Felix Küng

Structures on the category of N-complexes

§ ABSTRACT

The theory of $N$-complexes is a generalization of both ordinary chain complexes and graded objects. Hence it yields deeper insight in the structure of these and offers a broader range of applications.
This work generalizes the tensor product of chain complexes and graded objects to the case of $N$-complexes using the structures of $q$-binomial coefficients. We then study different approaches to realize the derived category of $N$-complexes. In particular we realize it as the Verdier quotient of the homotopy category of $N$-complexes, as the $\mathrm{h}$-projective objects and as the homotopy category of a category admitting a Quillen model structure.

§ INTRODUCTION

The theory of chain complexes is a very thoroughly studied field in mathematics. Since a defining property of these is $d^2=0$, an obvious generalization is the notion of an $N$-complex, which has the same structure as a chain complex but instead of $d^2=0$ we have $d^N=0$.

Objects of this form were first studied by W. Mayer <cit.> in 1942 as an alternative homology theory on topological spaces. This approach did not get significant attention since this $N$-homology theory fulfills the Eilenberg-Steenrod axioms and therefore has to be equivalent to ordinary homology. Later in 1991 Kapranov studied them from a purely algebraic point of view <cit.> which spiked great interest in that topic. In particular a considerable amount of homological framework for these was developed by M. Dubois-Violette <cit.>. One of the main motivations of these papers was that $N$-complexes can be used to study representations of quantum groups similarly to the way one uses chain complexes to study representations of groups. Furthermore the concept of cohomology of $N$-complexes also got attention in physics where it is referred to as amplitude cohomology, or homology depending on the convention <cit.>. After this initial work there have been papers about the relation of $N$-complexes to other topics such as Koszulity <cit.>, simplicial structures <cit.> or dualizing categories <cit.>. An especially interesting example of the work done in this field is the paper “Algebraic Model for $N$-connection” <cit.> since it applies the otherwise purely algebraic notions of $N$-complexes to geometry and suggests an application of $N$-complexes to non-flat connections. During the last few years the concept of $N$-complexes has also arrived in the field of  categorification: in particular categorification at roots of unity relies on the theory of $N$-complexes. For example M. Khovanov used hopfological algebra to model $N$-complexes in his paper “Hopfological algebra and categorification at a root of unity: the first steps” <cit.>.

Although there was already quite some theory on the applications and properties of $N$-complexes, a rigid foundation of the homological properties and constructions used in these papers was missing. This has been remedied by Iyama, Kato and Miyachi in <cit.>. However, some of the notions defined in their paper deserve a more thorough study, and although one has now this solid foundation of the theory involved, there are still some open questions on the structure of $N$-complexes.

For example:

* Can we construct a canonical monoidal structure on the category of $N$-complexes if the underlying category is monoidal?

* Can one characterize the derived category of $N$-complexes as well over h-projectives?

* Is the derived category really the Gabriel-Zismann localisation of the category of $N$-complexes along $N$-quasi-isomorphisms?

Throughout this work we will try to answer these questions and give very thorough study of the properties of the involved structures.

In the first three sections we will collect some general properties of $N$-complexes, study $q$-binomial coefficients and apply these to lift bifunctors using a twisted Leibniz rule. This involves the construction of certain standard $N$-complexes, proving that the category of $N$-complexes is complete, cocomplete, abelian or additive if the underlying category is, a weaker but more intuitive form of cyclic sieving and how we can lift which functors and bifunctors.

In the remaining section <ref> we then apply these structures of $q$-binomial coefficients and liftable functors to lift monoidal structures in most cases.

We continue in section <ref> and <ref> by studying the homotopy structure of the category of $N$-complexes, we start by constructing its homotopy category as stable category of a Frobenius exact sturcture and prove that it is triangulated.

Using this structure we imitate in section <ref> the three classical constructions for realizing the derived category as Verdier-quotient, $\mathrm{h}$-projective objects and prove that these can be identified by considering semifree objects as $\mathrm{h}$-projective resolutions.

Finally in section <ref> we finish by defining a Quillen model structure on the category of $N$-complexes realizing the derived category as homotopy category. This shows in particular that the derived category of $N$-complexes indeed is the Gabriel-Zismann localisation along $N$-quasi-isomorphisms.

Although the model structure constructed in section <ref> coincides with the model structure constructed by  J. Gillespie and M. Hovey in their paper <cit.>, our work is completely independent and also uses different techniques that are more hands on. Furthermore it is quite natural that both approaches yield the same model structure since it is the obvious generalization of the projective model structure on the category of chain complexes.

§ ACKNOWLEDGEMENT

I would like to express my gratitude to my advisor O. Schnürer for pointing me towards this topic and his helpful comments and engagement.
Furthermore I would like to thank G. Jasso for his lecture on dg-structures and insight on classical constructions of chain complexes.
Finally I want to thank G. Janssens for suggesting to finally publish this work and believing in it.

§ THE CATEGORY $\CC_N(\A)$

The aim of this section is to introduce the concept of $N$-complexes. In particular we will give the definition and first properties of the category of $N$-complexes in a pointed category. We will use a general definition of diagrams that look like $N$-complexes, without restriction on the differential, which we will refer to as $\infty$-complexes. This allows the use of some abstract arguments for the structure and existence of this category and later also simplifies certain proofs about properties of the category of $N$-complexes. After that we will define the category of $N$-complexes as a full subcategory and discuss certain “standard” $N$-complexes which resemble the standard projective chain complexes and admit similar adjunctions as in the chain-complex case. Later we will prove that the category of $N$-complexes and $\infty$-complexes inherits properties from their base category. In particular we will show that the category of $N$-complexes is abelian if the ground category is, and that it admits colimits and limits if the ground category does.

Let us begin by defining the category of $\infty$-complexes as the category of diagrams of a certain shape in an arbitrary category.

Let $\C$ be a category.

* An $\infty$-complex in $\C$ is a diagram

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (P0) at (0,0){$\cdot \cdot \cdot$};
\node (X1) at (1,0){$X^{n-1}$};
\node (X2) at (2,0){$X^n$};
\node (X3) at (3,0){$X^{n+1}$};

\node (P2) at (4,0){$\cdot \cdot \cdot$};

\draw[->]
(P0) edge node[above] {$d_X$} (X1)
(X1) edge node[above] {$d_X$} (X2)
(X2) edge node[above] {$d_X$} (X3)
(X3) edge node[above] {$d_X$} (P2);
\end{equation*}

in $\C$.

We will sometimes drop the $\C$ if it is clear from the context.

* A morphism $g:X \to Y$ of $\infty$-complexes in $\C$ is a family of morphisms $\left(g^i:X^i \to Y^i\right)_{i \in \Z}$, such that the following diagram in $\C$ commutes:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (P0) at (0,1){$\cdot \cdot \cdot$};
\node (X1) at (1,1){$X^{i-1}$};
\node (X2) at (2,1){$X^i$};
\node (X3) at (3,1){$X^{i+1}$};

\node (P2) at (4,1){$\cdot \cdot \cdot$};

\node (P1) at (0,0){$\cdot \cdot \cdot$};
\node (Y1) at (1,0){$Y^{i-1}$};
\node (Y2) at (2,0){$Y^i$};
\node (Y3) at (3,0){$Y^{i+1}$};
\node (P3) at (4,0){$\cdot \cdot \cdot$};
\node[overlay] (Point) at (4.5,0){.};

\draw[->]
(P0) edge node[above] {$d_X$} (X1)
(X1) edge node[above] {$d_X$} (X2)
(X2) edge node[above] {$d_X$} (X3)
(X3) edge node[above] {$d_X$} (P2)

(P1) edge node[above] {$d_Y$} (Y1)
(Y1) edge node[above] {$d_Y$} (Y2)
(Y2) edge node[above] {$d_Y$} (Y3)
(Y3) edge node[above] {$d_Y$} (P3)

(X1) edge node[right] {$g^{i-1}$} (Y1)
(X2) edge node[right] {$g^i$} (Y2)
(X3) edge node[right] {$g^{i+1}$} (Y3);
\end{equation*}

The collection of $\infty$-complexes in $\C$ together with morphisms of $\infty$-complexes in $\C$ defines a category, which we will refer to as $\CC_\infty (\C)$.

Let $\I$ be the category with objects $x \in \Z$ and $\I(x,y):=\{*\}$ if $x\le y$ and $\I(x,y):=\emptyset$ otherwise. Then $\CC_\infty (\C)$ is the category of functors from $\I$ to $\C$.

If $\C$ is  pre-additive, additive or abelian, so is $\CC_\infty(\C)$.

This is clear by the identification of $\CC_\infty(\C)$ as functors from  $\I$ to $\C$.

Now we can define the category of $N$-complexes as a full subcategory of the category of $\infty$-complexes.

Let $\C$ be a pointed category, i. e. a category with a zero object, and $N\in \N$.

* An $\infty$-complex $X$ in $\C$ is an $N$-complex in $\C$ or just an $N$-complex, if $d_X^N=0$.

* We denote the full subcategory of $\CC_\infty(\C)$ consisting of the $N$-complexes by $\CC_N(\C)$.

It is particularly interesting to observe that the notion of $N$-complexes generalizes the following cases.

There are three well known cases for the category of $N$-complexes in a pointed category $\C$:

$N=0$: In this case we have for every $n \in \Z$ that $d_X^0=\id_{X^n}=0$, hence $X^n\cong 0$ and so $\CC_0(\C)$ is the subcategory consisting of zero objects in $\CC_\infty(\C)$.

$N=1$: Here the above equation yields $d_X=0$. Since morphisms in a pointed category always commute with the zero map we get that the category $\CC_1(\C)$ is isomorphic to the category of graded objects in $\C$.

$N=2$: This case yields by definition the case of chain complexes in $\C$.

Since we will later imitate constructions from the theory of chain complexes, we will need to define the following objects introduced in <cit.> and prove that they give rise to certain adjunctions. This will be one of the main ingredients for our constructions regarding the homotopy category of $N$-complexes and the projective model structure.

Let $\C$ be a pointed category.

* Let $X \in \C$ and $i \in \Z$. The diagram

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=2,baseline]{

\node (P0) at (0,0){$\cdot \cdot \cdot$};
\node (X1) at (1,0){0};
\node (X2) at (2,0){$X$};
\node (X3) at (3,0){$X$};
\node (P1) at (4,0){$\cdot \cdot \cdot$};
\node (XN-2) at (5,0){$X$};
\node (XN-1) at (6,0){$X$};
\node (XN) at (7,0){0};
\node (P2) at (8,0){$\cdot \cdot \cdot$};

\node (P3) at (0,0.5){\tiny{$\cdot \cdot \cdot$}};
\node (i-n-1) at (1,0.5){\tiny{ $i-N$}};
\node (i-n) at (2,0.5){\tiny{$i-N+1$}};
\node (i-n+1) at (3,0.5){\tiny{$i-N+2$}};
\node (P4) at (4,0.5){\tiny{$\cdot \cdot \cdot$}};
\node (i-1) at (5,0.5){\tiny{$i-1$}};
\node (i) at (6,0.5){\tiny{$i$}};
\node (i+1) at (7,0.5){\tiny{$i+1$}};
\node (P5) at (8,0.5){\tiny{$\cdot \cdot \cdot$}};

\node[overlay] (Point) at (8.5,0){.};

\draw[->]
(P0) edge node[above] {0} (X1)
(X1) edge node[above] {0} (X2)
(X2) edge node[above] {$\id_{X}$} (X3)
(X3) edge node[above] {$\id_{X}$} (P1)
(P1) edge node[above] {$\id_{X}$} (XN-2)
(XN-2) edge node[above] {$\id_{X}$} (XN-1)
(XN-1) edge node[above] {0} (XN)
(XN) edge node[above] {0} (P2);
\end{equation*}

defines an $N$-complex that we denote by $\mu^i_N(X)$. Here the numbers in the top row denote the degree of the corresponding object.

* Let $g: M \to N$ be a morphism in $\C$, then $\mu_N^i(g):\mu_N^i(M) \to \mu_N^i (N)$ is the morphism of $N$-complexes in $\C$ which is levelwise $(\mu_N^i(g))^j:= g$ for $i-N < j\le i$ and the zero morphism otherwise.

Consider the $\CCC$-vectorspace  $\CCC$. Then we have that $\mu_4^7(\CCC)$ is the following $4$-complex

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=2,baseline]{

\node (P0) at (0,0){$\cdot \cdot \cdot$};
\node (X3) at (1,0){0};
\node (X4) at (2,0){$\CCC$};
\node (X5) at (3,0){$\CCC$};
\node (X6) at (4,0){$\CCC$};
\node (X7) at (5,0){$\CCC$};
\node (X8) at (6,0){$0$};
\node (P1) at (7,0){$\cdot \cdot \cdot$};

\node (P3) at (0,0.5){\tiny{$\cdot \cdot \cdot$}};
\node (i-n-1) at (1,0.5){\tiny{ $3$}};
\node (i-n) at (2,0.5){\tiny{$4$}};
\node (i-n+1) at (3,0.5){\tiny{$5$}};
\node (P4) at (4,0.5){\tiny{$6$}};
\node (i-1) at (5,0.5){\tiny{$7$}};
\node (i) at (6,0.5){\tiny{$8$}};
\node (P5) at (7,0.5){\tiny{$\cdot \cdot \cdot$}};

\draw[->]
(P0) edge node[above] {0} (X3)
(X3) edge node[above] {$0$} (X4)
(X4) edge node[above] {$\id_{X}$} (X5)
(X5) edge node[above] {$\id_{X}$} (X6)
(X6) edge node[above] {$\id_{X}$} (X7)
(X7) edge node[above] {$0$} (X8)
(X8) edge node[above] {$0$} (P1);
\end{equation*}

in the category of $\CCC$-vectorspaces.

Let $\C$ be a pointed category and $N\in \N$. Then the assignments $M \mapsto \mu_N^i(M)$ and $ g \mapsto \mu_N^i(g)$ define functors $\C \to \CC_N(\C)$ for each $i \in \Z$.

* the functor $(\_)^i:\CC_N(\C)\to \C$ is left adjoint to $\mu_N^i:\C \to \CC_N(\C)$,

* the functor $\mu_N^i : \C \to \CC_N(\C)$ is left adjoint to $(\_)^{i-N+1}:\CC_N(\C)\to \C$.

* Let $X \in \CC_N(\C)$ and $M \in \C$. If $f:X^i \to M$ is a morphism in $\C$, then

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (P10) at (0,1){$\cdot \cdot \cdot$};
\node (X1) at (1,1){$X^{i-N}$};
\node (X2) at (2,1){$X^{i-N+1}$};
\node (X3) at (3,1){$X^{i-N+2}$};
\node (P11) at (4,1){$\cdot \cdot \cdot$};
\node (XN-2) at (5,1){$X^{i-1}$};
\node (XN-1) at (6,1){$X^i$};
\node (XN) at (7,1){$X^{i+1}$};
\node (P12) at (8,1){$\cdot \cdot \cdot$};

\node (P00) at (0,0){$\cdot \cdot \cdot$};
\node (M1) at (1,0){$0$};
\node (M2) at (2,0){$M$};
\node (M3) at (3,0){$M$};
\node (P01) at (4,0){$\cdot \cdot \cdot$};
\node (MN-2) at (5,0){$M$};
\node (MN-1) at (6,0){$M$};
\node (MN) at (7,0){$0$};
\node (P02) at (8,0){$\cdot \cdot \cdot$};

\draw[->]
(P10) edge node[above] {$d_X$} (X1)
(X1) edge node[above] {$d_X$} (X2)
(X2) edge node[above] {$d_X$} (X3)
(X3) edge node[above] {$d_X$} (P11)
(P11) edge node[above] {$d_X$} (XN-2)
(XN-2) edge node[above] {$d_X$} (XN-1)
(XN-1) edge node[above] {$d_X$} (XN)
(XN) edge node[above] {$d_X$} (P12)

(P00) edge node[above] {0} (M1)
(M1) edge node[above] {0} (M2)
(M2) edge node[above] {$\id_{M}$} (M3)
(M3) edge node[above] {$\id_{M}$} (P01)
(P01) edge node[above] {$\id_{M}$} (MN-2)
(MN-2) edge node[above] {$\id_{M}$} (MN-1)
(MN-1) edge node[above] {0} (MN)
(MN) edge node[above] {0} (P02)

(X1) edge node[right] {0} (M1)
(X2) edge node[right] {$f \circ d^{N-1}$} (M2)
(X3) edge node[right] {$f \circ d^{N-2}$} (M3)

(XN-2) edge node[right] {$f\circ d$} (MN-2)
(XN-1) edge node[right] {$f$} (MN-1)
(XN) edge node[right] {0} (MN);
\end{equation*}

is a morphism $\widehat{f}:X \to \mu^i_N(M)$ in $\CC_N(\C)$ with $\widehat{f}^i=f$.

Conversely every $g:X  \to \mu^i_N(M)$ in $\CC_N(\C)$ is of this form for a unique $f:X^i \to M$, namely $f=g^i$. This establishes a bijection which is natural in $X$ and $M$.

* Let $X \in \CC_N(\C)$ and $M \in \C$. If $f:M \to X^{i-N+1}$ is a morphism in $\C$, then

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (P10) at (0,0){$\cdot \cdot \cdot$};
\node (X1) at (1,0){$X^{i-N}$};
\node (X2) at (2,0){$X^{i-N+1}$};
\node (X3) at (3,0){$X^{i-N+2}$};
\node (P11) at (4,0){$\cdot \cdot \cdot$};
\node (XN-2) at (5,0){$X^{i-1}$};
\node (XN-1) at (6,0){$X^i$};
\node (XN) at (7,0){$X^{i+1}$};
\node (P12) at (8,0){$\cdot \cdot \cdot$};

\node (P00) at (0,1){$\cdot \cdot \cdot$};
\node (M1) at (1,1){$0$};
\node (M2) at (2,1){$M$};
\node (M3) at (3,1){$M$};
\node (P01) at (4,1){$\cdot \cdot \cdot$};
\node (MN-2) at (5,1){$M$};
\node (MN-1) at (6,1){$M$};
\node (MN) at (7,1){$0$};
\node (P02) at (8,1){$\cdot \cdot \cdot$};

\draw[->]
(P10) edge node[above] {$d_X$} (X1)
(X1) edge node[above] {$d_X$} (X2)
(X2) edge node[above] {$d_X$} (X3)
(X3) edge node[above] {$d_X$} (P11)
(P11) edge node[above] {$d_X$} (XN-2)
(XN-2) edge node[above] {$d_X$} (XN-1)
(XN-1) edge node[above] {$d_X$} (XN)
(XN) edge node[above] {$d_X$} (P12)

(P00) edge node[above] {0} (M1)
(M1) edge node[above] {0} (M2)
(M2) edge node[above] {$\id_{M}$} (M3)
(M3) edge node[above] {$\id_{M}$} (P01)
(P01) edge node[above] {$\id_{M}$} (MN-2)
(MN-2) edge node[above] {$\id_{M}$} (MN-1)
(MN-1) edge node[above] {0} (MN)
(MN) edge node[above] {0} (P02)

(M1) edge node[right] {0} (X1)
(M2) edge node[right] {$ f  $} (X2)
(M3) edge node[right] {$d \circ f $} (X3)

(MN-2) edge node[right] {$d^{N-2} \circ f$} (XN-2)
(MN-1) edge node[right] {$d^{N-1} \circ f$} (XN-1)
(MN) edge node[right] {0} (XN);
\end{equation*}

is a morphism $\widehat{f}:\mu_N^{i}(M)\to X$ in $\CC_N(\C)$ with $\widehat{f}^{i-N+1}=f$.

Conversely every $g:\mu_N^{i}(M) \to X$ in $\CC_N(\C)$ is of this form for a unique morphism $f:M \to X^{i-N+1} $, namely $f=g^{i-N+1}$. This establishes a bijection which is natural in $X$ and $M$.

If $\A$ is an additive category, then $\mu^i_N:\A \to \CC_N(\A)$ defines an additive functor, and all the above adjunctions become adjunctions of additive functors.

For $f,g \in \A(X,Y)$ it is clear by the definition of $\mu^i_N$ that $\mu_N^i(f+g)=\mu_N^i(f)+ \mu_N^i(g)$. Moreover the bijections from Proposition <ref> are compatible with this addition.

After having defined these standard complexes we will inspect which properties the category of $N$-complexes inherits from the ground category. We are especially intrested in limits and colimits since we need coproducts for defining our monoidal structure and colimits for the model structure.

Let $\C$ be a category and assume that $\C$ admits all colimits (respectively limits). Then $\CC_\infty(\C)$ admits all colimits (respectively limits). If $\C$ is pointed, then $\CC_N(\C)$ admits colimits and limits if $\C$ does.

The claim for $\CC_\infty(\C)$ follows from the fact that $\CC_\infty(\C)$ is the category of functors from the small category $\I$ to the category $\C$ by Lemma <ref>. In particular colimits and limits in $\CC_\infty(\C)$ are computed pointwise.

