* M), i^! \iota_S(B^{\bullet})) \\
\iota_S \Homcal_S(i_* i^* M, B^{\bullet}) \isoarrow{uu} \arrow{r}{\star} & \Homcal_S(\iota_S(i_* i^* M), \iota_S(B^{\bullet})) \arrow{r}{\sim} & \Homcal_S(i_* \iota_Z(i^* M), \iota_S(B^{\bullet})) \isoarrow{u}
\end{tikzcd}
\end{equation*}

\begin{equation*}
\begin{tikzcd}[font=\small]
\iota_S j_* \Homcal_U(j^* M, j^* B^{\bullet}) \arrow{r}{\sim} & j_* \iota_U \Homcal_U(j^* M, j^* B^{\bullet}) \arrow{r}{\sim} & j_* \Homcal_U(\iota_U(j^* M), \iota_U(j^* B^{\bullet})) \isoarrow{d} \\
&& j_* \Homcal_U(\iota_U(j^* M), j^* \iota_S(B^{\bullet})) \\
\iota_S \Homcal_S(j_! j^* M, B^{\bullet}) \isoarrow{uu} \arrow{r}{\star} & \Homcal_S(\iota_S(j_! j^* M), \iota_S(B^{\bullet})) \arrow{r}{\sim} & \Homcal_S(j_! \iota_U(j^* M), \iota_S(B^{\bullet})) \isoarrow{u}
\end{tikzcd}
\end{equation*}

respectively. Now, taking into account <Ref>(4), we see that the result of <Ref> applies to our setting as well, and thus the same holds for <Ref>. Therefore the two natural diagrams above are commutative. Moreover, in each of them, the vertical arrow in the lower-right corner is invertible by <Ref>, while the horizontal arrow in the upper-right corner is invertible by assumption. We deduce that the marked arrow is invertible as well.
This concludes the proof.

We want to apply the criteria of <Ref> and <Ref>. To this end, fix an object $M \in \Mcal^0(S)$. Choose a smooth Zariski-dense affine open immersion $j: U \hookrightarrow S$ such that $j^* M \in \Mcal^{0,loc}(U)$: this is possible since the $\Ucal_S$-fibered subcategory $\Mcal^{0,loc}$ of $\Mcal^0$ is weakly cofinal by <Ref>.
Up to writing $U$ as the disjoint union of its connected components and treating each component separately, we may assume that $U$ is connected of dimension $d \geq 0$. We divide the proof into two main steps: first of all, we check that the hypotheses are satisfied if $j^* M$ lies in the image of $\mathbf{R}(\Dcal^{0,loc}(U))$; then, we explain how to deduce the general case. Here are the details:

(Step 1) Suppose first that the object $j^* M$ lies in the image of $\mathbf{R}(\Dcal^{0,loc}(U))$ under the quotient functor $\pi_U: \A(\Dcal^{0,loc}(U)) \rightarrow \Mcal^{0,loc}(U)$.
In this case, by <Ref>, we have a canonical adjunction of exact functors

\begin{equation}\label{adj:F-G-proofHomcal}
F: \Mcal^0(U) \rightleftarrows \Mcal^0(U): G,