In the pointed case we need to show that the pointwise construction of limits and colimits defines an object in $\CC_N(\C)$. Hence consider $X:=\colim_{j\in J}X_j$. Now $d_X= \colim_{j\in J} d_{X_j}$, but this implies by pointwise calculation that $d_X^N= \colim_{j\in J} d_{X_j}^N$ and by $d_{X_j}^N=0$ for all $j \in J$ we get $d_X^N=0$. So $\colim_{j\in J}X_j$ is again an object in $\CC_N(\C)$.
An analogous argument shows the claim for limits.

Let $\A$ be an abelian category, then $\CC_\infty(\A)$ and $\CC_N(\A)$ are abelian as well.

The category $\CC_\infty(\A)$ is abelian since it can be interpreted by Lemma <ref> as the functor category $\Fun(\I,\A)$ where kernels and cokernels are computed pointwise.
Now to show that $\CC_N(\A)$ is abelian it suffices to show that the kernels and cokernels in $\CC_\infty(\A)$ of maps in the subcategory $\CC_N(\A)$ are again objects in $\CC_N(\A)$. Consider for this a morphism $\varphi : X \to Y$ in $\CC_N(\A)$.

kernelsLet $f: K \to X$ be the kernel of $\varphi$ in $\CC_\infty(\A)$. Then $f$ is a pointwise monomorphism. Furthermore we have $f\circ d_K=d_X \circ f$, in particular one gets

\begin{equation*}
f \circ d_K^N=d_X^N \circ f =0 \circ f =0.
\end{equation*}

Since $f$ is a pointwise monomorphism this implies $d_K^N=0$. So $K$ is an object of $\CC_N(\A)$ as desired.

cokernels Let $g:Y\to C$ be the cokernel of $\varphi$ in $\CC_\infty(\A)$, then $g$ is a pointwise epimorphism and we have again that $d_C \circ g = g \circ d_Y$, which yields

\begin{align*}
d_C ^N \circ g = g \circ d_Y^N =g \circ 0 =0.
\end{align*}

Now we get by the cancelation property that $d_C^N=0$, and so $C$ is an object of $\CC_N(\A)$.

All together we have that $\CC_N(\A)$ is abelian.

§ INTERLUDE ON COMBINATORICS OF $Q$-BINOMIAL COEFFICIENTS

§.§ Background

Sign rules play a very important role in homological algebra, and hence also in the theory of chain complexes. In order to extend these to $N$-complexes for arbitrary natural numbers $N$ we study $q$-binomial coefficients. These are of interest, since they arise as coefficients if we apply the Leibniz rule with $-1$ replaced by a generic $q$. We also show that these vanish for primitive roots of unity, which is exactly what we desire to define a monoidal structure on $N$-complexes.

A lot of the results in this subsection are known in the general literature. However, we still show the proofs as these show the intimate relation to binomial coefficients and to supply an easy accessible reference for these results usually given as exercises.

In $\Z[q]$ define:

* The $q$-bracket of $n \in \N$ for $1 < n$ as

\begin{align*}[n]_q:=1+q+q^2+...+q^{n-1}\text{ and }[0]_q:=0,[1]_q:=1.
\end{align*}

* The $q$-factorial of $n\in \N$ as

\begin{align*}[n]!_q:=[1]_q[2]_q...[n]_q \text{ and } [0]!_q:=1.
\end{align*}

And in $\Q(q)$ define:

* The $q$-binomial coefficient or Gaussian binomial coefficient of $n$ over $k$ for $n,k \in \N$ where $k \le n$ as

\begin{align*}\binom{n}{k}_q:=\frac{[n]!_q}{[n-k]!_q[k]!_q}.
\end{align*}

We will prove in Lemma <ref>  that actually $\binom{n}{k}_q$ defines an element in $\N[q]$, i. e. a polynomial with only natural coefficients.

Some non-trivial examples for $q$-binomial coefficients are:

\begin{align*}
\binom{4}{2}_q&=q^4+q^3+2q^2+q+1,\\
\binom{5}{2}_q&=q^6 + q^5 + 2 q^4 + 2 q^3 + 2 q^2 + q + 1,\\
\binom{6}{3}_q&=q^9 + q^8 + 2 q^7 + 3 q^6 + 3 q^5 + 3 q^4 + 3 q^3 + 2 q^2 + q + 1.
\end{align*}

The following equations hold for all $n \in \N$ and $1 \le k \le n-1$:

\begin{align}
\binom{n}{k}_q&= q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q,\\
\binom{n}{k}_q&=  \binom{n-1}{k}_q + q^{n-k}\binom{n-1}{k-1}_q.
\end{align}

* We calculate explicitly:

\begin{align*}
q^k \binom{n-1}{k}_q +\binom{n-1}{k-1}_q &= q^k \frac{[n-1]!_q}{[n-k-1]!_q[k]!_q} +\frac{[n-1]!_q}{[n-k]!_q[k-1]!_q} \\
&= q^k \frac{[n-1]!_q}{[k]_q[n-k-1]!_q[k-1]!_q} +\frac{[n-1]!_q}{[n-k]_q[n-k-1]!_q[k-1]!_q}\\
&= \frac{q^k[n-k]_q[n-1]!_q + [k]_q[n-1]!_q}{[k]_q[n-k]_q[n-k-1]!_q[k-1]!_q}\\
&=\frac{\left( q^k[n-k]_q + [k]_q \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left( q^k\sum_{i=0}^{n-k-1} q^i + \sum_{j=0}^{k-1} q^j \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left( \sum_{i=k}^{n-1} q^i + \sum_{j=0}^{k-1} q^j \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left( \sum_{i=0}^{n-1} q^i \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{[n]_q[n-1]!_q}{[n-k]!_q[k]!_q} =\frac{[n]!_q}{[n-k]!_q[k]!_q}= \binom{n}{k}_q.
\end{align*}

* Again explicit calculation yields:

\begin{align*}
\binom{n-1}{k}_q +q^{n-k} \binom{n-1}{k-1}_q &= \frac{[n-1]!_q}{[n-k-1]!_q[k]!_q} +q^{n-k} \frac{[n-1]!_q}{[n-k]!_q[k-1]!_q} \\
&= \frac{[n-1]!_q}{[k]_q[n-k-1]!_q[k-1]!_q} +q^{n-k}\frac{[n-1]!_q}{[n-k]_q[n-k-1]!_q[k-1]!_q}\\
&= \frac{[n-k]_q[n-1]!_q + q^{n-k}[k]_q[n-1]!_q}{[k]_q[n-k]_q[n-k-1]!_q[k-1]!_q}\\
&=\frac{\left( [n-k]_q + q^{n-k}[k]_q \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left(\sum_{i=0}^{n-k-1} q^i + q^{n-k}\sum_{j=0}^{k-1} q^j \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left( \sum_{i=0}^{n-k-1} q^i + \sum_{j=n-k}^{n-1} q^j \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{\left( \sum_{i=0}^{n-1} q^i \right)[n-1]!_q}{[n-k]!_q[k]!_q}\\
&=\frac{[n]_q[n-1]!_q}{[n-k]!_q[k]!_q} =\frac{[n]!_q}{[n-k]!_q[k]!_q}= \binom{n}{k}_q. \qedhere
\end{align*}

As mentioned before the above lemma in particular proves by induction $\binom{n}{k}_q \in \N[q]$, since we have $\binom{n}{0}_q=\binom{1}{0}_q =1=\binom{1}{1}_q=\binom{n}{n}_q$ and $\binom{n}{1}_q=\binom{n}{n-1}_q=[n]_q$ for all $n \in \N$.

For $R$ a ring and $\xi \in R$, we denote the images of the above elements under the evaluation homomorphism

\begin{align*}
ev_\xi: \Z[q] &\to R
\\q &\mapsto \xi
\end{align*}

by $[n]_\xi, [n]!_\xi$ and $\binom{n}{k}_\xi$.

The following Lemma will be of use in multiple calculations and generalizes the ordinary binomial formula to a $q$-binomial formula:

The following formula holds in $\Z[x,y,q]$:

$$\prod_{k=0}^{n-1}(x+q^ky)=\sum_{k=0}^{n}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{n-k}y^k .$$

We use induction. For the induction base consider the case $n=1$, which leads

\begin{align*}
\prod_{k=0}^{0}(x+q^ky)&= (x+y)=\sum_{k=0}^{1}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{n-k}y^k .
\intertext{Now the claim follows by the following calculation:}
\prod_{k=0}^{n}\left(x+q^k y \right)&=\left(\prod_{k=0}^{n-1}\left(x+q^k y \right)\right)\left(x+q^ny\right)\\
&=\left(\sum_{k=0}^{n}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{n-k}y^k\right)\left(x+q^n y\right)\\
&=\sum_{k=0}^{n}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{n-k+1}y^k + \sum_{k=0}^{n}q^{\frac{k(k-1)}{2}+n}\binom{n}{k}_q x^{n-k}y^{k+1}\\
&=\sum_{k=0}^{n}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{n-k+1}y^k + \sum_{k=1}^{n+1}q^{\frac{(k-1)(k-2)}{2}+n}\binom{n}{k-1}_q x^{n-k+1}y^{k}\\
&=x^{n+1} + q^{\frac{n(n-1)}{2}+n}y^{n+1} +\sum_{k=1}^{n}\left(q^{\frac{k(k-1)}{2}}\binom{n}{k}_q + q^{\frac{(k-1)(k-2)}{2}+n}\binom{n}{k-1}_q\right) x^{n-k+1}y^{k}\\
&=x^{n+1} + q^\frac{(n+1)n}{2}y^{n+1} +\sum_{k=1}^{n}\left(q^{\frac{k(k-1)}{2}}\binom{n}{k}_q + q^{\frac{k(k-1)}{2}+n-k+1}\binom{n}{k-1}_q\right)  x^{n-k+1}y^{k}\\
&=x^{n+1} + q^\frac{(n+1)n}{2}y^{n+1}+\sum_{k=1}^{n}q^{\frac{k(k-1)}{2}}\left(\binom{n}{k}_q + q^{n-k+1}\binom{n}{k-1}_q\right) x^{n-k+1}y^{k} \\
&=x^{n+1} + q^\frac{(n+1)n}{2}y^{n+1}+\sum_{k=1}^{n}q^{\frac{k(k-1)}{2}}\binom{n+1}{k}_q x^{n-k+1}y^{k} \\
&=\sum_{k=0}^{n+1}q^{\frac{k(k-1)}{2}}\binom{n+1}{k}_q x^{n-k+1}y^{k}. \\
\end{align*}

Here we used in the third to last equality formula (2) from Lemma <ref>.

§.§ Vanishing at roots of unity

Now we will show that the $q$-binomial coefficients vanish at certain primitive roots of unity. A very powerful approach for this is to use cyclic sieving. Since this needs a more involved argument and we only need a vanishing argument for $q$-binomial coefficients, we will present a less technical proof. However our proof is strongly inspired by cyclic sieving and suffices for our purposes. For the cyclic sieving argument we refer to <cit.>.

We first need to review some standard representation theory notions to apply them in the vanishing theorem.

Let $n,k \in \N$ such that $0 \le k \le n$. Then

* let the set $\left\{1,2,...,n \right\}$ be denoted by $[n]$,

* let $\langle(1,2,...,n)\rangle :=C_n\subset S_n$ denote the cyclic group of order $n$ identified as the subgroup of the symmetric group generated by $(1,2,...,n)$.

Let $G$ be a group, $V$ a finite dimensional vector space over a field $\kk$ and $\varphi: G \to \Aut(V)$ an action of $G$ on $V$. Then the character $\chi_\varphi: G \to \kk $ of $\varphi$ is defined by $\chi_\varphi(g)=\tr(\varphi( g))$ for $g \in G$.

This is well defined since the trace is basis independent

We will denote for a ring $R$ and a set $I$ the free $R$-module over $I$ by $\langle I \rangle_R$.

Let $n \in \N$. Then define the action $\varphi_{n}$ of $C_n:= \langle (1,2,...,n)\rangle \subset S_n$  on $\langle[n]\rangle_\CCC$ as

$$ \varphi_{n}(1,2,...,n)(i)=(1,2,...,n)(i). $$

Let $V$ be a vector space over a field $\kk$. Then denote for $k \in \N$ the $k$-th exterior product of $V$ by $\Lambda^kV$.

Furthermore define for a group representation $\varphi:G \to \Aut({V,V})$ the induced representation

\begin{align*}
\Lambda^k\varphi: G &\to \Aut_\kk(\Lambda^kV,\Lambda^kV)\\
g &\mapsto  (\Lambda^k\varphi(g) :\Lambda^kV \to \Lambda^kV)
\end{align*}

Let $n,k \in \N$ such that $1\le k <n$. Then we have

Consider the basis $B:=\{e_{\{i_1,i_2,...,i_k\}}:= i_1\wedge i_2 \wedge ... \wedge i_k \mid 1 \le i_1 < i_2 <...<i_k \le n\}$ of $\Lambda^k \langle[n]\rangle_\CCC $. Then we get by definition

\begin{align*}
\Lambda^k \varphi_{n}(1,2,...,n)(e_{\{i_1,i_2,...,i_k\}}) &= \Lambda^k\varphi_{n}(1,2,...,n)(i_1\wedge i_2 \wedge ... \wedge i_k)\\
&=(1,2,...,n)i_1\wedge (1,2,...,n)i_2 \wedge ... \wedge (1,2,...,n)i_k  \\
&= \delta e_{\{(1,2,...,n)i_1,(1,2,...,n)i_2,...,(1,2,...,n)i_k\}},
\end{align*}

where $\delta = \pm 1$. In particular the matrix of  $\Lambda^k \varphi_{n}(1,2,...,n)$ with respect to $B$ has only zeroes on the diagonal. So we get

$$\chi_{\Lambda^k\varphi_{n}}(1,2,...,n)=\tr\left(\Lambda^k \varphi_{n}(1,2,...,n)\right)=0$$

as claimed.

Let $V$ be an $n$-dimensional vector space over a field $\kk$ and $f: V \to V$ a diagonalizable endomorphism with eigenvalues $x_1,x_2,...,x_n$. Then we have for $k \in \N$ that the endomorphism
$$\Lambda^k(f): \Lambda^k(V) \to \Lambda^k(V)$$
induced by $f$ is diagonalizable and we have

\begin{equation*}
\tr(\Lambda^k(f))= \sum_{\mathclap{1 \le i_1 <i_2<...<i_k \le n}} x_{i_1} x_{i_2}...x_{i_k}.
\end{equation*}

We have by assumption a basis $B:=\{b_1,b_2,...,b_n\}$ of $V$ consisting of eigenvectors for the respective eigenvalues of $f$. This basis induces a basis $\Lambda^k B$ on  $\Lambda^k(V)$. We may consider $\Lambda^k(f)$ with respect to this basis and get

\begin{align*}
\Lambda^k(f)\left(b_{i_1}\wedge b_{i_2} \wedge ... \wedge  b_{i_k}\right)=x_{i_1}b_{i_1}\wedge  x_{i_2}b_{i_2}\wedge  ... \wedge  x_{i_k}b_{i_k}=x_{i_1}x_{i_2} ...  x_{i_k}b_{i_1}\wedge  b_{i_2}\wedge  ... \wedge  b_{i_k}.
\end{align*}

So the vectors in $\Lambda^k B$ are eigenvectors of $\Lambda^k(f)$ as well. Using this basis to calculate $\tr(\Lambda^k(f))$ gives

\begin{align*}
\tr(\Lambda^k(f))= \sum_{\mathclap{1\le i_1 <i_2<...<i_k \le n}} x_{i_1} x_{i_2}...x_{i_k}.
\end{align*}

This finishes the proof.

Let $n,k \in \N$, such that $0\le k \le n$. We denote by $e_{k,n}(x_1,...,x_n)$ the $k$-th elementary symmetric polynomial in  $n$ variables. This is defined as

\begin{align*}
e_{k,n}(x_1,...,x_n):=&\sum_{\mathclap{1 \le i_1 <i_2<...<i_k \le n}} x_{i_1} x_{i_2}...x_{i_k}\in \Z[x_1,x_2,...,x_n],
\intertext{for $0<k$ and for $k=0$ as}e_{0,n}(x_1,...,x_n):=&1.
\end{align*}

After this reminder on some linear algebra and representation theory the only thing missing is a connection to $q$-binomials. For this we consider the following relation between $e_{k,n}(1,...,q^{n-1})$ and $\binom{n}{k}_q$.

Let $n,k\in \N$ such that $k\le n$ and $1 \le n$. Then the equation

\begin{equation*}
\end{equation*}

holds in $\Z[q]$.

This follows from Lemma <ref> by considering $x=1$ and $y=t$. More precisely we get the following sequence of identities, using for the second identity Vieta's formula:

\begin{align*}
\sum_{k=0}^{n}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q t^k = \prod_{k=0}^{n-1}\left( 1 + q^k t\right)= \sum_{k=0}^{n}e_{k,n}(1,q,...,q^{n-1})t^k.
\end{align*}

This gives the desired equation by comparing coefficients.

Now we just need to put the collected statements together and get:

Let $n,k \in \N$ such that $0<k<n$ and let $\xi\in \CCC$ be a primitive $n$-th root of unity. Then we have in $\CCC$

First of all we know $0 = \chi_{\Lambda^k\varphi_{n}}(1,2,...,n)$ by Proposition <ref>. Now $\varphi_{n}(1,2,...,n)$ has characteristic polynomial $x^n-1$ and so it has a basis of eigenvectors $\{v_1,...,v_n\}$, where $v_i$ has the eigenvalue $\xi^i$.  In particular it is diagonalizable. Now by Proposition <ref> also $\Lambda^k(\varphi_{n})(1,2,...,n)$ is diagonalizable and we get:

\begin{align*}
0 &= \chi_{\Lambda^k(\varphi_{n})}(1,2,...,n)=\sum_{\mathclap{1 \le i_1 <i_2<...<i_k \le n}} \xi^{i_1} \xi^{i_2}...\xi^{i_k}=e_{k,n}(\xi,\xi^2,...,\xi^{n-1},\xi^n).
\intertext{We get by $\xi^n=1=\xi^0$, Proposition~\ref{2.17} and the symmetry of the elementary symmetric polynomials that}
\end{align*}

Since $\CCC $ is a field and $\xi \neq 0$ we get $\binom{n}{k}_\xi=0$ as claimed.

The previous statement about $q$-binomial coefficients relies heavily on the particular structure of complex numbers.
Importantly, standard arguments, such as the cyclotomic polynomial <cit.>, allow to carry it over to arbitrary integral domains:

Let $R$ be an integral domain, $n \in \N$ and $\xi \in R$ a primitive $n$-th root of unity i. e. an element such that $\xi^n=1$ and $\xi^j\neq 1$ for all $0<j<n$. Then we have $\binom{n}{k}_\xi=0$ for all $0 < k < n$.

We will prove this statement in two steps, firstly showing the statement for $x \in \Z[x]/\left(\Phi_n\right)$, where $\Phi_n$ is the $n$-th cyclotomic polynomial, and in a second step that the evaluation morphism $\Z[x] \to R$ defined by $x \mapsto \xi$ factors through $\Z[x]/\left(\Phi_n\right)$.

Consider the map $\Z[x]/\left(\Phi_n\right) \to \CCC$ defined by $x \mapsto \xi$ for $\xi$ a primitive $n$-th root of unity. This is a well defined monomorphism since $\Phi_n$ is the $n$-th cyclotomic polynomial. We have $\binom{n}{k}_\xi=0$ for all $0< k < n$ by Theorem <ref> since this is just the image of $\binom{n}{k}_x \in \Z[x]/\left(\Phi_n\right) $ we get the desired formula in

Now consider for an arbitrary integral domain the following composition denoting by $\Quot(R)$ and $\overline{\Quot(R)}$ the quotient field of $R$ and its algebraic closure:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{
\node (ZX) at (0,1){$\Z[x]$};
\node (ZXmodphi) at (0,0){$\Z[x]/\left(\Phi_n\right)$};
\node (R) at (1.1,1){$R$};
\node (QR) at (2.1,1){$ \Quot(R)$};
\node (QRclosure) at (3.45,1){$\overline{\Quot(R)},$};

\draw[->]
(ZX) edge node[above]{$X \mapsto \xi$}(R);
\draw[right hook->]
(R) edge (QR)
(QR) edge (QRclosure);
\draw[->>]
(ZX)edge (ZXmodphi);
\draw[dashed, ->]
(ZXmodphi) edge (R);
\end{equation*}

where we only need to show that the dashed arrow exists such that the diagram commutes. By the homomorphism theorem it suffices to check that $\Phi_n(\xi)=0$ in $R$. This is equivalent to $\Phi_n(\xi)=0$ in $\overline{\Quot(R)}$, which holds since $\Phi_n$ is the $n$-th cyclotomic polynomial and $\xi$ is an $n$-th primitive root of unity in $R$ and hence also in $\overline{\Quot(R)}$.

This statement heavily uses the cyclotomic polynomial $\Phi_n$ and its properties. In particular that for every field $\kk$ with $\mathrm{char}(\kk)\nmid n$ the $n$-th primitive roots of unity are roots of $\Phi_n$, and that $\Phi_n$ can be defined in $\Z[x]$ and hence is independent of the field. For the precise arguments that show this we refer to Lang's book <cit.>.

§ LIFTING FUNCTORS

In this section we will apply the tools developed in the previous sections to define functors on $\CC_N(\C)$ using functors on $\C$. This is of particular interest since a lot of structures on a category can be encoded using functors. First we will consider ordinary functors, but later we will discuss how to lift bifunctors which will be essential for constructing the monoidal structure on the category of $N$-complexes.

Let $\C,\D$ be categories and $F: \C \to \D$ a functor. Then define for an $\infty$-complex $X$ in $\C$ the $\infty$-complex $F(X)$ in $\D$ as

\begin{align*}
F(X)^i &:= F(X^i) \\
(d_{F(X)} :F(X^i) \to F(X^{i+1}) ) &:=  F(d_X:X^i \to X^{i+1}).
\end{align*}

Furthermore define for a morphism of $\infty$-complexes $f:X \to Y$, the family of morphisms $(F(f)^i)_{i \in \Z}$ as

\begin{align*}
F(f)^i= F(f^i).
\end{align*}

The above assignments define a functor $F: \CC_\infty(\C) \to \CC_\infty(\D)$.

By construction $F(X)$ is an $\infty$-complex in $\D$, so it just remains to be checked that for a morphism of $\infty$-complexes $f:X \to Y$, the family $F(f)$ is indeed a morphism of $\infty$-complexes, since functoriality follows by levelwise functoriality. Hence we need to show that the diagram

\begin{equation*}
\tikz[heighttwo,xscale=3,yscale=2,baseline]{
\node (Xi) at (0,1){$ F(X^i)=F(X)^i$};
\node (Xi+1) at (0,0){$F(X^{i+1})=F(X)^{i+1}$};
\node (Yi+1) at (2,0){$F(Y^{i+1})=F(Y)^{i+1}$};
\node (Yi) at (2,1){$F(Y^{i})=F(Y)^i$};

\draw[->]
(Xi) edge node[left]{$d_{F(X)} =F(d_X)$}(Xi+1)
(Yi) edge node[right]{$d_{F(Y)} =F(d_Y)$}(Yi+1)
(Xi) edge node[above]{$ F(f)^i=F(f^i)$}(Yi)
(Xi+1) edge node[below]{$ F(f)^{i+1}=F(f^{i+1}) $}(Yi+1);
\end{equation*}
commutes for all $i \in \Z$. Since $F: \C \to \D$ is a functor, it suffices to show that

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{
\node (Xi) at (0,1){$ X^i$};
\node (Xi+1) at (0,0){$X^{i+1}$};
\node (Yi+1) at (2,0){$Y^{i+1}$};
\node (Yi) at (2,1){$Y^{i}$};

\draw[->]
(Xi) edge node[left]{$d_X$}(Xi+1)
(Yi) edge node[right]{$d_Y$}(Yi+1)
(Xi) edge node[above]{$f^i$}(Yi)
(Xi+1) edge node[below]{$f^{i+1} $}(Yi+1);
\end{equation*}
commutes for all $i \in \Z$, but this holds since $f$ is a morphism of $\infty$-complexes in $\C$.

Let $\C,\D$ be pointed categories, $N \in \N$ and $F:\C \to \D$ a pointed functor, i. e. $F(0)\cong 0$ for $0$ a zero object in $\C$. Then the functor $F: \CC_\infty(\C) \to \CC_\infty(\D)$ induces a functor $\CC_N(\C) \to \CC_N(\D)$.

By Proposition <ref> it suffices to check that $F(X)$ is an $N$-complex in $\D$ if $X$ is an $N$-complex in $\C$, but this is clear, since

and so $F(X)$ is an $N$-complex.

Let $\C$,$\D$ be categories and $F:\C \rightleftarrows \D:G$ a pair of adjoint functors. Then $F:\CC_\infty(\C) \rightleftarrows \CC_\infty(\D): G$ is also a pair of adjoint functors.

Since $F,G$ are adjoint functors we have natural transformations $\eta:\id_\C \to G\circ F$ and  $\epsilon: F \circ G \to \id_\D$ such that the following diagrams commute:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{
\node (F1) at (0,1){$F$};
\node (FGF) at (1,1){$FGF,$};
\node (F2) at (1,0){$F$};
\node (G1) at (3,1){$ G$};
\node (GFG) at (4,1){$GFG$};
\node (G2) at (4,0){$G.$};

\draw[->]
(F1) edge node[above]{$F \eta $}(FGF)
(FGF) edge node[right]{$\epsilon F$}(F2)
(G1) edge node[above]{$\eta G$}(GFG)
(GFG) edge node[right]{$G \epsilon $}(G2)
(F1) edge node[below left]{$\id$}(F2)
(G1) edge node[below left]{$\id $}(G2);
\end{equation*}
Now since $\eta$ and $\epsilon$ are natural transformations, we get that the following squares commute for all $X,Y \in \CC_\infty(\C)$ and $j \in \Z$:

\begin{equation*}
\tikz[heighttwo,xscale=3,yscale=2,baseline]{
\node (X1) at (0,1){$X^j$};
\node (GFX) at (1,1){$G\circ F(X^j)$};
\node (GFY) at (1,0){$G\circ F(X^{j+1})$};
\node (Y1) at (0,0){$X^{j+1}$};
\node (FGY) at (3,0){$ F\circ G (Y^{j+1})$};
\node (FGX) at (3,1){$ F\circ G (Y^j)$};
\node (X2) at (4,1){$Y^j$};
\node (Y2) at (4,0){$Y^{j+1}.$};
\draw[->]
(X1) edge node[above]{$ \eta_{X^j} $}(GFX)
(Y1) edge node[below]{$\eta_{X^{j+1}} $}(GFY)
(X1) edge node[left]{$d_X$}(Y1)
(GFX) edge node[right]{$G\circ F (d_X) $}(GFY)
(FGY) edge node[below]{$ \epsilon_{Y^{j+1}} $}(Y2)
(FGX) edge node[above]{$\epsilon_{Y^{j}} $}(X2)
(FGX) edge node[left]{$F \circ G (d_Y)$}(FGY)
(X2) edge node[right]{$d_Y $}(Y2);
\end{equation*}

Hence $(\eta_{X^j}:X^j \to G\circ F (X^j))_{j \in \Z}$ and $(\eta_{Y^j}:F \circ G(Y^j) \to Y^j)_{j \in \Z}$ define morphisms of $\infty$*complexes. These define by construction natural transformations $\eta_{\CC_\infty}:\id_{\CC_\infty(\C)} \to G \circ F$ and $\epsilon_{\CC_\infty(\C)}: F \circ G \to \id_{\CC_\infty(\D)}$. Furthermore the diagrams

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{
\node (F1) at (0,1){$F$};
\node (FGF) at (1,1){$FGF$};
\node (F2) at (1,0){$F$};
\node (G1) at (3,1){$ G$};
\node (GFG) at (4,1){$GFG$};
\node (G2) at (4,0){$G$};
\draw[->]
(F1) edge node[above]{$F \eta_{\CC_\infty} $}(FGF)
(FGF) edge node[right]{$\epsilon_{\CC_\infty} F$}(F2)
(G1) edge node[above]{$\eta_{\CC_\infty} G$}(GFG)
(GFG) edge node[right]{$G \epsilon_{\CC_\infty} $}(G2)
(F1) edge node[below left]{$\id$}(F2)
(G1) edge node[below left]{$\id $}(G2);
\end{equation*}

commute levelwise by construction and so they commute in general. This means that $\eta_{\CC_\infty}$ and $\epsilon_{\CC_\infty}$ define a counit and unit for $F$ and $G$ such that the triangle identities hold, which proves the claim.

Let $\C,\D$ be pointed categories, $N\in \N$ and $F:\C \rightleftarrows \D:G$ an adjunction of functors. Then $F:\CC_N(\C) \rightleftarrows \CC_N(\D): G$ is also a pair of adjoint functors.

By Proposition <ref> we get an adjunction $F:\CC_\infty(\C) \rightleftarrows \CC_\infty(\D): G$. Now by Corollary <ref> it suffices to show that $F:\C \to \D$ and $G:\D \to \C$ are actually pointed functors. Consider the following isomorphisms induced by the adjunction for all $X \in \C$ and $Y \in \D$:

\begin{align*}
\{*\}&\cong \C(0,F(Y))\cong \D(G(0),Y)\\
\{*\}&\cong \D(X,F(0))\cong \D(G(X),0).
\end{align*}

So $G(0)$ is an initial object and $F(0)$ is a final object. Now since $\C$ and $\D$ are pointed we get $G(0)\cong 0$ and $F(0)\cong 0$ which finishes the claim.

The following definition is a general construction which can be used to encode the property of bifunctors respecting an enriched structure in the sense that they commute with the operations coming from this structure.

Let $(\D,\otimes_\D,\unit_\D,\alpha_\D,\lambda_\D,\rho_\D)$ be a symmetric monoidal category and $\C,\C'$ be categories enriched over it.
We define $\C \otimes_\D \C'$ as the category consisting of objects of the form $(X , Y)$, where $X \in \C,Y \in \C'$, and morphism objects
$$\C \otimes_\D \C'\left((X_1,Y_1),(X_2,Y_2)\right):= \C(X_1,X_2)\otimes_\D \C'(Y_1, Y_2).$$

Now we will need some kind of lifting for bifunctors of the form $\C \times \C' \to \D$  since we want to construct a monoidal structure on the category of $N$-complexes. Similar to the ordinary case of chain complexes we need some kind of twisted Leibniz rule. We say twisted, because the ordinary Leibniz rule just works for $2$-complexes or $N$-complexes over rings with characteristic $2$. This is owing to the fact that $-1$ is a primitive second root of unity, which will be clear by the proof of Proposition <ref>. Hence we will define our lifted functors using a Leibniz rule which involves arbitrary elements of a ring. Furthermore we will show that if these elements are suitable roots of unity, we indeed get functors of $N$-complexes and not just $\infty$-complexes.

Let $R$ be a commutative ring, $\C,\C'$ and $\D$ be categories enriched over $\Mod_R$ such that $\D$ admits countable coproducts, $\xi \in R$, and let
$$F:\C \otimes_{\Mod_R} \C' \to \D$$
be an additive $R$-linear functor. Then define for objects $X \in \CC_\infty(\C)$ and $ Y \in \CC_\infty(\C')$ the $\infty$-complex $F_\xi(X,Y)$ in $\D$ by

$$F_\xi(X,Y)^n := \bigoplus_{i+j=n}F(X^i, Y^j)$$

with the differential defined on summands by

$$d_{F_\xi(X,Y)}\circ \iota_{i,j}:=\iota_{i+1,j}\circ F(d_X,\id)+ \xi^i \iota_{i,j+1} \circ F(\id,d_Y).$$

Here $\iota_{k,l}$ denotes the canonical inclusion
$$F(X^k,Y^l) \to \bigoplus_{i+j=n}F(X^i,Y^j)$$
for $k+l=n$.

Furthermore for morphisms of $\infty$-complexes $f:X \to X'$ and $g:Y \to Y'$ define the family $\left( F_\xi(f,g)^n \right)_{n\in \Z}$ of morphisms in $\D$ by

$$F_\xi(f,g)^n :=\bigoplus_{i+j=n} F(f^i,g^j).$$

Let $R$ be a commutative ring, $\C,\C'$ and $\D$ be categories enriched over $\Mod_R$ such that $\D$ admits countable coproducts, $\xi \in R$, and let
$$F:\C \otimes_{\Mod_R} \C' \to \D$$
be an additive $R$-linear functor. Then the assignments from Definition <ref> define a functor

$$F_\xi(\_,\_): \CC_\infty(\C) \otimes \CC_\infty (\C') \to \CC_\infty (\D).$$

It suffices to check for $f:X \to Y$ and $g:X' \to Y'$ that
$$F_\xi(f,g)\circ d_{F_\xi(X,X')}=d_{F_\xi(Y,Y')} \circ F_\xi(f,g).$$
Since by definition $F_\xi(\_,\_)$ is functorial as well as it has the right codomain. Hence we check the statement on direct summands of the form $F(X^{i}, X'^{j})$:

\begin{align*}
F_\xi(f,g)^n\circ d_{F_\xi(X,X')}\circ \iota_{i,j}&=\bigoplus_{i'+j'=n}F(f^{i'} , g^{j'})\circ \left(\iota_{i+1,j}\circ F(d_X, \id)  + \xi^{i} \iota_{i,j+1} \circ F(\id , d_{X'})\right)\\
&= \iota_{i+1,j}\circ F(f^{i+1}\circ d_X , g^{j})  + \xi^{i} \iota_{i,j+1} \circ F(f^{i} , g^{j+1} \circ d_{X'})\\
&=\iota_{i+1,j}\circ F(d_Y \circ f^{i} , g^{j})+ \xi^{i} \iota_{i,j+1} \circ  F(f^{i} , d_{Y'} \circ g^{j} )\\
&=\left(\iota_{i+1,j}\circ F(d_Y, \id) + \xi^{i} \iota_{i,j+1} \circ F(\id , d_{Y'})\right)\circ F(f^{i} , g^{j})\\
&=d_{F_\xi(X,X')} \circ\iota_{i,j} \circ  F(f^{i} , g^{j})\\
&=d_{F_\xi(Y,Y')} \circ \left( \bigoplus_{i'+j'=n}F(f^{i'} , g^{j'})\right) \circ \iota_{i,j}\\
&=d_{F_\xi(Y,Y')} \circ F_\xi(f,g) \circ \iota_{i,j}. \qedhere
\end{align*}

Having this general definition we can use our work on $q$-binomial coefficients to prove that $F_\xi(\_,\_)$ indeed defines a bifunctor on the category of $N$-complexes for $\xi$ a suitable root of unity. To make precise what suitable means we define the following:

Let $(\C,N,\xi)$ be a triple consisting of an additive category $\C$ enriched over $\Mod_R$ for $R$ a ring, $N \in \N$ and $\xi \in R$. Then we say that $(\C,N,\xi)$ allows the twisted Leibniz rule if one of the following holds:

* Every morphism in $\C$ is $p$-torsion for $p$ a prime number i. e. $pf=0$ for all morphisms $f$, $\xi=1$ and $N=p^n$ for some natural number $n\in \N$.

* The element $\xi \in R$ is a primitve $N$-th root of unity.

* Every morphism in $\C$ is $p$-torsion for $p$ a prime number, the element $\xi \in R$ is a primitive $k$-th root of unity and $N=p^nk$ for some natural number $n\in \N$.

In the above definition, case (iii) specializes to case (i) by setting $\xi=1$ and $k=1$.
Furthermore one can think of case (iii) as an enhancement of case (ii), by applying the structure of case (i) to case (ii).

Let $R$ be an integral domain, $\C,\C'$ and $\D$ be additive categories enriched over $\Mod_R$ such that $\D$ admits countable coproducts, $N \in \N$, $\xi \in R$ and let $F:\C \otimes_{\Mod_R} \C' \to \D$ be an additive bifunctor. Then the functor defined in Definition <ref> restricts to a functor

$$F_\xi(\_,\_): \CC_N(\C) \otimes \CC_N (\C) \to \CC_N (\C)$$
if $(\D,N,\xi)$ allows the twisted Leibniz rule.

Since we already know by Proposition <ref> that $F_\xi(\_,\_):\CC_\infty(\C) \times \CC_\infty(\C') \to \CC_\infty(\D)$ is a functor we only need to show in all cases that for $N$-complexes $X,Y$ we get that $F_\xi(X,Y)$ is an $N$-complex as well. For this we consider the three cases in Definition <ref> separately and calculate omitting inclusions:

* Since every morphism in $\D$ is a $p$-torsion for a prime number $p$ and $N=p^n$, we may use that for all morphisms $f$ in $\D$ we have $p f=f+f+...+f=0 \in \D$, hence it suffices to consider $\D$ as enriched over $R/pR$. Now we may use that $\binom{p}{k}=0$ in $R/pR$ for all $0<k<p$. First we generalize this to $\binom{N}{k}=0$ if $0<k < N$.
For this consider the following calculation in $R/pR[x,y]$

\begin{align*}
\sum_{k=1}^{N} \dbinom{N}{k} x^{N-k}y^k&=(x+y)^N=(((x+y)^p)^p...)^p=(((x^p+y^p)^p)^p...)^p\\
&=x^{p^n} + y^{p^n}=x^N+y^N.
\end{align*}

So the claim holds by comparing coefficients. Now using this we get

\begin{align*}
d_{F_1(X,Y)}^N=& d^{N-1}(F(d_X, \id_Y) + F(\id_X, d_Y))\\
=&d^{N-2}(F(d_X^2, \id_Y) + F(d_X, d_Y) + F(d_X, d_Y) + F(\id_X, d^2_Y))\\
=& d^{N-2}(F(d_X^2 , \id_Y) + 2 F( d_X , d_Y)  + F(\id_X , d^2_Y))\\
=&d^{N-3}(F(d_X^3, \id_Y) + 3 F(d_X^2, d_Y) + 3 F(d_X, d_Y^2) + F(\id_X, d^3_Y)\\
=&\sum_{k=0}^{N} \dbinom{N}{k} F(d^{N-k}_X, d^k_Y)\\
=&F(d_X^N, \id_Y) + F(\id_X , d_Y^N)\\
\end{align*}

and so $F_\xi(X, Y)$ is an $N$-complex. Here we used in the second to last equality $\binom{N}{k}=0$ if $0<k < N$.

* In this case we will use the two formulas from Proposition <ref> and Corollary <ref>.
Here we need for $\binom{N}{k}_\xi=0$ that $\xi$ is a primitive $N$-th root of unity.
Analogously to Lemma <ref> we calculate for each summand (cf. Remark <ref>):

\begin{align*} &\left. d_{F_\xi (X ,Y)}^N \right|_{F(X^i, Y^j)}\\
=& d^{N-1} \left(F(d_X, \id_Y) + \xi^i F(\id_X, d_Y)\middle) \right|_{F(X^i, Y^j)}\\
=&d^{N-2}\left( F(d_X^2, \id_Y) + \xi^{i+1} F(d_X, d_Y) + \xi^i F(d_X , d_Y) + \xi^{2i}F(\id_X , d^2_Y) \middle)\right|_{F(X^i, Y^j)}\\
=&d^{N-2}\left( \binom{2}{2}_{\xi}F(d_X^2, \id_Y)  + \xi^{i} \binom{2}{1}_{\xi}F(d_X , d_Y) +\xi^{2i}\binom{2}{0}_{\xi}F(\id_X, d^2_Y) \middle) \right|_{X^i\otimes Y^j}\\
=&d^{N-3} \left( \binom{2}{2}_{\xi}F(d_X^3 , \id_Y)  +\xi^{i+2}\binom{2}{2}_{\xi}F(d_X^2 , d_Y) +\xi^{i} \binom{2}{1}_{\xi}F(d_X^2 , d_Y) \right.\\
+& \left. \xi^{2i+1} \binom{2}{1}_{\xi}F(d_X , d_Y^2)+\xi^{2i}\binom{2}{0}_{\xi}F(d_X d^2_Y)+\xi^{3i}\binom{2}{0}_{\xi}F(\id_X , d^3_Y) \middle)\right|_{F(X^i, Y^j)}\\
=& d^{N-3} \left( \binom{3}{3}_{\xi}F(d_X^3 , \id_Y)  +\xi^{i}\binom{3}{2}_{\xi}F(d_X^2 , d_Y)  + \xi^{2i} \binom{3}{1}_{\xi}F(d_X ,d_Y^2)+\xi^{3i}\binom{3}{0}_{\xi}F(d_X , d^3_Y )\middle)\right|_{F(X^i, Y^j)}\\
&\vdots \\
=&\left. \sum_{k=0}^N \xi^{\frac{k(k-1)}{2}} \binom{N}{N-k}_{\xi} F(d_X^{N-k}, d_Y^{k})\right|_{F(X^i, Y^j)}\\
=& F(d_X^{N} , \id_Y) + F(\id_X , d_Y^N)\\
\intertext{Hence}
d_{F_\xi(X, Y)}^N=&F(d_X^{N}, \id_Y) +  F(\id_X , d_Y^N)=0
\end{align*}

on all of $X\otimes_\xi Y$ and so it is an $N$-complex as desired.

* Since this case is a combination of the above cases we will use both of the above calculations. First we will use that for $\xi\in R$ a primitive $k$-th root of unity and $l \in \N$ we have

\begin{align*}
d_{F_\xi(X, Y)}^k=& F(d_X^{k}, \id_Y) + F(\id_X , d_Y^k).
\end{align*}

Using this we observe that

\begin{align*}
d_{F_\xi(X, Y)}^N&=d_{F_\xi(X, Y)}^{p^nk}= d_{F_\xi(X, Y)}^{k}\circ    d_{F_\xi(X, Y)}^{k} \circ ...\circ  d_{F_\xi(X, Y)}^{k},
\end{align*}

with $p^n$ factors in the factorization. Now consider $d_{F_\xi(X, Y)}^k$ instead of $d_{F_\xi(X, Y)}$. The morphism $d_{F_\xi(X, Y)}^k$ now behaves the same way as $d_{F_\xi(X, Y)}$, with $\xi$ replaced by $1$ and $N$ replaced by $p^n$, yielding an analogous calculation to case (1):

\begin{align*}
d_{F_\xi(X, Y)}^N=&d_{F_\xi(X, Y)}^{N-k}(F(d_X^{k} , \id_Y) +  F(\id_X, d_Y^k))\\
=&d_{F_\xi(X, Y)}^{N-2k}(F(d_X^{2k} , \id_Y) + 2F(d_X^{k} , d_Y^{k})   + F(\id_X, d_Y^{2k}))\\
=&F_\xi(d_{X, Y}^{N-3k}(F(d_X^{3k} , \id_Y) + 3 F(d_X^{2k} , d_Y^{k}) + 3F(d_X^{k} d_Y^{2k})   + F(\id_X ,d_Y^{3k}))\\
&\vdots \\
=&\sum_{l=0}^{p^n} \dbinom{p^n}{l} F(d^{(p^n-l)k}_X, d^{lk}_Y)\\
=&F(d_X^{p^nk},\id_Y) + F(\id_X , d_Y^{p^nk})\\
=&F(d_X^N, \id_Y) + F(\id_X , d_Y^N)\\
\end{align*}

so $F_\xi(X,Y)$ is again an $N$-complex.

The above proof can be also seen as an application of the formula

$$\prod_{k=0}^{N-1}(x+q^ky)=\sum_{k=0}^{N}q^{\frac{k(k-1)}{2}}\binom{n}{k}_q x^{N-k}y^k $$

from Lemma <ref> by setting $x:=F(d_X,\id_Y), y:= F(\id_X, d_Y)$ and $q:= \xi$. Observing that $\xi$ is a $k$-th primitive root of unity with $k \mid N$ yields that every distinct power of $\xi$ appears precisely $\frac{N}{k}$ times and so we get the desired result. However, we proved this $q$-binomial formula in the ring $\Z[x,y,q]$, so we would need a general argument to apply it here, which we do not think is necessary, since we only applied it in full strength in case (2).

Another important type of functors we need to lift are functors which are covariant in one variable and contravariant in the other since these arise as internal homs. In particular they are of interest if we want to construct a closed monoidal structure on the category of $N$-complexes.

Let $R$ be a commutative ring, $\C,\C'$ and $\D$ be categories enriched over $\Mod_R$ such that $\D$ admits countable products, $\xi \in R$ and $F(\_,\_):\C^{op} \otimes_{\Mod_R} \C' \to \D$ an $R$-linear bifunctor. Then define for two $\infty$-complexes $X,Y$ the $\infty$-complex $F_\xi(X,Y)$ as

\begin{align*}
F_\xi(X,Y)^i&:=\prod_{j\in \Z} F(X^j,Y^{i+j})
\end{align*}

with the differential defined on factors by

\begin{align*}
\pi_{j,i+j} \circ d_{F_\xi(X,Y)}:= F(X^j,d_Y)\circ \pi_{j,i+j-1} - \xi^{i-1} F(d_X,Y^{i+j})\circ \pi_{j+1,i+j}.
\end{align*}

Here $\pi_{k,l}$ denotes the canonical morphism
$$\pi_{k,l}: \prod_{j\in \Z} F(X^j,Y^{i+j})\to F(X^k,Y^{l})$$
for $i=l-k$.

Furthermore define for a morphism of $\infty$-complexes $f:X \to X'$ the family of morphisms in $\C$:

\begin{align*}
F_\xi(f,Y)&:=\left( \prod_{j\in \Z}F(f^j,Y^{i+j}):F_\xi(X',Y)^i \to F_\xi(X,Y)^i\right)_{i \in \Z},
\end{align*}

and for $f:Y \to Y'$ the family of morphisms in $\C'$:

\begin{align*}
F_\xi(X,f)^i&:=\left( \prod_{j\in \Z}F(X^j,f^{i+j}): F_\xi(X,Y)^i \to F_\xi(X,Y')^i \right)_{i \in \Z} .
\end{align*}

Let $R$ be a commutative ring, $\C,\C'$ and $\D$ be additive categories enriched over $\Mod_R$ such that $\D$ admits countable products, $\xi \in R$ and $F(\_,\_):\C^{op} \otimes_{\Mod_R} \C' \to \D$ an $R$-linear bifunctor. Then
the assignments from Definition <ref> define a functor

\begin{align*}
F_\xi(\_,\_):\CC_\infty(\C)^{op} \times \CC_\infty(\C') \to \CC_\infty(\D).
\end{align*}

One just needs to check that for $f:X \to X'$ in $\C$ and $Y \in \C'$, respectively $Y\in \C$ and $f:X \to X'$ in $\C'$, the families $F_\xi(f,Y)$ and $F_\xi(Y,f)$ indeed define morphisms of $\infty$-complexes since the assignments are levelwise products of functors. Consider first
$$F_\xi(f,Y): F_\xi(X',Y) \to F_\xi(X,Y)$$
for $f:X\to X' \in \C(X,X')$ and $Y \in \C'$. So we need to show that

\begin{equation*}
\tikz[heighttwo,xscale=4,yscale=2,baseline]{
\node (1) at (0,1){$F_\xi(X',Y)^{i-1}$};
\node (2) at (1,1){$F_\xi(X,Y)^{i-1}$};
\node (3) at (0,0){$F_\xi(X',Y)^{i}$};
\node (4) at (1,0){$F_\xi(X,Y)^{i}$};
\draw[->]
(1) edge node[left] {$d_{F_\xi(X',Y)}$}(3)
(2) edge node[right] {$d_{F_\xi(X,Y)}$}(4)
(1) edge node[above] {$F_\xi(f,Y)^{i-1}$} (2)
(3) edge node[below] {$F_\xi(f,Y)^{i}$} (4)
\end{equation*}

commutes. This we check on factors:

\begin{align*}
&\pi_{j,i+j}\circ \left( F_\xi(f,Y)^{i} \circ d_{F_\xi(X',Y)}\right)  \\
&=  \pi_{j,i+j}\circ \prod_{j'\in \Z} F(f^{j'},Y^{i+j'}) \circ d_{F_\xi(X',Y)}\\
&=F(f^j,Y) \circ \pi_{j,i+j} \circ  d_{F_\xi(X',Y)}\\
&= F(f^j,Y) \circ (F({X'}^j,d_Y)\circ \pi_{j,i+j-1} - \xi^i F(d_{X'},Y^{i+j})\circ \pi_{j+1,i+j})  \\
&= F(f^{j},d_Y)\circ \pi_{j,i+j-1} -\xi^i F(f^{j} \circ d_{X'} , Y^{j+i})\circ \pi_{j+1,i+j}  \\
&=  F(f^{j},d_Y)\circ \pi_{j,i+j-1} - \xi^i F((d_{X'} \circ f^{j})^{op}   , Y^{j+i})\circ \pi_{j+1,i+j}  \\
&= F(f^{j},d_Y)\circ \pi_{j,i+j-1}- \xi^i F((f^{j+1} \circ d_{X} )^{op} , Y^{j+i})\circ \pi_{j+1,i+j}  \\
&= F(f^{j},d_Y)\circ \pi_{j,i+j-1} - \xi^i F(d_{X}  \circ f^{j+1} , Y^{j+i})\circ \pi_{j+1,i+j}   \\
&=  F(X,d_Y)\circ F(f^{j},Y^{i+j-1}) \circ\pi_{j,i+j-1 }- \xi^i F(d_{X} , Y^{j+i})\circ F(f^{j+1} , Y^{j+i})\circ \pi_{j+1,i+j} \\
&= \left(F(X,d_Y) \circ \pi_{j,i+j-1}- \xi^i F(d_{X} , Y^{j+i})\circ  \pi_{j+1,i+j}  \right) \circ \prod_{j'\in\Z}F(f^{j'},Y^{i+j'-1}) \\
&= \pi_{j,i+j} \circ (d_{F_\xi(X,Y)} \circ F_\xi(f,Y)^{i-1}).
\end{align*}

Now we show for $F_\xi(Y,f): F_\xi(Y,X) \to F_\xi(Y,X')$ for $Y \in \C$ and $f:X \to X'$ a morphism in $\C'$ that the diagram

\begin{equation*}
\tikz[heighttwo,xscale=4,yscale=2,baseline]{
\node (1) at (0,1){$F_\xi(Y,X)^{i-1}$};
\node (2) at (1,1){$F_\xi(Y,X')^{i-1}$};
\node (3) at (0,0){$F_\xi(Y',X)^{i}$};
\node (4) at (1,0){$F_\xi(Y,X')^{i}$};
\draw[->]
(1) edge node[left] {$d_{F_\xi(Y,X)}$}(3)
(2) edge node[right] {$d_{F_\xi(Y,X')}$}(4)
(1) edge node[above] {$F_\xi(Y,f)^{i-1}$} (2)
(3) edge node[below] {$F_\xi(Y,f)^{i}$} (4)
\end{equation*}

indeed commutes. Again we check on factors:

\begin{align*}
&\pi_{j,i+j}\circ \left( F_\xi(Y,f)^{i} \circ d_{F_\xi(Y,X)}\right) \\
&= \pi_{j,i+j}\circ \left(\prod_{j'\in \Z}F(Y^{j'},f^{i+j'})\right) \circ d_{F_\xi(Y,X)} \\
&=F(Y^j,f^{i+j})\circ \pi_{j,i+j}\circ d_{F_\xi(Y,X)}\\
&=F(Y^j,f^{i+j})\circ  \left( F(Y^j,d_X)\circ \pi_{j,i+j-1}- \xi^i F(d_Y,X^{i+j})\circ \pi_{j+1,i+j} \right)\\
&=  F(Y^j,f^{i+j})\circ F(Y^j,d_X)\circ \pi_{j,i+j-1} - \xi^i F(Y^j,f^{i+j})\circ F(d_Y,X^{i+j})\circ \pi_{j+1,i+j} \\
&= F(Y^j,f^{i+j}\circ d_X)\circ \pi_{j,i+j-1}- \xi^i F(d_Y,f^{i+j})\circ \pi_{j+1,i+j}\\
&=   F(Y^j,d_X \circ  f^{i+j-1})\circ \pi_{j,i+j-1}- \xi^i F(d_Y,X^{i+j}) \circ  F(Y^{j+1},f^{i+j})\circ \pi_{j+1,i+j} \\
&=  F(Y^j,d_X) \circ  F(Y^j,f^{i+j-1})\circ \pi_{j,i+j-1}- \xi^i F(d_Y,X^{i+j}) \circ  F(Y^{j+1},f^{i+j})\circ \pi_{j+1,i+j}  \\
&= \left( F(Y^j,d_X) \circ \pi_{j,i+j-1}- \xi^i F(d_Y,X^{i+j}) \circ  \pi_{j+1,i+j}\right) \circ \left(\prod_{j'\in \Z}F(Y^{j'},f^{i+j'-1})\right)\\
&=\pi_{j,i+j} \circ d_{F_\xi(X,Y)} \circ F_\xi(Y,f)^{i-1}.
\end{align*}

Here the shift in $Y$ between the fourth and sixth line occurs since $F(\_,X)$ is contravariant.

Similarly to the case of additive bifunctors, we want that $F_\xi(\_,\_)$ again defines a functor on the category of $N$-complexes for certain $\xi$ and $F(\_,\_)$ a contravariant/covariant bifunctor.

Let $R$ be an integral domain, $\C,\C'$ and $\D$ be additive categories enriched over $\Mod_R$, $\xi \in R$ such that $\D$ admits countable products, $N \in \N$ such that $(\D,N,\xi)$ allows the twisted Leibniz rule and let $F(\_,\_):\C^{op} \otimes_{\Mod_R} \C' \to \D$ be an $R$-linear bifunctor. Then the functor

\begin{align*}
F_\xi(\_,\_):\CC_\infty(\C)^{op} \times \CC_\infty(\C') &\to \CC_\infty(\D)
\intertext{induces a functor}
F_\xi(\_,\_):\CC_N(\C)^{op} \times \CC_N(\C') &\to \CC_N(\D).
\end{align*}

By Proposition <ref> we have that $F_\xi(\_,\_)$ induces a functor $\CC_\infty(\C)^{op} \times \CC_\infty(\C') \to \CC_\infty(\D)$, so it suffices to check in all cases, that for $X\in \CC_N(\C)$ and $Y \in \C'$ the complex $F_\xi(X,Y)$ is again an $N$-complex. This is analogous to the proof of Theorem <ref>, still we give for completeness reasons the proof. Similarly to Theorem <ref> we will check all three cases in Definition <ref> separately. Furthermore we omit for legibility the projections in the calculations.

* Since every morphism in $\D$ is a $p$-torsion for a prime number $p$ and $N=p^n$, we may use that for all morphisms $f$ in $\D$ we have $p f=0 \in \D$, hence it suffices to consider $\D$ as enriched over $R/pR$. Now we may use in $R/pR$ that $\binom{N}{k}=0$ for all $0<k<N$ just like in Theorem <ref>.
Using this one gets

\begin{align*}
\pi_{j,i+j} \circ  d_{F_1(X,Y)}^N =& (F(d_X,Y^{i+j}) -  F(X^j,d_Y))\circ d_{F_\xi (X,Y)}^{N-1}\\
=&(F(d^2_X,Y^{i+j}) - F(d_X,d_Y) - F(d_X,d_Y) + F(X^j,d^2_Y))\circ d_{F_\xi (X,Y)}^{N-2}\\
=& (F(d^2_X,Y^{i+j}) - 2 F(d_X,d_Y) + F(X^j,d^2_Y))\circ d_{F_\xi (X,Y)}^{N-2}\\
=&(F(d^3_X,Y^{i+j}) - 3F(d^2_X,d_Y) + 3 F(d_X,d^2_Y) -F(X^j,d^3_Y))\circ d_{F_\xi (X,Y)}^{N-3}\\
=&\sum_{k=1}^{N} (-1)^k \dbinom{N}{k} F(d^{N-k}_X,d^k_Y)\\
=&F(d^N_X,Y^{i+j}) + (-1)^N F(X^j,d^N_Y))\\
\end{align*}

and so $F_\xi(X,Y)$ is an $N$-complex.

* In this case we will use the following two formulas from Proposition <ref> and Corollary <ref>:
Where we need for $\binom{N}{k}_\xi=0$ that $\xi$ is an $N$-th root of unity.
Now we calculate analogously to the formula of Lemma <ref> $d^N_{F_\xi(X,Y)}$ for each summand, keeping track of the respective degrees and setting $x=F(\id_X,d_Y),\; y=-F(d_X,\id_Y)$ and $q=\xi$:

\begin{align*} &\pi_{j,i+j} \circ  d_{F_\xi (X,Y)}^N \\
=&\left(  F(X^j,d_Y) - \xi^i F(d_X,Y^{i+j})\right)\circ d_{F_\xi (X,Y)}^{N-1}\\
=&\left(F(X^j,d^2_Y) - \xi^i F(d_X,d_Y) -\xi^{i+1} F(d_X,d_Y)+ \xi^{2i} F(d^2_X,Y^{i+j})    \right)\circ d_{F_\xi (X,Y)}^{N-2}\\
=&\left( \binom{2}{0}_{\xi}F(X^j,d^2_Y)  - \xi^{i} \binom{2}{1}_{\xi} F(d_X,d_Y) +\xi^{2i} \binom{2}{2}_{\xi}F(d^2_X,Y^{i+j}) \right)\circ d_{F_\xi (X,Y)}^{N-2}\\
=& \left( \binom{2}{0}_{\xi}F(X^j,d^3_Y)   -\xi^{i+2}\binom{2}{0}_{\xi}F(d_X,d^2_Y)  -\xi^{i} \binom{2}{1}_{\xi}F(d_X,d^2_Y) \right.\\
& + \left.\xi^{2i+1} \binom{2}{1}_{\xi}F(d^2_X,d_Y)+\xi^{2i}\binom{2}{2}_{\xi}F(d^2_X,d_Y)- \xi^{3i} \binom{2}{2}_{\xi}F(d^3_X,Y^{i+j}) \right)\circ d_{F_\xi (X,Y)}^{N-3}
\\
=&\left(\binom{3}{0}_{\xi}F(X^j,d^3_Y)-  \xi^{i} \binom{3}{1}_{\xi}F(d_X,d^2_Y)+\xi^{2i}\binom{3}{2}_{\xi}F(d^2_X,d_Y)- \xi^{3i} \binom{3}{3}_{\xi}F(d^3_X,Y^{i+j}) \right)\circ d_{F_\xi (X,Y)}^{N-3}\\
&\vdots \\
=& \sum_{k=0}^N \xi^{\frac{k(k-1)}{2}} \binom{N}{N-k}_{\xi}(-1)^k F(d^k_X,d^{N-k}_Y)\\
=& F(d^N_X,Y^{i+j}) + (-1)^k F(X^j,d^N_Y)\\
\end{align*}

\begin{equation*}
d_{F_\xi (X,Y)}^N=F(d^N_X,Y^{i+j}) + F(X^j,d^N_Y)=0
\end{equation*}

on all factors of $F_\xi (X,Y)$ and so it is an $N$-complex as desired.

* Since this case is a combination of the above two we will use both of the above calculations. First we will use that for $\xi\in R$ a primitive $k$-th root of unity and $l \in \N$ we have

\begin{align*}
d_{F_\xi(X, Y)}^k=& F(d_X^{k}, \id_Y) + (-1)^k F(\id_X , d_Y^k).
\end{align*}

Using this we observe that the following holds:

\begin{align*}
d_{X\otimes_\xi Y}^N&=d_{X\otimes_\xi Y}^{p^nk}= d_{X\otimes_\xi Y}^{k}\circ    d_{X\otimes_\xi Y}^{k} \circ ...\circ  d_{X\otimes_\xi Y}^{k},
\end{align*}

with $p^n$ factors in the composition. Now consider $d_{F_\xi(X, Y)}^k$ instead of $d_{F_\xi(X, Y)}$. The morphism $d_{F_\xi(X, Y)}^k$ now behaves the same way as $d_{F_\xi(X, Y)}$, with $\xi$ replaced by $1$ and $N$ replaced by $p^n$, yielding an analogous calculation to case (1):

\begin{align*}
\pi_{j,i+j} \circ d_{F_\xi (X,Y)}^N=&\left(F(d^k_X,Y^{i+j}) +  (-1)^k F(X^j,d^k_Y)\right) \circ d_{F_\xi (X,Y)}^{N-k}\\
=&\left(F(d^{2k}_X,Y^{i+j}) + (-1)^k 2F(d^k_X,d^k_Y)   + (-1)^{2k} F(X^j,d^{2k}_Y)\right) \circ d_{F_\xi (X,Y)}^{N-2k}\\
=&\left((F(d^{3k}_X,Y^{i+j}) + (-1)^k 3F(d^{2k}_X,d^k_Y) + (-1)^{2k} 3F(d^k_X,d^{2k}_Y)   + (-1)^{3k} F(X^j,d^{3k}_Y)\right) \circ d_{F_\xi (X,Y)}^{N-3k}\\
=&\sum_{l=1}^{p^n} \dbinom{p^n}{l}(-1)^{lk}F(d^{(p^n-l)k}_X,d^{lk}_Y)\\
=&F(d^{p^nk}_X,Y^{i+j}) + (-1)^{nk} F(X^j,d^{p^nk}_Y)\\
=&F(d^{N}_X,Y^{i+j}) + (-1)^{nk} F(X^j,d^{N}_Y)\\
\end{align*}

So $F_\xi (X,Y)$ is again an $N$-complex.

Having established these tools for the lifting of functors, we can finally construct the monoidal structure on the category of $N$-complexes and show that it is closed if the underlying category is the category of modules over an integral domain.

§ MONOIDAL STRUCTURE

In this section we will generalize the tensor product on the category of chain complexes to arbitrary $N$-complexes. Since we want to do this in a very general context we need to introduce first the notions of countable distributive monoidal categories and monoidal categories enriched over a ring $R$. This will allow us to use liftings of bifunctors from the previous section to define a monoidal structure.

§.§ Background on distributive monoidal structures

First we will discuss general distributivity properties for monoidal categories in order to use the liftings from the previous section.

Let $(\C,\otimes,\unit,\alpha,\lambda,\rho)$ be a monoidal category that admits finite coproducts. Then define the following morphisms:

* For $X,Y,Z\in \C$ the left distributivity morphism

\begin{align*}
{}_{X,Y,Z}\delta : (X \otimes Y) \oplus (X \otimes Z) \to X \otimes (Y \oplus Z),
\end{align*}

is the morphism induced by the universal property of the coproduct in the following diagram:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{
\node (1) at (0,1){$ X\otimes Y$};
\node (2) at (2,2){$ X\otimes Z$};
\node (3) at (2,1){$(X \otimes Y)\oplus (X \otimes Z)$};
\node (4) at (4,0){$ X \otimes (Y \oplus Z)$};
\node[overlay] (Point) at (4.7,-0.05){.};
\draw[->]
(1) edge node[above]{$  \iota_{X\otimes Y}$}(3)
(2) edge node[left]{$\iota_{X \otimes Z} $}(3)
(1) edge node[below left]{$ \id_X \otimes \iota_Y$}(4)
(2) edge node[above right]{$\id_X \otimes \iota_Z$}(4)
\draw[dashed,->]
(3) edge (4);
\end{equation*}

* For $X,Y,Z\in \C$ the right distributivity morphism

\begin{align*}
\delta_{Y,Z,X} : (Y \otimes X) \oplus (Z \otimes X) \to  (Y \oplus Z) \otimes X,
\end{align*}

is the morphism induced by the universal property of the coproduct and the morphisms

\begin{align*}
\iota_Y \otimes \id_X &: (Y \otimes X)\to(Y \oplus Z) \otimes X \\
\iota_Z \otimes \id_X &: (Z \otimes X)\to(Y \oplus Z) \otimes X .
\end{align*}

By construction the left distributivity morphism defines a natural transformations between the two functors

\begin{align*}
\C \times \C \times \C &\to \C \\
(X,Y,Z) &\mapsto (X\otimes Y)\oplus (X \otimes Z) \\
(X,Y,Z) &\mapsto X\otimes (Y\oplus Z).
\end{align*}

The same holds for the right distributivity morphism and

\begin{align*}
\C \times \C \times \C &\to \C \\
(X,Y,Z) &\mapsto (Y\otimes X)\oplus (Z \otimes X) \\
(X,Y,Z) &\mapsto  (Y\oplus Z)\otimes X .
\end{align*}

A monoidal category $(\C,\otimes,\unit,\alpha,\lambda,\rho)$, is called

* distributive if it is additive, the morphisms ${}_{X,Y,Z}\delta$ and $\delta_{Y,Z,X}$ are isomorphisms for all $X,Y,Z \in \C$ and $X \otimes 0 \cong 0 \cong 0 \otimes X$ for all $X \in \C$.

* countable distributive if $\C$ is distributive, admits countable coproducts and for all $X,Y_i \in \C$, for $i \in \Z$ the canonical morphisms

\begin{align*}
\tau_{X,Y_i} &: \bigoplus_{i \in \Z} (X \otimes Y_i)\to  X \otimes \bigoplus_{i \in \Z} Y_i \\
\sigma_{Y_i,X} &: \bigoplus_{i \in \Z} ( Y_i \otimes X)\to  \left(\bigoplus_{i \in \Z} Y_i \right) \otimes X
\end{align*}
are isomorphisms.

Let $(\C,\otimes_\C,\unit_\C,\alpha_\C,\lambda_\C,\rho_\C)$ be a monoidal category enriched over a symmetric monoidal category $(\D,\otimes_\D,\unit_\D,\alpha_\D,\lambda_\D,\rho_\D)$.
We call $\C$ a $\D$-monoidal category if the bifunctor $\otimes_\C : \C \times \C  \to \C$ and structure morphisms factor through the canonical functor $\C \times \C \to\C \otimes_\D \C$, which is the identity on objects and $\otimes_\D$ on morphism spaces.

Definition <ref> implies that $\_ \otimes_\C Y$ , $X\otimes_\C \_$ and the natural transformations involved in the monoidal structure of $\C$ are in fact $\D$-enriched functors and natural transformations of $\D$-enriched functors.

Now we have a good setup for applying the constructions from Section 3 and we can define the monoidal structure on $\CC_N(\C)$. However it is not yet clear that this indeed defines a monoidal structure since we are in a very general setup at the moment.

Let $(\C,\otimes, \unit,\alpha,\lambda,\rho)$ be a countable distributive $\Mod_R$-monoidal category, for $R$ a commutative ring, and $\xi \in R$. Then define the following  on $\CC_\infty(\C)$:

* A functor $\_ \otimes_\xi \_ : \CC_\infty(\C) \times \CC_\infty(\C) \to \CC_\infty(\C)$ just as in Definition <ref> using the bifunctor $\_ \otimes \_ :\C \otimes_\D \C \to \C$.

* The $\infty$-complex $\unit_\xi:= \mu^0_1(\unit)$.

* The morphism $\alpha_{\xi,X,Y,Z} : (X\otimes Y) \otimes Z \to X\otimes (Y \otimes Z)$ levelwise defined as (cf. Remark <ref>)

$$(\alpha_{\xi,X,Y,Z})^j:=\left(\bigoplus_{l+m=n}\sigma_{X^i\otimes Y^k,Z^l}\right)\circ \gamma_{l,i,k}^{-1} \circ \left(\bigoplus_{i+k+l=n}\alpha_{X^i,Y^k,Z^l}\right) \circ \gamma_{i,k,l} \circ \left( \bigoplus_{i+j=n}\tau^{-1}_{X^i,Y^k \otimes Z^l}\right).$$

Here $\gamma_{i,k,l}$ is the canonical natural isomorphism
$$\bigoplus_{i+j=n} \left(\bigoplus_{k+l=j}M_{i,k,l} \right) \to \bigoplus_{i+k+l=n}M_{i,k,l} .$$

* The morphism $\lambda_{\xi,X} :\unit_\xi \otimes_\xi X  \to X$ levelwise by $(\lambda_\xi(X))^j:=\lambda(X^j)$.

* And the morphism $\rho_{\xi,X} : X\otimes_\xi \unit_\xi \to X$  levelwise by $(\rho_\xi(X))^j:=\rho(X^j)$.

Regarding Definition <ref> the following remarks are to be made:

* The construction of $\alpha_{\xi,X,Y,Z}$ can be seen more intuitively in the following diagram in $\C$, which also explains the numerous assumptions on the monoidal structure of $\C$:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=3,baseline]{
\node (1) at (0,2){$ \bigoplus\limits_{i+j=n}\left( X^i \otimes \left(\bigoplus\limits_{k+l=j}Y^k\otimes Z^l\right)\right)$};
\node (2) at (0,1){$ \bigoplus\limits_{i+j=n}\left(\bigoplus\limits_{k+l=j}X^i \otimes \left( Y^k\otimes Z^l \right)\right)$};
\node (3) at (0,0){$ \bigoplus\limits_{i+k+l=n}X^i \otimes \left( Y^k\otimes Z^l \right)$};
\node (4) at (4,0){$  \bigoplus\limits_{i+k+l=n} \left(X^i \otimes Y^k \right) \otimes Z^l$};
\node (5) at (4,1){$  \bigoplus\limits_{m+l=n}\left(\bigoplus\limits_{i+k=m}\left(X^i \otimes  Y^k\right)  \otimes Z^l \right) $};
\node (6) at (4,2){$ \bigoplus\limits_{m+l=n}\left( \left(\bigoplus\limits_{i+k=m}X^i \otimes Y^k\right)\otimes Z^l\right)$};

\node[overlay] (Point) at (5.2,-0.05){.};

\draw[->]
(1) edge node[left]{$  \bigoplus\limits_{i+j=n}\tau^{-1}_{X^i,Y^k \otimes Z^l}$}(2)
(2) edge node[left]{$\gamma_{i,k,l} $}(3)
(3) edge node[below]{$ \bigoplus\limits_{i+k+l=n}\alpha_{X^i,Y^k,Z^l}$}(4)
(4) edge node[right]{$\gamma_{l,i,k}^{-1} $}(5)
(5) edge node[right]{$ \bigoplus\limits_{m+l=n}\sigma_{X^i\otimes Y^k,Z^l}$}(6)
\draw[dashed,->]
(1) edge (6);
\end{equation*}

* By Definition <ref> the morphisms $\lambda_{\xi,X}$ and $\rho_{\xi,X}$ are well defined.

* The restriction on $\C$ being enriched over some ring can be easily dropped, since if $\C$ is just an additive category, it is already enriched over $\Z$. In this case however, one restricts oneself to $\xi \in \Z$.

Using these remarks we can prove that we obtain a monoidal structure on the category of $\infty$-complexes. This will already give us a monoidal structure on the category of $N$-complexes if $(\C,N,\xi)$ allows the twisted Leibniz rule.

Let $(\C,\otimes, \unit,\alpha,\lambda,\rho)$ be a countable distributive $\Mod_R$-monoidal category for $R$ a commutative ring and $\xi \in R$. Then the morphisms from Definition <ref> define natural isomorphisms

\begin{align*}
\alpha_\xi :(\_ \otimes_\xi \_ ) \otimes_\xi \_ &\xrightarrow[]{\sim} \_ \otimes_\xi (\_ \otimes_\xi \_ )\\
\lambda_\xi :(\unit_\xi \otimes_\xi \_ ) &\xrightarrow[]{\sim} \id_\C \\
\rho_\xi :  (\_ \otimes_\xi \unit_\xi) &\xrightarrow[]{\sim} \id_\C .
\end{align*}

Every morphism involved in Definition <ref> is a natural isomorphism, so all of the above morphisms define levelwise natural isomorphisms. Since $\C$ is a $\Mod_R$-monoidal category, these natural isomorphisms commute with $\xi$ and hence also with the respective differentials, so they define natural isomorphisms.

The proof of the following theorem is a bit sketchy. However, it is really straight forward, but since we have to deal with countable coproducts combined with the pentagon identity and the triangle identity, a rigid proof would take some pages of diagrams which do not give additional insight.

Let $(\C,\otimes, \unit,\alpha,\lambda,\rho)$ be a countable distributive $\Mod_R$-monoidal category for $R$ a commutative ring and $\xi \in R$. Then $(\CC_\infty(\C),\otimes_\xi,\unit_\xi ,\alpha_\xi, \lambda_\xi,\rho_\xi)$ defines a monoidal structure on $\CC_\infty(\C)$ which is countable distributive.

All of the above structure morphisms are natural isomorphisms between the corresponding functors by Proposition <ref>. Hence the only thing left to check are the triangle and the pentagon identities. By Proposition <ref> and Remark <ref> we have natural isomorphisms

\begin{align*}
\bigoplus\limits_{i+j=n}\left( X^i \otimes \left(\bigoplus\limits_{k+l=j}Y^k\otimes Z^l\right)\right) &\cong \bigoplus\limits_{\mathclap{i+k+l=n}}X^i \otimes \left( Y^k\otimes Z^l \right)\\ \bigoplus\limits_{\mathclap{i+k+l=n}} \left(X^i \otimes Y^k \right) \otimes Z^l &\cong  \bigoplus\limits_{m+l=n}\left( \left(\bigoplus\limits_{i+k=m}X^i \otimes Y^k\right)\otimes Z^l\right).
\end{align*}

So it suffices if the pentagon and triangle identities are satisfied on summands of the form $X^i\otimes \left( Y^k\otimes Z^l \right)$. There they are clearly satisfied, since $(\C,\otimes, \unit,\alpha,\lambda,\rho)$ is a monoidal category and
$$\unit_\xi^j=\begin{cases} \unit \text{ if } j=0, \\ 0 \text{ else.}\end{cases}$$

Now for countable distributivity observe first of all that by Proposition <ref> the category $\CC_\infty(\C)$ admits countable coproducts and is additive by Corollary <ref>. Furthermore, since $\_\otimes_\xi \_$ is defined levelwise as a coproduct in a countable distributive monoidal category, we get that $\_ \otimes_\xi \_$ is levelwise countable distributive and hence countable distributive since coproducts are computed levelwise.

Let $(\C,\otimes, \unit,\alpha,\lambda,\rho)$ be a countable distributive $\Mod_R$-monoidal category, for $R$ an integral domain, $\xi \in R$ and $N\in \N$. Then $(\CC_N(\C),\otimes_\xi,\unit_\xi, \alpha_\xi, \lambda_\xi,\rho_\xi)$ is a countable distributive monoidal category if the triple $(\C,N,\xi)$ allows the twisted Leibniz rule (cf. Definition <ref>).

Since by Proposition <ref> $(\CC_\infty(\C),\otimes_\xi,\unit_\xi , \alpha_\xi, \lambda_\xi,\rho_\xi)$ is a countable distributive monoidal category it suffices to check in all cases, that for $X,Y \in \CC_N(\C)$ the complex $X\otimes_\xi Y$ is again an $N$-complex. But this is just the statement of Proposition <ref>, since $\_\otimes \_ $ is in all cases a bifunctor and $(\C,N,\xi)$ allows the twisted Leibniz rule.

§.§ Closed monoidal structure

Since most interesting categories are not just monoidal but actually closed monoidal, we also want to lift the adjunctions needed for the structure of a closed monoidal category and an analogous structure on $\CC_N(\C)$ for $N \in \N \cup \{\infty\}$.
For this we will use again the liftings we constructed in Section 3 . However, before we can define the internal hom we need to make the following general observation.

Let $F:\C \times \C' \to \D$ be a bifunctor, such that for all $X \in \C'$ the functor $F(\_,X)$ admits a right adjoint $G(X,\_): \D \to \C$. Then $G(X,\_)$ induces a bifunctor
$$G(\_,\_):\C'^{op} \times \D \to \C.$$

The functoriality in the right variable is given by assumption, hence it just remains to be shown that a morphism $f:X \to X'$ in $\C'$ induces naturally a morphism
$$G(f,Y):G(X',Y)\to G(X,Y).$$
For this consider the following composition:

\begin{align*}
F(G(X',Y),X) \xrightarrow{F(\id,f)} F(G(X',Y),X')  \xrightarrow{\epsilon} Y,
\end{align*}

where $\epsilon$ is the counit of the adjunction. This composition is adjoint to a morphism
$$G(f,Y):G(X',Y) \to G(X,Y),$$
which is by construction compatible with the structure of $G(X,\_):\C \to \D$.

Now using this observation we can lift both functors involved in the structure of a closed monoidal structure, and show that these are still compatible in the desired way if we consider modules over a ring.

We conjecture that the following statement holds in general for a closed countable distributive $\Mod_R$-monoidal category. However the constructions involving $\id_X\otimes d_Y$ and $[d_Y,\id_Z]$ blow up in the general case. Hence we will only prove the case of modules over a ring $R$, since there the proof is less confusing and one has good control of the adjoints.

Let $R$ be  a commutative ring and $\xi \in R$. Then we have an adjunction
$$\_ \otimes_\xi X : \CC_\infty(\Mod_R) \leftrightarrow \CC_\infty(\Mod_R) : [X,\_]_\xi.$$

$$[\_,\_]_\xi:\CC_\infty(\Mod_R)^{op} \times \CC_\infty(\Mod_R) \to \CC_\infty(\Mod_R)$$
is the functor induced by $[\_,\_]:\Mod_R^{op} \otimes_{\Mod_R} \Mod_R \to \Mod_R$ and Definition <ref>.

Throughout this proof we denote by $\Gr(\C)$ the category of graded objects in a category $\C$.

First observe that we already have in $\Gr(\Mod_R)$ an isomorphism

$$\Gr(\Mod_R)(X\otimes_\xi Y,Z) \xrightarrow{\sim} \Gr(\Mod_R)(X,[Y,Z]_\xi)$$

by Definition <ref> and the tensor-hom adjunction in $\Gr(\Mod_R)$.
So it suffices to check that this isomorphism preserves and reflects that a morphism commutes with the differentials.
For this consider a morphism $f \in \CC_\infty(\C)(X\otimes_\xi Y,Z)$. In particular $f\in \Gr(\Mod_R)(X\otimes_\xi Y,Z)$ and $f$ has the property
$$ f \circ d_{X \otimes Y} -d_Z \circ f =0.  $$

Evaluation of this map at an element $x \otimes y \in X^i \otimes Y^{n-i} \subset X \otimes Y$ yields now

\begin{align*}
(f\circ d_{X \otimes Y} -d_Y \circ f) (x\otimes y) &= f(d_X(x)\otimes y + \xi^i x \otimes d_Y(y))- d_Z(f(x\otimes y))\\
&=f(d_X(x)\otimes y) + \xi^i f(x \otimes d_Y(y))- d_Z(f(x\otimes y)).
\end{align*}

This is adjoint to the map defined levelwise as

\begin{align*}
X^i \to& [Y,Z]_\xi^{i+1}\\
x \mapsto& \begin{pmatrix} Y^n \to Z^{n+i}  \hfill\\
\; \; \; y \mapsto f(d_X(x)\otimes y) + \xi^i f(x \otimes d_Y(y))- d_Z(f(x\otimes y))
\end{pmatrix}
\end{align*}

since the adjunction in $\Gr(\Mod_R)$ is an adjunction of additive functors.
Now consider the adjoint map $\widehat{f}:X \to [Y,Z]_\xi$ of $f$. Levelwise this is the map

\begin{align*}
X^i &\to [Y,Z]_\xi^{i+1}\\
x &\mapsto  \begin{pmatrix}
Y \to Z  \hfill\\
\; y \mapsto  f(x \otimes y)
\end{pmatrix}.
\end{align*}

Precomposing with $d_X$ yields that the map $\widehat{f}\circ d_X$ is levelwise of the form

\begin{align*}
X^i \to& [Y,Z]_\xi^{i+1}\\
x \mapsto& \begin{pmatrix} Y^n \to Z^{n+i}  \hfill\\
\; \; \; y \mapsto f(d_X(x) \otimes y)\end{pmatrix},
\end{align*}

and postcomposing with $d_{[Y,Z]}$ gives that the levelwise structure of $d_{[Y,Z]}\circ \widehat{f}$ is

\begin{align*}
X^i \to& [Y,Z]_\xi^{i+1}\\
x \mapsto& \begin{pmatrix} Y^n \to Z^{n+i}  \hfill\\
\; \; \; y \mapsto  d_Z(f(x\otimes y)) - \xi^i f(x \otimes d_Y(y))\end{pmatrix}.
\end{align*}

Now just as above, $\widehat{f}$ is actually a morphism of $\infty$-complexes if and only if

$$0=\widehat{f}\circ d_X -d_{[Y,Z]} \circ \widehat{f}.$$

After evaluating at an $x \in X^i$, this is just the map

\begin{align*}
X^i \to& [Y,Z]_\xi^{i+1}\\
x \mapsto& \begin{pmatrix} Y^n \to Z^{n+i} \hfill\\
\; \; \;y \mapsto f(d_X(x)\otimes y) + \xi^i f(x \otimes d_Y(y))- d_Z(f(x\otimes y))\end{pmatrix}.
\end{align*}

Hence the adjoint map of $d_Z \circ f - f \circ d_{X \otimes_\xi Y}$ and $\widehat{f}\circ d_Z - d_{[Y,Z]_\xi} \circ \widehat{f}$ have to coincide.
So we may conclude, using that the adjunction is an adjunction between pointed functors, that $f$ is a morphism of $\infty$-complexes  if and only if $\widehat{f}$ is a morphism of $\infty$-complexes.

Let $N \in \N$, $R$ be an integral domain and $\xi \in R$, then we have an adjunction
$$\_ \otimes_\xi X : \CC_N(\Mod_R) \leftrightarrow \CC_N(\Mod_R) : [X,\_]_\xi$$
if $(\Mod_R,N,\xi)$ allows the twisted Leibniz rule.

In particular we get that $(\CC_N(\Mod_R),\otimes_\xi,\unit_\xi, \alpha_\xi, \lambda_\xi,\rho_\xi)$ is a closed monoidal category if $(\Mod_R,N,\xi)$ allows the twisted Leibniz rule.

By Theorem <ref> it just remains to show that $\_\otimes_\xi\_$ and $[\_,\_]_\xi$ induce functors of $N$-complexes in $\Mod_R$, which is just Corollary <ref> and Theorem <ref>.

§ FROBENIUS EXACT STRUCTURE ON $\CC_N(\A)$

Similar to the the chain complex case one can construct a Frobenius exact structure on the category of $N$-complexes in an additive category $\A$. We will use this to define the homotopy category of $N$-complexes over $\A$ as the stable category with respect to this exact structure. This yields immediately that the homotopy category of $N$-complexes is triangulated by Happel's Theorem. In particular the work done here repeats arguments by Iyama, Kato and Miyashi from their paper <cit.> thoroughly. For the notions of exact and Frobenius exact  categories we refer to <cit.> and Appendix A in the paper by B. Keller <cit.>.

Let $\A$ be an additive category and $N  \in \N$. Then define $\Se_\oplus$ to be the collection of all levelwise split short exact sequences in $\CC_N(\A)$, i. e. sequences which are levelwise isomorphic to a sequence of the form

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0.3){$0$};
\node (X1) at (1,0.3){$X$};
\node (X2) at (2,0.3){$X\oplus Y$};
\node (X3) at (3,0.3){$Y$};
\node (02) at (3.7,0.3){$0$};

\draw[->]
(01) edge (X1)
(X3) edge (02)
(X1) edge node[above] {$\begin{bmatrix}
\id \\ 0
\end{bmatrix}$} (X2)
(X2) edge node[above] {$\begin{bmatrix}
0 & \id
\end{bmatrix}$} (X3);

\end{equation*}

We will show that $S_\oplus$ defines a Frobenius exact structure, hence we need to show the following three properties of $\Se_\oplus$:

* $\Se_\oplus$ is an exact structure,

* $\Se_\oplus$ has enough $\Se_\oplus$-projective objects and $\Se_\oplus$-injective objects,

* the $\Se_\oplus$-projective objects are precisely the $\Se_\oplus$-injective objects.

First we show that it indeed is an exact structure:

Let $\A$ be an additive category and $N  \in \N$. Then $\Se_\oplus$ defines an exact structure on $\CC_N(\A)$.

Observe that $\Se_\oplus$ defines levelwise an exact structure. Now direct sums, pushouts and pullbacks in $\CC_\infty(\A)$ are computed pointwise. Hence $\Se_\oplus$ defines an exact structure on $\CC_\infty(\A)$ and also on $\CC_N(\A)$, since it is a full subcategory.

To see that $\Se_\oplus$ has enough $\Se_\oplus$-projective objects and $\Se_\oplus$-injective objects we will construct for an $N$-complex $X$ an $\Se_\oplus$-injective object $\II_N(X)$ and an $\Se_\oplus$-projective object $\PP_N(X)$ such that we have an admissible monomorphisms $X \to \II_N(X)$ and an admissible epimorphism $\PP_N(X) \to X$. The construction of these is motivated by the case of chain complex, which can be recovered from the given construction by setting $N=2$.

Let $\A$ be an additive category, $N \in \N$ and $X \in \CC_N(\A)$. Then define

* the $N$-complex

\begin{equation*}
\II_N(X):=\bigoplus_{i\in\Z}\mu^{i}_N(X^i),
\end{equation*}

* the $N$-complex

\begin{equation*}
\PP_N(X):=\bigoplus_{i\in\Z}\mu^{i+N-1}_N(X^i),
\end{equation*}

* and the morphisms $\lambda_X^{\II}:X \to \II_N(X) $ and $\lambda_X^{\PP}:\PP_N(X)  \to X $ by setting levelwise

\begin{equation*}
(\lambda^\II_X)^i:= \begin{bmatrix} \id_{X^i} & d_X & d_X^2 & \cdots & d_X^{N-2}& d_X^{N-1} \end{bmatrix}^T
\end{equation*}

\begin{equation*}(\lambda_X^{\PP})^i:=\begin{bmatrix} d_X^{N-1} & d_X^{N-2} &\cdots & d_X^2 & d_X & \id_{X^{i-N+1}} \end{bmatrix}.
\end{equation*}

Since the data of an $\Se_\oplus$-injective hull and an $\Se_\oplus$-projective cover contain the morphisms as well, we check that these are morphisms of $N$-complexes.

Let $\A$ be an additive category and $N  \in \N$. Then the maps $\lambda_X^\II : X \to \II_N(X)$ and $\lambda_X^\PP :\PP_N(X) \to X$ define morphisms of $N$-complexes.

We check separately:

$\lambda_X^\II$: One needs to check that $d_{\II_N(X)}\circ \lambda_X^\II = \lambda_X^\II \circ d_X$. we calculate

\begin{align*}
(d_{\II_N(X)} \circ \lambda_X^\II)^i &= d_{\II_N(X)}\circ \begin{bmatrix} \id_{X^i} & d_X & d_X^2 & \cdots & d_X^{N-2}& d_X^{N-1} \end{bmatrix}^T\\
&= \begin{bmatrix} d_X & d_X^2 &d_X^{3}& \cdots & d_X^{N-1} & 0 \end{bmatrix}^T
\end{align*}

\begin{align*}
(\lambda_X^\II \circ d_X)^i =& \begin{bmatrix} \id_{X^{i+1}} & d_X & d_X^2 & \cdots & d_X^{N-2}& d_X^{N-1} \end{bmatrix}^T \circ d_X\\
=&  \begin{bmatrix}  d_X & d_X^2 & d_X^3 \cdots & d_X^{N-1}& d_X^N \end{bmatrix}^T.
\end{align*}

So by $d_X^N=0$ the claim holds.

One needs to check again that $d_X\circ \lambda_X^\PP = \lambda_X^\PP \circ d_{\PP(X)}.$ Now calculation yields

\begin{align*}
(d_X \circ \lambda_X^\PP)^i &= d_X\circ \begin{bmatrix} d_X^{N-1} & d_X^{N-2} &\cdots & d_X^2 & d_X & \id_{X^{i-N+1}} \end{bmatrix}\\
&= \begin{bmatrix} d_X^N & d_X^{N-1} &\cdots & d_X^3 & d_X^2 & d_X  \end{bmatrix}
\end{align*}

\begin{align*}
(\lambda_X^\PP \circ d_{\PP_N(X)})^i &=\begin{bmatrix} d_X^{N-1} & d_X^{N-2} &\cdots & d_X^2 & d_X & \id_{X^{i-N+1}} \end{bmatrix} \circ d_{\PP_N(X)} \\
&=\begin{bmatrix} 0 &d_X^{N-1}  &\cdots & d_X^3 &d_X^2 & d_X \end{bmatrix}.
\end{align*}

This again concludes the proof since $d_X^N=0$.

Now we also check that $\PP_N(X)$ and $\II_N(X)$ define an $\Se_\oplus$-cover and $\Se_\oplus$-hull respectively.

There are sequences in $\Se_\oplus$ of the form

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0.3){$0$};
\node (X1) at (1,0.3){$X$};
\node (X2) at (2,0.3){$\II_N(X)$};
\node (X3) at (3,0.3){$Y$};
\node (02) at (3.7,0.3){$0$};

\draw[->]
(01) edge  (X1)
(X3) edge (02)
(X1) edge node[above] {$\lambda^\II_X$}(X2)
(X2) edge  (X3);
\end{equation*}

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0.3){$0$};
\node (X1) at (1,0.3){$Y'$};
\node (X2) at (2,0.3){$\PP_N(X)$};
\node (X3) at (3,0.3){$X$};
\node (02) at (3.7,0.3){$0$};

\draw[->]
(01) edge  (X1)
(X3) edge (02)
(X1) edge node[above] {$\lambda^\PP_Y$}(X2)
(X2) edge  (X3);
\end{equation*}

This becomes obvious if we consider $Y=\bigoplus_{i\in\Z}\mu^{i-1}_{N-1}(X^i)$ in the first sequence and $Y'=\bigoplus_{i\in\Z}\mu^{i+N}_{N-1}(X^i)$ in the second.

To check that $\II_N(X)$ and $\PP_N(X)$ really are $\Se_\oplus$-injective (resp. $\Se_\oplus$-projective) objects, we need to show by the definition of $\Se_\oplus$-injective (resp. $\Se_\oplus$-projective) objects that every sequence in $\Se_\oplus$ such that the rightmost term is $\Se_\oplus$-projective, or the leftmost term is $\Se_\oplus$-injective admits a splitting in the category of $N$-complexes. For this we will explicitly construct the splittings and prove that they indeed are morphisms of $N$-complexes.

Let $\A$ be an additive category and $N\in \N$. Then we have

* $\II_N(X)$ is $\Se_\oplus$-injective,

* $\PP_N(X)$ is $\Se_\oplus$-projective.

We will prove in both cases  that if we have a levelwise splitting morphism $\II_N(X) \to Y$ respectively $Y \to \PP_N(X)$ that this morphism already admits a splitting in the $\CC_N(\A)$.

* Let $f: \II_N(X) \to Y$ be a morphism of $N$-complexes such that there exists for every $i \in \Z$ a levelwise split $h^i: Y^i \to \II_N(X)^i$, satisfying $h^i \circ f^i =\id_{\II_N(X)}$. We need to provide a morphism of $N$-complexes $g:Y \to \II_N(X)$ such that $g \circ f = \id_{\II_N(X)}$.

The $f^i$ and $h^i$ morphisms are by definition of the form

\begin{equation*}
f^i= \begin{bmatrix}f_1^i & f_2^i& \cdots & f_N^i\end{bmatrix} \; , \;h^i=\begin{bmatrix}
h_1^i & h_2^i& \cdots &h_N^i\end{bmatrix}^T,
\end{equation*}

now the fact that $h^i \circ f^i = \id_{\II_N(X)^i}$ just means $h_j^i\circ f_j^i = \id_{X^{i+j}}$ and $h_j^i\circ f_k^i = 0$ if $k \neq j$. Since $f$ is a morphism of $N$-complexes, we know that $d_Y \circ f^i=f^{i+1} \circ d_{\II_N(X)}$, but this just implies, that $d_Y \circ f_j^i= f_{j-1}^{i+1}$ if $1 < j \le N$ and zero otherwise. Furthermore, observe that by the definition of $\II_N(X)$ the following equation holds for $2\le j \le N $:

\begin{equation*}
h_j^i\circ f_j^i = \id_{X^{i+j}}=h_{j-1}^{i+1} \circ f_{j-1}^{i+1}.
\end{equation*}

Now define the morphism $g: Y \to \II_N(X)$ levelwise as

$$g^i:=\begin{bmatrix}h_1^i&h_1^{i+1}\circ d_Y& \cdots &h_1^{i+N-1}\circ d_Y^{N-1}\end{bmatrix}^T.$$

Note that this $g$ is well defined by the definition of $\II_N(X)$. Furthermore this indeed defines a morphism of $N$-complexes since

\begin{align*}
(g\circ d_Y)^i& = \begin{bmatrix} h_1^{i+1}\circ d_Y & h_1^{i+2} \circ d_Y^2 &\cdots & h_1^{i+N-1} \circ d_Y^{N-1}& h_1^{i+N} \circ d_Y^N\end{bmatrix}^T \\
&=\begin{bmatrix}h_1^{i+1}\circ d_Y & h_1^{i+2} \circ d_Y^2&\cdots& h_1^{i+N-1} \circ d_Y^{N-1}& 0\end{bmatrix}^T \\
&= (d_{\II_N(X)} \circ g)^i.
\end{align*}

The morphism $g$ is a section in $\CC_N(\A)$, since

\begin{align*}
g \circ f &=
\begin{bmatrix}
h_1^i \\
h_1^{i+1}\circ d_Y\\
h_1^{i+2}\circ d_Y^2\\
\vdots \\
h_1^{i+N-1}\circ d_Y^{N-1}]^T
\end{bmatrix}
\circ
\begin{bmatrix}
f_1^i & f_2^i & f_3^i & ... & f_N^i
\end{bmatrix}
\\
\begin{bmatrix}
h_1^i\circ f_1^i & h_1^i \circ f_2^i & h_1^i \circ f_3^i & \cdots & h^i_1\circ f^i_N \\
h_1^{i+1}\circ d_Y \circ f_1^i & h_1^{i+1}\circ d_Y \circ f_2^i & h_1^{i+1} \circ d_Y \circ f_3^i & \cdots & h_1^{i+1} \circ d_Y \circ f_N^i \\
h_1^{i+2}\circ d_Y^2 \circ f_1^i & h_1^{i+2}\circ d_Y^2 \circ f_2^i & h_1^{i+2} \circ d_Y^2 \circ f_3^i & \cdots & h_1^{i+2} \circ d_Y^2 \circ f_N^i
\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
h_1^{i+N-1}\circ d_Y^{N-1} \circ f_1^i & h_1^{i+N-1}\circ d_Y^{N-1} \circ f_2^i & h_1^{i+N-1} \circ d_Y^{N-1} \circ f_3^i & \cdots & h_1^{i+N-1} \circ d_Y^{N-1} \circ f_N^i
\end{bmatrix}\\
\displaybreak[3]
\begin{bmatrix}
h_1^i\circ f_1^i & h_1^i \circ f_2^i & h_1^i \circ f_3^i & \cdots & h^i_1\circ f^i_N \\
h_1^{i+1}\circ 0 & h_1^{i+1}\circ f_1^{i+1} & h_1^{i+1} \circ f_2^{i+1} & \cdots & h_1^{i+1} \circ f_{N-1}^{i+1} \\
h_1^{i+2}\circ 0 & h_1^{i+2} \circ 0 & h_1^{i+2} \circ f_3^{i+2} & \cdots & h_1^{i+2} \circ f_{N-2}^{i+2}
\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
h_1^{i+N-1}\circ 0 & h_1^{i+N-1}\circ 0 & h_1^{i+N-1} \circ 0 & \cdots & h_1^{i+N-1} \circ f_1^{i+N-1}
\end{bmatrix}\\
\begin{bmatrix}
h_1^i\circ f_1^i & 0 & 0 & \cdots & 0 \\
0 & h_1^{i+1}\circ f_1^{i+1} &0 & \cdots & 0\\
0 &0 & h_1^{i+2} \circ  f_1^{i+2} & \cdots &0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & h_1^{i+N-1} \circ f_1^{i+N-1}
\end{bmatrix}\\
\begin{bmatrix}
\id_{X^i} & 0 & 0 & \cdots & 0 \\
0 & \id_{X^{i+1}} &0 & \cdots & 0\\
0 &0 &\id_{X^{i+2}} & \cdots &0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \id_X{^{i+N-1}}
\end{bmatrix}\\
&= \id_{\II_N(X)}^i,
\end{align*}

where we use in the third to last row that $d_Y \circ f_j^i= f_{j-1}^{i+1}$ and $d_Y \circ f_1^i = 0$. In the second to last row we used $f_j^i\circ h_k^i = 0$ for $k \neq j$.

* This case is just the dual of (1), but for completeness we give the proof as well.

Let $f: Y \to \PP_N(X)$ be a morphism of $N$-complexes such that there exists for all $i \in \Z$ a levelwise split $h^i: \PP_N(X)^i \to Y^i$, satisfying $f^i \circ h^i =\id_{\PP_N(X)}^i$, we now need to provide a morphism of $N$-complexes $g:\PP_N(X) \to Y$ such that $f \circ g = \id_{\PP(X)}$.

Again the morphisms $f^i$ and $h^i$ are of the form

\begin{equation*}
f^i=\begin{bmatrix}f^i_1&f_2&f^i_3&\cdots &f^i_N\end{bmatrix}^T \; , \; h^i=\begin{bmatrix}h^i_1&h^i_2&h^i_3&\cdots &h^i_N\end{bmatrix}.
\end{equation*}

The fact that $f^i \circ h^i = \id_{\PP_N(X)^i}$ yields $f_j^i\circ h_j^i = \id_{X^{i-N+j+1}}$ and $f_j^i\circ h_k^i = 0$ if $k \neq j$. Since $f$ is a morphism of $N$-complexes, we know that $f^i \circ d_Y=d_{\PP_N(X)} \circ f^{i+1} $, but this implies that $ f_j^i \circ d_Y= f_{j-1}^{i+1}$ if $1 < j \le N$ and zero otherwise. Furthermore, observe that by the definition of $\PP_N(X)$ the following equation holds for $2\le j \le N $:

\begin{equation*}
f_j^i  \circ h_j^i= \id_{X^{i-N+j}}=f_{j-1}^{i+1} \circ h_{j-1}^{i+1}.
\end{equation*}

Hence define $g: \PP_N(X) \to Y$ dually to (1) as

$$g^i:=\begin{bmatrix} h_1^i&d_Y \circ h_1^{i+1} & \cdots &d_Y^{N-1} \circ h_1^{i+N-1}\end{bmatrix}.$$

Note that this $g$ is again well defined by the definition of $\PP_N(X)$, and defines a morphism of $N$-complexes since

\begin{align*}
(d_Y \circ g)^i& = \begin{bmatrix} d_Y \circ h_1^{i-1} &d_Y^2 \circ h_1^i &\cdots & d_Y^{N-1} \circ h_1^{i+N-3}&d_Y^N \circ h_1^{i+N-2}\end{bmatrix} \\
&=\begin{bmatrix} d_Y \circ h_1^{i-1} &d_Y^2 \circ h_1^i &\cdots & d_Y^{N-1} \circ h_1^{i+N-3}&0\end{bmatrix} \\
&= (g \circ d_{\PP_N(X)})^i.
\end{align*}

the morphism $g$ is a section in $\CC_N(\A)$, since

\begin{align*}
f \circ g &=
\begin{bmatrix}
f_1^i \\
\vdots \\
\end{bmatrix}
\circ
\begin{bmatrix}
h_1^i & d_Y \circ h_1^{i+1} & d_Y^2 \circ h_1^{i+2} & ... & d_Y^{N-1} \circ h_1^{i+N-1}
\end{bmatrix}
\\
\begin{bmatrix}
f_1^i\circ h_1^i & f_1^i \circ d_Y \circ h_1^{i+1} & f_1^i \circ  d_Y^2 \circ h_1^{i+2} & \cdots & f^i_1\circ d_Y^{N-1}\circ h_1^{i+N-1}\\
f_2^i \circ h_1^i & f_2^i \circ d_Y \circ h_1^{i+1}& f_2^i \circ  d_Y^2 \circ h_1^{i+2} & \cdots & f_2^i \circ  d_Y^{N-1} \circ h_1^{i+N-1} \\
f_3^i\circ h_1^i  & f_3^i \circ d_Y \circ h_1^{i+1} &f_3^i \circ  d_Y^2 \circ h_1^{i+2} & \cdots & f_3^i \circ  d_Y^{N-1} \circ h_1^{i+N-1}
\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
f_N^i\circ h_1^i  & f_N^i \circ d_Y \circ h_1^{i+1} & f_N^i \circ  d_Y^2 \circ h_1^{i+2} & \cdots & f_N^i \circ  d_Y^{N-1} \circ h_1^{i+N-1}
\end{bmatrix}\\
\begin{bmatrix}
f_1^i\circ h_1^i & 0 \circ h_1^{i+1} & 0 \circ h_1^{i+2} & \cdots & 0 \circ h_1^{i+N-1}\\
f_2^i \circ h_1^i & f_1^{i+1}  \circ h_1^{i+1}& 0 \circ h_1^{i+2} & \cdots & 0 \circ h_1^{i+N-1} \\
f_3^i\circ h_1^i  & f_2^{i+1}  \circ h_1^{i+1} &f_1^{i+2} \circ h_1^{i+2} & \cdots & 0 \circ h_1^{i+N-1}
\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
f_N^i\circ h_1^i  & f_N^i \circ d_Y \circ h_1^{i+1} & f_N^i \circ  d_Y^2 \circ h_1^{i+2} & ... & f_N^i \circ  d_Y^{N-1} \circ h_1^{i+N-1}
\end{bmatrix}
\displaybreak[3]
\\
\begin{bmatrix}
f_1^i \circ h_1^i & 0 & 0 & \cdots & 0 \\
0 &  f_1^{i+1} \circ h_1^{i+1} &0 & \cdots & 0\\
0 &0 & f_1^{i+2} \circ h_1^{i+2}& \cdots &0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & ... & f_1^{i+N-1} \circ h_1^{i+N-1}
\end{bmatrix}\\
\begin{bmatrix}
\id_{X^i-N+j} & 0 & 0 & \cdots & 0 \\
0 & \id_{X^{i-N+1}} &0 & \cdots & 0\\
0 &0 &\id_{X^{i-N+2}} & \cdots &0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \id_X{^{i}}
\end{bmatrix}
\\&= \id_{\PP_N(X)}^i,
\end{align*}

where we use in the third to last row $ f_j^i \circ d_Y= f_{j-1}^{i+1}$ and $ f_1^i \circ d_Y= 0$. In the second to last row we use $f_j^i\circ h_k^i = 0$ for $k \neq j$.

Now the last thing we need for $\Se_\oplus$ to be a Frobenius exact structure on $\CC_N(\A)$ is that $\Se_\oplus$-injectives and $\Se_\oplus$-projectives coincide.

Let $\A$ be an additive category and $N  \in \N$. Then an object in $\CC_N(\A)$ is $S_\oplus$-projective if and only if it is $S_\oplus$-injective.

Let $P$ be $\Se_\oplus$-projective, then we have a sequence in $\Se_\oplus$ of the form:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0.3){$0$};
\node (X1) at (1,0.3){$X$};
\node (X2) at (2,0.3){$\PP_N(P)$};
\node (X3) at (3,0.3){$P$};
\node (02) at (3.7,0.3){$0$};

\draw[->]
(01) edge (X1)
(X3) edge (02)
(X1) edge node[above] {} (X2)
(X2) edge node[above] {$\lambda^\PP_P$} (X3)

\end{equation*}

Now since $P$ is $\Se_\oplus$-projective, this short exact sequence splits, hence $\PP_N(P)\cong P \oplus X$. But since $\PP_N(P)=\II_N(P[-N+1])$ we have that $P$ is a direct summand of an $\Se_\oplus$-injective object, and hence $\Se_\oplus$-injective itself. Here $P[-N+1]$ denotes the $N$-complex of $P$ shifted by $-N+1$.

Conversely let $I$ be an $\Se_\oplus$-injective object. Then we have the following short exact sequence in $\Se_\oplus$:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0.3){$0$};
\node (X1) at (1,0.3){$I$};
\node (X2) at (2,0.3){$\II_N(I)$};
\node (X3) at (3,0.3){$X$};
\node (02) at (3.7,0.3){$0$};

\draw[->]

(01) edge (X1)
(X3) edge (02)
(X1) edge node[above] {$\lambda^\II_I$} (X2)
(X2) edge node[above] {} (X3)

\end{equation*}

Now again the short exact sequence splits since $I$ is $\Se_\oplus$-injective, and we get $\II_N(I)\cong I \oplus X$. Now $\II_N(I)=\PP_N(I[N-1])$ and hence $I$ is a summand of an $\Se_\oplus$-projective object and hence itself $\Se_\oplus$-projective.

Now we can put together the above properties of $\Se_\oplus$ to get:

For an additive category $\A$ and $N  \in \N$ the exact structure $\Se_\oplus$ defines a Frobenius exact structure on the category $\CC_N(\A)$.

By Proposition <ref>, The collection $\Se_\oplus$ defines an exact structure. It has enough $\Se_\oplus$-projectives and $\Se_\oplus$-injectives, as any object admits an admissible monomorphism $\lambda^\II_X:X \to \II_N(X)$ and an admissible epimorphism $\lambda_X^\PP :\PP_N(X) \to X$ by Proposition <ref> and Proposition <ref>. Finally $\Se_\oplus$-injective objects are precisely the $\Se_\oplus$-projective objects by Proposition <ref>. So all defining criteria of Frobenius exactness are fulfilled.

Having established this Frobenius exact structure we define the homotopy category of $N$-complexes and use Happel's Theorem I.2.6 <cit.> to conclude that the homotopy category is well defined and triangulated in a canonical way.

We would like to remind that by 13.4 in <cit.> the stable category $\underline{\C}_\Se$ of a Frobenius exact category $\C_\Se$ is defined to have the same objects as $\C$ and for $X,Y \in \C$ the morphism space $\underline{\C}_\Se(X,Y):=\C(X,Y)/I$, where $I$ is the two-sided ideal of morphisms factoring through an $\Se$-injective-projective object.

Let $\A$ be an additive category and $N \in \N$. Then denote by $\KK_N(\A)$ the stable category of $\CC_N(\A)_{S_\oplus}$.

The category $\KK_N(A)$ is triangulated for $\A$ an additive category and $N \in \N$. Furthermore every $\Se_\oplus$-short exact sequence

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0.3,0){$0$};
\node (X1) at (1,0){$X$};
\node (X2) at (2,0){$Y$};
\node (X3) at (3,0){$Z$};
\node (02) at (3.7,0){$0$};

\draw[->]

(01) edge (X1)
(X3) edge (02)
(X1) edge node[above] {$f$} (X2)
(X2) edge node[above] {$g$} (X3)

\end{equation*}

induces a triangle

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (X1) at (1,0){$X$};
\node (X2) at (2,0){$Y$};
\node (X3) at (3,0){$Z$};
\node (X4) at (4,0){$\Sigma X$};

\draw[->]

(X1) edge node[above] {$f$} (X2)
(X2) edge node[above] {$g$} (X3)
(X3) edge node[above] {$h$} (X4)

\end{equation*}
and every triangle is isomorphic to a triangle of this form.

This is a direct application of Happel's Theorem I.2.6 <cit.>.

From now on we will often need that every morphism can be completed into a triangle in $\KK_N(\A)$. We will refer to a completion of a morphism as its cone. Since we do not need any specific structure of these cones we just fix any choice of completion.

Let  $f:X \to Y$ be a morphism in $\CC_N(\A)$ for $\A$ an additive category and $N\in \N$. Then we refer to the cone of $f$ in $\KK_N(\A)$ as $\Cone_{K_N(\A)}(f)$, or just $\Cone(f)$ if it is clear from context. This is just any choice of completion of the morphism into a triangle

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (X1) at (1,0.3){$X$};
\node (X2) at (2,0.3){$Y$};
\node (X3) at (3,0.3){$Z$};
\node (X4) at (4,0.3){$\Sigma X$};

\draw[->]

(X1) edge node[above] {$f$} (X2)
(X2) edge node[above] {$g$} (X3)
(X3) edge node[above] {$h$} (X4)

\end{equation*}

For a precise construction of the cone objects together with the defining morphisms arising by Happel's Theorem we refer the reader to “Derived categories of $N$-complexes” <cit.>.

By the definition of a triangulated category these completions are unique up to non-unique isomorphism.

§ PROPERTIES OF THE HOMOTOPY CATEGORY OF $N$-COMPLEXES

Since a goal of this work is to imitate the definition of the derived category of an abelian category, we will need some general properties of the homotopy category $\KK_N(\A)$. This will give us more control over the objects, morphisms and coproducts in $\KK_N(\A)$.

First of all we will need to be able to factor through specific $\Se$-injective and $\Se$-projective objects. This is possible for any exact structure which has enough projectives and injectives respectively. Using this property we will be able to show that coproducts in $\KK_N(\A)$ can be calculated naively and prove a more calculation based approach towards the homotopy category.

Let $\C_\Se$ be an exact category. Then the following statements hold:

* If $X\in \C$ admits an $\Se$-injective hull $X \xrightarrow{\lambda_X} I_X$ we have that a morphism $f:X \to Y$ factors as $f:X \to I \to Y$, where $I$ is an $\Se$-injective object, if and only if $f$ factors as $f:X  \xrightarrow{\lambda_X} I_X \to Y$.

* If $Y\in \C$ admits an $\Se$-projective resolution $P_Y \xrightarrow{\lambda_Y} Y$ we have that a morphism $f:X \to Y$ factors as $f:X \to P \to Y$, where $P$ is an $\Se$-projective object, if and only if $f$ factors as $f:X \to P_Y \xrightarrow{\lambda_Y} Y$.

* The if part is clear, since if $f:X \to Y$ factors over a chosen $\Se$-injective, it especially factors over an $\Se$-injective object.

For the only if part, let $f:X \xrightarrow{\lambda} I \xrightarrow{g} Y$ be a factorisation of a morphism $f$ such that $I$ is $\Se$-injective. Now consider the following diagram in $\C$:

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Y) at (0,0){$Y$};
\node (I) at (0,1){$I$};
\node (X) at (0,2){$X$};
\node (IX) at (1,2){$I_X$};
\node (Z1) at (2,2){$Z$};
\node (I') at (1,1){$M$};
\node (Z2) at (2,1){$Z$};
\node (po) at (0.75,1.25){$\ulcorner$};

\node[overlay] (Point) at (2.2,0){,};

\draw[->]
(X) edge [bend right=30] node[left] {$f$} (Y)
(X) edge  node[right]{$\lambda$}(I)
(X) edge node[above]{$\lambda_X$}(IX)
(IX) edge (Z1)
(I) edge  node[right]{$g$} (Y)
(IX) edge node[right]{$h$}(I')
(I) edge node[above]{$j$}(I')
(I') edge (Z2);
\draw[double distance = 2pt]
\end{equation*}

where the top sequence is induced by the $\Se$-injective hull $I_X$ for $X$. The left square is a pushout, which we can form since $(X \xrightarrow{\lambda_X} I_X \to Z)\in \Se$.
Since the top sequence is in $\Se$ we get by the stability of exact structures under pushouts that also the bottom one is in $\Se$. The bottom sequence now splits since $I$ is $\Se$-injective by assumption. Call this split $s:M \to I$. Hence we get the diagram in $\C$:

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Y) at (0,0){$Y$};
\node (I) at (0,1){$I$};
\node (X) at (0,2){$X$};
\node (IX) at (1,2){$I_X$};
\node (Z1) at (2,2){$Z$};
\node (I') at (1,1){$M$};
\node (Z2) at (2,1){$Z$};
\node (po) at (0.75,1.25){$\ulcorner$};

\node[overlay] (Point) at (2.2,0){.};

\draw[->]
(X) edge [bend right=30] node[left] {$f$} (Y)
(X) edge  node[right]{$\lambda$}(I)
(X) edge node[above]{$\lambda_X$}(IX)
(IX) edge (Z1)
(I) edge  node[right]{$g$} (Y)
(IX) edge node[right]{$h$}(I')
(I) edge node[above]{$j$}(I')
(I') edge (Z2);
\draw[double distance = 2pt]
\draw[ ->]
(I') edge [bend left=30] node[below] {$s$} (I);
\end{equation*}

Using this to factor $f$ we get the claimed factorization:
$$f=g \circ \lambda =g \circ s \circ j \circ \lambda =g \circ s  \circ h \circ \lambda_X.$$

* Again the if part is clear since a specific choice of an $\Se$-projective object in particular gives us an $\Se$-projective object.
For the only if part consider again a morphism $f:X \to Y$ which factors as $f:X \xrightarrow{g} P \xrightarrow{\lambda} Y$ for some $\Se$-projective object $P$. Now consider the $\Se$-projective resolution $P_Y \xrightarrow{\lambda_Y} Y$, and the following diagram in $\C$:

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Y) at (2,0){$Y$};
\node (P) at (2,1){$P$};
\node (X) at (2,2){$X$};
\node (PY) at (1,0){$P_Y$};
\node (Z1) at (0,0){$Z$};
\node (P') at (1,1){$M$};
\node (Z2) at (0,1){$Z$};
\node (pb) at (1.25,0.75){$\lrcorner$};

\node[overlay] (Point) at (2.2,0){,};

\draw[->]
(X) edge [bend left=30] node[right] {$f$} (Y)
(P) edge  node[left]{$\lambda$}(Y)
(PY) edge node[below]{$\lambda_Y$}(Y)
(Z1) edge (PY)
(X) edge  node[left]{$g$} (P)
(P') edge node[left]{$h$}(PY)
(P') edge node[below]{$j$}(P)
(Z2) edge (P');
\draw[double distance = 2pt]
\end{equation*}

where the right square is a pullback, which we can form since the lower sequence is in $\Se$. Now we get again using that $\Se$ is an exact structure, that the top row is in $\Se$. So the fact that $P$ is $\Se$-projective yields that $j$ splits. Call this split $s:P \to M$. This means that we get the diagram:

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Y) at (2,0){$Y$};
\node (P) at (2,1){$P$};
\node (X) at (2,2){$X$};
\node (PY) at (1,0){$P_Y$};
\node (Z1) at (0,0){$Z$};
\node (P') at (1,1){$M$};
\node (Z2) at (0,1){$Z$};
\node (pb) at (1.25,0.75){$\lrcorner$};

\node[overlay] (Point) at (2.2,0){.};

\draw[->]
(X) edge [bend left=30] node[right] {$f$} (Y)
(P) edge  node[left]{$\lambda$}(Y)
(PY) edge node[below]{$\lambda_Y$}(Y)
(Z1) edge (PY)
(X) edge  node[left]{$g$} (P)
(P') edge node[left]{$h$}(PY)
(P') edge node[below]{$j$}(P)
(P) edge [bend right=30]node[above]{$s$}(P')
(Z2) edge (P');
\draw[double distance = 2pt]
\end{equation*}

Using this diagram we can calculate
$$f=\lambda \circ g = \lambda \circ j \circ s \circ g= \lambda_Y \circ h \circ s \circ g,$$

which concludes the proof.

As mentioned before we can use this argument to show that $\KK_N(\A)$ admits coproducts and give a precise description of them.

Let $\F_\Se$ be a Frobenius exact category such that $\F$ admits $I$-indexed coproducts for $I$ a set. Then the stable category $\underline{\F}_\Se$ also admits $I$-indexed coproducts and we have that the canonical functor $\F \to \underline{\F}_\Se$ preserves coproducts,  i. e. the morphisms exhibiting the coproduct in $\F$ are representatives of the morphisms exhibiting the coproduct in $\underline{\F}_\Se$.

We will show that for a coproduct $\left(\bigoplus\limits_{i \in I}X_i,\iota_i\right)$ in $\F$ the maps $\left[\iota_i\right]$ exhibit $\bigoplus\limits_{i\in I} X_i$ as a coproduct in $\underline{\F}_\Se$.

First consider the coproduct $\left(\bigoplus\limits_{i \in I}X_i,\iota_i\right)$ in $\F$. The $\iota_i$ induces morphisms $[\iota_i]:X_i \to \bigoplus\limits_{i\in I}X_i$ in $\underline{\F}_\Se$. Now consider for each $i \in I$ a morphism $\psi_i:X_i \to T$ in $\underline{\F}_\Se$ and pick for each of these a representative $\varphi_i$ such that $[\varphi_i]=\psi_i$, which are morphisms $\varphi_i:X_i \to T$ in $\F$. So we get by the universal property of the coproduct a unique morphism $\varphi: \bigoplus\limits_{i\in I} X_i \to T$, such that the following diagram commutes for each $i \in I$:

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (Xi) at (0,0.8){$X_i$};
\node (X) at (1,0.8){$\bigoplus X_i$};
\node (T)  at (0.5,0) {$T.$};

\draw[->]
(Xi) edge node[above] {$\iota_i$} (X)
(Xi) edge node[left] {$\varphi_i$} (T)
(X) edge node[right] {$\varphi$} (T);
\end{equation*}

This implies by the definition of the stable category that the same diagram commutes in $\underline{\F}_\Se$.

Hence it just remains to show uniqueness of $\varphi$. Since $\F$ is additive it actually suffices to show that if we have a morphism $f:\bigoplus\limits_{i\in I}X_i \to T$ in $\underline{\F}_\Se$ such that all $f \circ [\iota_i]$ vanish, so does $f$. Hence assume $f \circ [\iota_i]=0$ and choose representatives $ [f_i]=f \circ [\iota_i]$. These $f_i$ now factor by definition of the stable category and Proposition <ref> as $f_i = \lambda_T \circ \psi_i$ for some $\psi_i:X_i \to P_T$ and $\lambda_T: P_T \to T$ some $\Se$-projective resolution of $T$. So we get for all $i \in I$ a commutative diagram in $\F$

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Xi) at (0,1.6){$X_i$};
\node (X) at (2,1.6){$\bigoplus X_i$};
\node (T)  at (1,0) {$T.$};
\node (P) at (1,1){$P_T$};
\draw[->]
(Xi) edge node[above right] {$\psi_i$}(P)
(P) edge node[above right] {$\lambda_T$} (T)
(Xi) edge node[above] {$\iota_i$} (X)
(Xi) edge node[left] {$f_i$} (T)
(X) edge node[right] {$f$} (T);
\end{equation*}

By the universal property of the coproduct this induces a map $\psi: \bigoplus\limits_{i\in I} X_i \to P_T$, such that the top, the left and the outer triangle commute in the following diagram in $\F$:

\begin{equation*}
\tikz[heighttwo,xscale=1.7,yscale=1.7,baseline]{

\node (Xi) at (0,1.6){$X_i$};
\node (X) at (2,1.6){$\bigoplus X_i$};
\node (T)  at (1,0) {$T.$};
\node (P) at (1,1){$P_T$};
\draw[->]
(X) edge node[above left] {$\psi$}(P)
(Xi) edge node[above right] {$\psi_i$}(P)
(P) edge node[above right] {$\lambda_T$} (T)
(Xi) edge node[above] {$\iota_i$} (X)
(Xi) edge node[left] {$f_i$} (T)
(X) edge node[right] {$f$} (T);
\end{equation*}

This implies that the whole diagram commutes. The commutativity of this diagram now just means $[f] =0$ in $\underline{\F}_\Se$, which concludes the proof.

One can show an analogous statement for products. There one just needs to use $\Se$-injective hulls instead of $\Se$-projective resolutions. However this is not important for the rest of our constructions, and so we omit it.

We are also interested in having a more calculation based approach towards the structure of the homotopy category.

Let $\A$ be an additive category, $N \in \N$ and let $f:X\to Y$ be a morphism of $N$-complexes in $\A$. Then $f$ is said to be nullhomotopic if there exists for all $j \in \Z$ an $s^j \in \A(X^j,Y^{j-N+1})$ such that for all $i \in \Z$ we get:

\begin{equation*}
f^i=\sum_{j=0}^{N-1} d^{N-j-1}\circ s^{i+j} \circ d^{j}.
\end{equation*}

This definition should of course be equivalent to the factoring of a morphism over an $\Se_\oplus$-injective-projective object.

Let $\A$ be an additive category and $f:X\to Y$ be a morphism in $\CC_N(\A)$. Then the following statements are equivalent:

* the morphism $f$ is nullhomotopic,

* the morphism $f$ factors through an $\Se_\oplus$-injective-projective object.

We show the implications separately:

By definition of a nullhomotopic morphism there exist $s^t \in \A (X^t,Y^{t-N+1})$ for all $t \in \Z$ such that

\begin{equation*}
f^i=\sum_{j=0}^{N-1} d^{N-j-1}\circ s^{i+j} \circ d^{j}.
\end{equation*}

Consider the map $\lambda^I_N: X \to \II_N(X)$. The maps $s^i: X^i \to Y^{i-N+1}$ induce maps $\hat{s}^i: \mu^i_N(X^i)^i \to Y$ via the adjunction $\CC_N(\A)(\mu_N^i (X^i), Y) \cong \A (X^i,Y^{i-N+1})$ from Proposition <ref>. But this gives a map $\hat{s}: \II_N(X) \to Y$ since the source is just a direct sum of $\mu_N^i(X^i)$. Unravelling the definition of $\II_N (X)$, $\lambda_X^\II$ and the adjunction one gets:

\begin{align*}
(\lambda^\II_X)^i&=\begin{bmatrix}\id & d & d^2 & \cdots & d^{N-2}& d^{N-1} \end{bmatrix}^T :
X^i \to X^i\oplus X^{i+1}\oplus \cdots \oplus X^{i+N-2}\oplus X^{i+N-1})  \\
\hat{s}^i&= \begin{bmatrix} d^{N-1} \circ s^i & d^{N-2} \circ s^{i+1} & d^{N-3} \circ s^{i+2} & \cdots  & s^{i+N-1} \end{bmatrix} :
X^i\oplus X^{i+1}\oplus  \cdots \oplus X^{i+N-1}  \to  Y^i ).
\end{align*}

Composing these morphisms yields now:

\begin{equation*}
(\hat{s} \circ \lambda_X^\II)^i = \sum_{j=0}^{N-1} d^{N-j-1}s^{i+j}d^{j} = f^i.
\end{equation*}

So since $\II_N(X)$ is $\Se_\oplus$-injective-projective this proves the claim.

Let $f: X \to Y$ factor as $f=s\circ g$ where $g: X \to P$, $s:P \to Y$ and $P$ is $\Se_\oplus$-injective-projective. Then we may assume by Lemma <ref> that $P \cong \II_N(X)$ and $g= \lambda^\II_X$. Now the claim follows cf. Remark <ref>, since we have that $f$ factors as $\lambda^\II_X \circ s$ where $\lambda^\II_X:X \to \II(X)$ and $s: \II_N(X) \to Y$. More precisely these $s^i$ now define levelwise maps $s^i: X^i\oplus X^{i+1}\oplus X^{i+2}...\oplus X^{i+N-1}  \to  Y^i$, the fact that these commute with the differential of $\II_N(X)$ yields

$$s^i\circ \iota_{X^j} = s^{i+1} \circ \iota_{X^{j}} \text{ for }0<j-i < N.$$

Here $\iota_{X^j}:X^j \hookrightarrow \bigoplus\limits_{l=i}^{\mathclap{i+N-1}}X^l$ is the canonicalinclusion for $i \le j \le i+N-1$.

This means that $s: \II_N(X) \to Y$ is equivalently defined by $\widehat{s^i}: X^i \to Y^{i-N+1}$ and

\begin{equation*}
s^i= \begin{bmatrix} d^{N-1} \circ \widehat{s}^i & d^{N-2} \circ \widehat{s}^{i+1} & d^{N-3} \circ \widehat{s}^{i+2} & \cdots  & \widehat{s}^{i+N-1} \end{bmatrix} :
X^i\oplus X^{i+1}\oplus X^{i+2} \cdots \oplus X^{i+N-1}  \to  Y^i .
\end{equation*}
So we get

\begin{equation*}
(s \circ \lambda_X^\II)^i = \sum_{j=0}^{N-1} d^{N-j-1}\circ \widehat{s}^{i+j}\circ d^{j} = f^i,
\end{equation*}

as claimed.

The construction in terms of adjunctions can be seen more intuitively using the following diagram in $\A$, in which we only consider one summand of $\mu_N^i (X)$:

\begin{equation*}
\tikz[heightone,xscale=2.5,yscale=2.5,baseline]{

\node (P00) at (-1,0){$\cdot \cdot \cdot$};
\node (P10) at (3,0){$\cdot \cdot \cdot$};
\node (P20) at (7,0){$\cdot \cdot \cdot$};
\node (P01) at (-1,1){$\cdot \cdot \cdot$};
\node (P11) at (3,1){$\cdot \cdot \cdot$};
\node (P21) at (7,1){$\cdot \cdot \cdot$};
\node (P02) at (-1,2){$\cdot \cdot \cdot$};
\node (P12) at (3,2){$\cdot \cdot \cdot$};
\node (P22) at (7,2){$\cdot \cdot \cdot$};

\node (X_i-N-1) at (0,0){$X^{i-N}$};
\node (X_i-N) at (1,0){$X^{i-N+1}$};
\node (X_i-N+1) at (2,0){$X^{i-N+2}$};

\node (X_i-1) at (4,0){$X^{i-1}$};
\node (X_i) at (5,0){$X^{i}$};
\node (X_i+1) at (6,0){$X^{i+1}$};

\node (Xi_N-i-1) at (0,1){$0$};
\node (Xi_N-i) at (1,1){$X^i$};
\node (Xi_N-i+1) at (2,1){$X^i$};

\node (Xi_i-1) at (4,1){$X^i$};
\node (Xi_i) at (5,1){$X^i$};
\node (Xi_i+1) at (6,1){$0$};

\node (Y_i-N-1) at (0,2){$X^{i-N}$};
\node (Y_i-N) at (1,2){$X^{i-N+1}$};
\node (Y_i-N+1) at (2,2){$X^{i-N+2}$};

\node (Y_i-1) at (4,2){$X^{i-1}$};
\node (Y_i) at (5,2){$X^{i}$};
\node (Y_i+1) at (6,2){$X^{i+1}$};

\draw[->]

(Y_i-N-1) edge node[right] {$0$} (Xi_N-i-1)
(Y_i-N) edge node[right] {$d^{N-1}$} (Xi_N-i)
(Y_i-N+1) edge node[right] {$d^{N-2}$} (Xi_N-i+1)
(Y_i-1) edge node[right] {$d$} (Xi_i-1)
(Y_i) edge node[right] {$\id_{X^i}$} (Xi_i)
(Y_i+1) edge node[right] {$0$} (Xi_i+1)

(Xi_N-i-1) edge node[right] {$0$} (X_i-N-1)
(Xi_N-i) edge node[right] {$s^i$} (X_i-N)
(Xi_N-i+1) edge node[right] {$d\circ s^i$} (X_i-N+1)
(Xi_i-1) edge node[right] {$d^{N-2}\circ s^i$} (X_i-1)
(Xi_i) edge node[right] {$d^{N-1}\circ s^i$} (X_i)
(Xi_i+1) edge node[right] {$0$} (X_i+1)

(P02) edge node[above] {$d$} (Y_i-N-1)
(Y_i-N-1) edge node[above] {$d$} (Y_i-N)
(Y_i-N) edge node[above] {$d$} (Y_i-N+1)
(Y_i-N+1) edge node [above] {$d$} (P12)

(P12) edge node[above] {$d$} (Y_i-1)
(Y_i-1) edge node[above] {$d$} (Y_i)
(Y_i) edge node[above] {$d$} (Y_i+1)
(Y_i+1) edge node[above] {$d$} (P22)

(P01) edge node[above] {$0$} (Xi_N-i-1)
(Xi_N-i-1) edge node[above] {$0$} (Xi_N-i)
(Xi_N-i) edge node[above] {$\id_{X^i}$} (Xi_N-i+1)
(Xi_N-i+1) edge node[above] {$\id_{X^i}$} (P11)

(P11) edge node[above] {$\id_{X^i}$} (Xi_i-1)
(Xi_i-1) edge node[above] {$\id_{X^i}$} (Xi_i)
(Xi_i) edge node[above] {$0$} (Xi_i+1)
(Xi_i+1) edge node[above] {$0$} (P21)

(P00) edge node[above] {$d$} (X_i-N-1)
(X_i-N-1) edge node[above] {$d$} (X_i-N)
(X_i-N) edge node[above] {$d$} (X_i-N+1)
(X_i-N+1) edge node[above] {$d$} (P10)

(P10) edge node[above] {$d$} (X_i-1)
(X_i-1) edge node[above] {$d$} (X_i)
(X_i) edge node[above] {$d$} (X_i+1)
(X_i+1) edge node[above] {$d$} (P20)

%% \draw[dashed, ->]
%% (Y_i-N-1) edge [bend right=30] node[below left] {$0$} (X_i-% N-1)
%% (Y_i-N) edge[bend right=30] node[below left] {$d^{N-1}s^i$}  (X_i-N)
%% (Y_i-N+1) edge[bend right=30] node[below left] {$d^{N-1}s^i$} (X_i-N+1)
%% (Y_i-1) edge[bend right=30] node[below left] {$d^{N-1}s^i$} (X_i-1)
%%(Y_i) edge[bend right=30] node[below left] {$d^{N-1}s^i$} (X_i)
%%(Y_i+1) edge[bend right=30] node[below left] {$d^{N-1}s^i$} (X_i+1)

\end{equation*}

By considering the composition of the two vertical morphisms one gets the desired summands for the homotopy equation.

§ TWO APPROACHES TO THE DERIVED CATEGORY OF $N$-COMPLEXES

After this work on the homotopy category of $N$-complexes and its properties we are finally able to inspect the notion of cohomology of $N$-complexes. In particular this yields the notions of $N$-quasi-isomorphisms and the derived category of $N$-complexes. As in the case of chain complexes this category is defined as a Verdier quotient of the homotopy category of $N$-complexes. However, in chain complexes over a ring one can realize the derived category already as a subcategory of the homotopy category, namely as the full subcategory of h-projectives.

The aim of this section is to understand the notion of $N$-cohomology and $N$-quasi-isomorphisms and use this in the definition of the derived category of $N$-complexes and the property of being h-projective. Finally we will show that for the category of modules over a ring the full subcategory of the homotopy category consisting of h-projective $N$-complexes already realizes the derived category. To prove this we will use a construction of semifree complexes which is inspired by an argument of Drinfeld <cit.>.

Each of the approaches mentioned above has its advantages. For example the construction as a Verdier quotient is more straightforward, intuitive and comes with a canonical functor. On the other hand the construction using h-projectives allows better control over the morphisms in the derived category. This is due to the fact that all morphisms are already contained in the homotopy category and calculations using roof diagrams are not needed.

In Section 8 we will construct another realization of the derived category of $N$-complexes using model structures. The advantage of this is that it shows that the derived category is indeed the Gabriel-Zismann localisation along certain morphisms.

In order to get a first notion of what the derived category of $N$-complexes should be we need to define the cohomology of an $N$-complex. Furthermore we will show that it still defines a functor just as in the usual case. A problem however is that the cohomology of $N$-complexes consists of more than one object in each degree, but we will see that we loose less than one might think due to that.

Let $\A$ be an abelian category, $N \in \N$ and $X \in \CC_N(\A)$. Then the $r$-th cohomology in degree $i$, for $1 \le r \le N-1$ and $i \in \Z$, is the following object in $\A$:

\begin{equation*}
H_{(r)}^i(X):= \ker(d^r)^i/\im(d^{N-r})^i.
\end{equation*}

Definition <ref> can be extended to an additive functor
$$H_{(r)}^i(\_):\CC_N(\A) \to \A.$$

Let $f:X \to Y$ be a morphism in $\CC_N(\A)$. By definition $f$ commutes with the differentials, hence $f\circ d_X^r=d_Y^r \circ f$. So $f$ induces a unique map $\ker(d^r)^i(f) :\ker(d_X^r)^i \to \ker(d^r_Y)^i $ by the universal property of $\ker(d^r_Y)$ such that the diagram

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (K) at (1,0){$\ker(d_Y^r)^i$};
\node (Yi) at (2,0){$Y^i$};
\node (Yi+r) at (3,0){$Y^{i+r}$};
\node (T)  at (2,-1) {$\ker(d_X^r)^i$};

\draw[->]

(T) edge node[right] {$f^i|_{\ker(d_Y^r)}$} (Yi)
(Yi) edge node[above] {$d^r$} (Yi+r);
\draw[right hook ->]
(K) to (Yi);
\draw[dashed, ->]
(T) edge node[below left] {$\exists ! \ker(d^r)^i(f)$}(K);
\end{equation*}
commutes, since

\begin{equation*}
d^r \circ f|_{\ker(d_X^r)} = f \circ d_X^r |_{\ker(d_X^r)}=0.
\end{equation*}

Hence to get a functor from $N$-complexes to the ground category it suffices to check that $f(\im(d_X^{N-r})) \subset \im(d_Y^{N-r})$, but this is satisfied, since by definition $f$ commutes with the differential.

Now it remains to check that this indeed is an additive functor. But for this we only need to check that the assignment $f \mapsto \ker(d^r)^i(f)$ is additive. Hence let $f=g +g'$. In this case both $\ker(d^r)^i(f)$ and $\ker(d^r)^i(g)+\ker(d^r)^i(g')$ would make the diagram commute, and hence they have to coincide by the universal property of the kernel.

Having defined the notion of cohomology, it is natural to define the notion of an $N$-quasi-isomorphism:

Let $\A$ be an abelian category and $N \in \N$. We call a morphism $f: X\to Y$ in $\CC_N(\A)$ an $N$-quasi-isomorphism if $H_{(r)}^i(f)$ is an isomorphism for all $i \in \Z$ and $1 \le r \le N-1$.

With these basic definitions we will now show that cohomology defines a functor on the category $\KK_N(\A)$ as well.

Let $\A$ be an abelian category, $N \in \N$ and $X\in \CC_N(\A)$ such that $X\cong 0$ in the category $\KK_N(\A)$. Then $H_{(r)}^i(X)=0$ for all $i \in \Z$ and $1 \le r \le N-1$.

The assumption $X \cong 0$ in $\KK_N(\A)$ implies that $\id_X$ factors through $\PP_N(X)$. Hence $X$ is a direct summand of $\PP_N(X)$ in $\CC_N(\A)$. So by Proposition <ref> it is enough to show for all $X \in \CC_N(\A)$ that $H^i_{(r)}(\PP_N(X))=0$. Now by the definition of $\PP_N(X)$ we have

\begin{equation*}
\PP_N(X):=\bigoplus_{j \in \Z} \mu_N^j(X^j).
\end{equation*}

So it suffices to check the claim for $\mu_N^j(X^j)$. Now observe that $\ker(d_{\mu_N^j(X^j)}^r)^i=X^j$ if $j-r < i \le j$, and zero else. So we just need to consider the case $j-r < i \le j$, but then
and hence $H_{(r)}^i(\mu_N^j(X^j))=0$. This finishes the proof since $H^i_{(r)}(\_)$ is additive.

Let $\A$ be an abelian category, $N \in \N$, $i \in \Z$ and $1 \le r \le N-1$. Then $H_{(r)}^i(\_)$ induces an additive functor $H_{(r)}^i(\_):\KK_N(\A) \to \A$.

By Proposition <ref> and Proposition <ref> the functor $H_{(r)}^i(\_):\CC_N(\A)\to \A$ is additive and sends all morphisms to zero that factor through an $\Se_\oplus$-injective-projective object. But $\KK_N(\A)$ is the stable category of $\CC_N(\A)_{\Se_\oplus}$ and hence $H_{(r)}^i(\_)$ factors through $\KK_N(\A)$.

Let $\A$ be an abelian category, $N \in \N$ and $f: X \to Y$ a morphism in $\CC_N(\A)$, such that $f$ induces an isomorphism in $\KK_N(\A)$. Then $f$ is an $N$-quasi-isomorphism.

Obvious by Corollary <ref>.

In the case of $N=2$ the long exact sequence in cohomology is a powerful tool to calculate cohomology.

Let $\A$ be an abelian category and

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0,0.3){$0$};
\node (X) at (1,0.3){$X$};
\node (Y) at (2,0.3){$Y$};
\node (Z) at (3,0.3){$Z$};
\node (02) at (4,0.3){$0$};

\draw[->]
(01) edge (X)
(X) edge node[above]{$f$} (Y)
(Y) edge node[above]{$g$} (Z)
(Z) edge (02);
\end{equation*}

a short exact sequence in $\CC_N(\A)$. Then there exists for every $1 \le r \le N-1$ a long exact sequence of the form:

\begin{equation*}
\tikz[heighttwo,xscale=4,yscale=2,baseline]{

\node (P1) at (2,3){$\cdots$};
\node (HZi-N) at (3,3){$H^{i-N}_{(r)}(Z)$};
\node (HXi-N+r) at (1,2) {$H^{i-N+r}_{(N-r)}(X)$};
\node (HYi-N+r) at (2,2) {$H^{i-N+r}_{(N-r)}(Y)$};
\node (HZi-N+r) at (3,2) {$H^{i-N+r}_{(N-r)}(Z)$};
\node (HXi) at (1,1){$H^i_{(r)}(X)$};
\node (HYi) at (2,1){$H^i_{(r)}(Y)$};
\node (HZi) at (3,1){$H^i_{(r)}(Z)$};
\node (HXi+r) at (1,0){$H^{i+r}_{(r)}(X)$};
\node (P2) at (2,0){$\cdots$};

\node[overlay] (Point) at (3.2,0){.};

\draw[overlay,->]
(P1) edge (HZi-N)
(HXi-N+r) edge node[above]{$H^{i-N+r}_{(N-r)}(f)$} (HYi-N+r)
(HYi-N+r) edge node[above]{$H^{i-N+r}_{(N-r)}(g)$} (HZi-N+r)
(HXi) edge node[above]{$H^i_{(r)}(f)$} (HYi)
(HYi) edge node[above]{$H^i_{(r)}(g)$} (HZi)
(HXi+r) edge (P2)
(HZi-N) edge[out=355,in=175]  (HXi-N+r)
(HZi-N+r) edge[out=355,in=175] (HXi)
(HZi) edge[out=355,in=175]  (HXi+r);
\end{equation*}

Fix $1 \le r \le N-1$. Then we may build a short exact sequence

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (01) at (0,0){$0$};
\node (X) at (1,0){$\widehat{X}$};
\node (Y) at (2,0){$\widehat{Y}$};
\node (Z) at (3,0){$\widehat{Z}$};
\node (02) at (4,0){$0$};

\draw[->]
(01) edge (X)
(X) edge node[above]{$\widehat{f}$} (Y)
(Y) edge node[above]{$\widehat{g}$} (Z)
(Z) edge (02);
\end{equation*}

in $\CC_2(\A)$ by considering the following diagram in $\A$:

\begin{equation*}
\tikz[heighttwo,xscale=3,yscale=1.9,baseline]{

\node(d-1)[overlay] at (0,2){\tiny(0)};
\node(d0)[overlay] at (0,3){\tiny(-1)};
\node(d1)[overlay] at (0,1){\tiny(1)};

\node (PX1) at (1,4){$\vdots$};
\node (PY1) at (2,4){$\vdots$};
\node (PZ1) at (3,4){$\vdots$};
\node (Xi-N+r) at (1,3){$X^{i-N+r}$};
\node (Yi-N+r) at (2,3){$Y^{i-N+r}$};
\node (Zi-N+r) at (3,3){$Z^{i-N+r}$};
\node (Xi) at (1,2){$X^i$};
\node (Yi) at (2,2){$Y^i$};
\node (Zi) at (3,2){$Z^i$};
\node (Xi+r) at (1,1){$X^{i+r}$};
\node (Yi+r) at (2,1){$Y^{i+r}$};
\node (Zi+r) at (3,1){$Z^{i+r}$};
\node (PX2) at (1,0){$\vdots$};
\node (PY2) at (2,0){$\vdots$};
\node (PZ2) at (3,0){$\vdots$};

\draw[overlay,->]
(PX1) edge node[left] {}(Xi-N+r)
(Xi-N+r) edge node[left] {$d_X^{N-r}$}(Xi)
(Xi) edge node[left] {$d_X^{r}$}(Xi+r)
(Xi+r) edge node[left] {}(PX2)
(PY1) edge node[left] {}(Yi-N+r)
(Yi-N+r) edge node[left] {$d_Y^{N-r}$}(Yi)
(Yi) edge node[left] {$d_Y^{r}$}(Yi+r)
(Yi+r) edge node[left] {}(PY2)
(PZ1) edge node[left] {}(Zi-N+r)
(Zi-N+r) edge node[left] {$d_Z^{N-r}$}(Zi)
(Zi) edge node[left] {$d_Z^{r}$}(Zi+r)
(Zi+r) edge node[left] {}(PZ2)
(Xi-N+r) edge node[above]{$f^{i-N+r}$} (Yi-N+r)
(Yi-N+r) edge node[above]{$g^{i-N+r}$} (Zi-N+r)
(Xi) edge node[above]{$f^i$} (Yi)
(Yi) edge node[above]{$g^i$} (Zi)
(Xi+r) edge node[above]{$f^{i+r}$} (Yi+r)
(Yi+r) edge node[above]{$g^{i+r}$} (Zi+r);
\end{equation*}

Now we get a long exact sequence in ordinary cohomology of the form

\begin{equation*}
\tikz[heighttwo,xscale=4,yscale=2,baseline]{

\node (P1) at (2,3){$\cdots$};
\node (HZi-N) at (3,3){$H^{-2}_{}(\widehat{X})$};
\node (HXi-N+r) at (1,2) {$H^{-1}_{}(\widehat{X})$};
\node (HYi-N+r) at (2,2) {$H^{-1}_{}(\widehat{Y})$};
\node (HZi-N+r) at (3,2) {$H^{-1}_{}(\widehat{Z})$};
\node (HXi) at (1,1){$H^0_{}(\widehat{X})$};
\node (HYi) at (2,1){$H^0_{}(\widehat{Y})$};
\node (HZi) at (3,1){$H^0_{}(\widehat{Z})$};
\node (HXi+r) at (1,0){$H^{1}_{}(\widehat{X})$};
\node (P2) at (2,0){$\cdots$};

\node[overlay] (Point) at (3.2,0){.};

\draw[overlay,->]
(P1) edge (HZi-N)
(HXi-N+r) edge node[above]{$H^{-1}_{}(\widehat{f})$} (HYi-N+r)
(HYi-N+r) edge node[above]{$H^{-1}_{}(\widehat{g})$} (HZi-N+r)
(HXi) edge node[above]{$H^0_{}(\widehat{f})$} (HYi)
(HYi) edge node[above]{$H^0_{}(\widehat{g})$} (HZi)
(HXi+r) edge (P2)
(HZi-N) edge[out=355,in=175]  (HXi-N+r)
(HZi-N+r) edge[out=355,in=175]  (HXi)
(HZi) edge[out=355,in=175](HXi+r);
\end{equation*}

But this is just the desired sequence by the definition of $\widehat{X}$, $\widehat{Y}$ and $\widehat{Z}$.

Let $\A$ be an abelian category, $N \in \N$  and

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=1,baseline]{

\node (X) at (1,0){$X$};
\node (Y) at (2,0){$Y$};
\node (Z) at (3,0){$Z$};
\node (X1) at (4,0) {$\Sigma(X)$};

\draw[->]
(X) edge node[above]{$f$} (Y)
(Y) edge node[above]{$g$} (Z)
(Z) edge node[above]{$h$} (X1);
\end{equation*}

be a triangle in $\KK_N(\A)$. Then this triangle induces long exact sequences like in Lemma <ref>.

The triangle comes by Corollary <ref> from an $\Se_\oplus$-exact sequence, hence we get the long exact sequence for this short exact sequence by Lemma <ref>. Since all objects in this short exact sequence are isomorphic in $\KK_N(\A)$ to the objects in our triangle we get the statement by Corollary <ref>.

One of the most important features of the long exact sequence in ordinary cohomology is a characterisation of a $2$-quasi-isomorphism by its cone. This is still possible in the general case.

Let $\A$ be an abelian category, $N \in \N$ and let $f: X\to Y$ be a morphism in $\CC_N(\A)$. Then $f$ is an $N$-quasi-isomorphism if and only if $H_{(r)}^i(\Cone(f))=0$ for all $i\in \Z$ and $1 \le r \le N-1$.

We have a triangle of the form

\begin{equation*}
\tikz[heighttwo,xscale=2,yscale=2,baseline]{

\node (X) at (1,0.3){$X$};
\node (Y) at (2,0.3){$Y$};
\node (Z) at (3,0.3){$\Cone(f)$};
\node (X1) at (4,0.3) {$\Sigma(X)$};

\draw[->]
(X) edge node[above]{$f$} (Y)
(Y) edge node[above]{$g$} (Z)
(Z) edge node[above]{$h$} (X1);
\end{equation*}

hence we get long exact sequences

\begin{equation*}
\tikz[heighttwo,xscale=1,yscale=2,baseline]{

\node (P1) at (4,3){$\cdots$};
\node (HZi-N) at (8,3){$H^{i-N}_{(r)}(\Cone(f))$};
\node (HXi-N+r) at (0,2) {$H^{i-N+r}_{(N-r)}(X)$};
\node (HYi-N+r) at (4,2) {$H^{i-N+r}_{(N-r)}(Y)$};
\node (HZi-N+r) at (8,2) {$H^{i-N+r}_{(N-r)}(\Cone(f))$};
\node (HXi) at (0,1){$H^i_{(r)}(X)$};
\node (HYi) at (4,1){$H^i_{(r)}(Y)$};
\node (HZi) at (8,1){$H^i_{(r)}(\Cone(f))$};
\node (HXi+r) at (0,0){$H^{i+r}_{(r)}(X)$};
\node (P2) at (4,0){$\cdots$};

\node[overlay] (Point) at (9.5,0){.};

\draw[overlay,->]
(P1) edge (HZi-N)
(HXi-N+r) edge node[above]{$H^{i-N+r}_{(N-r)}(f)$} (HYi-N+r)
(HYi-N+r) edge node[above]{$H^{i-N+r}_{(N-r)}(g)$} (HZi-N+r)
(HXi) edge node[above]{$H^i_{(r)}(f)$} (HYi)
(HYi) edge node[above]{$H^i_{(r)}(g)$} (HZi)
(HXi+r) edge (P2)
(HZi-N) edge[out=355,in=175](HXi-N+r)
(HZi-N+r) edge[out=355,in=175] (HXi)
(HZi) edge[out=355,in=175] (HXi+r);
\end{equation*}

But by the exactness of these sequences we get that $H^i_{(r)}(f)$ is an isomorphism for all $i \in \Z$ and $1 \le r \le N-1$, if and only if $H^i_{(r)}(\Cone(f))=0$ for all $i \in \Z$ and $1 \le r \le N-1$.

§.§ Verdier quotient