I\to\mathcal{C}$, the left Kan extension $f^{k}G\colon J\to\mathcal{C}$ is
constant on subposets of $J$ that are not only convex but also have global
minima.

######  7.10.

Assume $I$ is a consistent upper semilattice of finite type. According to 6.7,
its realisation $\mathcal{R}(I)$ is also an upper semilattice. Choose an
element $d$ in $I$ and a finite subset $V\subset(-1,0)$. According to
Corollary 6.8, the inclusion $\alpha\colon\mathcal{R}_{I\leq
d}(I,V)\subset\mathcal{R}(I)$ is a sublattice (see 5.7). It consists of these
pairs $(a,f)$ for which $a\leq d$, and non-zero values of $f$ are in $V$.

We are going to describe the transfer
$\alpha^{!}\colon\mathcal{R}(I)_{\ast}\to\mathcal{R}_{I\leq d}(I,V)_{\ast}$.
Let $(a,f)$ be in $\mathcal{R}(I)$. By definition (see 7.1):

$\alpha^{!}(a,f)=\begin{cases}-\infty&\text{ if }\mathcal{R}_{I\leq
d}(I,V)\leq(a,f)\text{ is empty}\\\ \bigvee\left(\mathcal{R}_{I\leq
d}(I,V))\leq(a,f)\right)&\text{ otherwise }\end{cases}$

Thus, to describe $\alpha^{!}(a,f)$, the set $\mathcal{R}_{I\leq
d}(I,V)\leq(a,f)$ needs to be discussed. We claim that it is non-empty if and
only if $d$ and $\text{supp}(f)$ have a common ancestor. Here is a proof. If
there is $(b,g)$ in $\mathcal{R}_{I\leq d}(I,V)$ for which $(b,g)\leq(a,f)$,
then, according to Proposition 4.4, any ancestor of $\text{supp}(g)$ is also
an ancestor of $\text{supp}(f)$. The relation $b\leq d$ implies therefore that
any common ancestor of $b$ and $\text{supp}(g)$ is also a common ancestor of
$d$ and $\text{supp}(f)$.

Assume $d$ and $\text{supp}(f)$ have a common ancestor. If $V$ is empty, set
$v_{0}:=0$, if $V$ is non-empty, set $v_{0}$ to be the minimal element in $V$.
Define $S:=\\{x\in\mathcal{P}(a)\ |\ f(x)<v_{0}\\}$. Since $S$ is a subset of
$\text{supp}(f)$, it has a common ancestor with $d$, and consequently the
following definition of an element $c$ in $I$ makes sense as the necessary
products exist due to $I$ being an upper semilattice of finite type:

$c:=\begin{cases}(\bigwedge S)\wedge d&\text{if }S\neq\emptyset\\\ a\wedge
d&\text{if }S=\emptyset\end{cases}$

Any ancestor of $\\{d,\text{supp}(f),a\\}$ is also an ancestor of $c$. Let $y$
be a parent of $c$. Consider the subset
$(y\leq\mathcal{P}(a))\subset\mathcal{P}(a)$. There are two possibilities,
either $c$ is an ancestor of the set $y\leq\mathcal{P}(a)$ or not. In the
second case, for every $x$ in $(y\leq\mathcal{P}(a))\setminus(c\leq I)$, two
things happen: first $f(x)\geq v_{0}$ ($x$ does not belong to $S$) and second
there is an equality $y=c\wedge x$. We use the first inequality to define a
function $h\colon\mathcal{P}(c)\to(-1,0]$ by the following formula:

$h(y):=\begin{cases}0&\text{if $c$ is an ancestor of }y\leq\mathcal{P}(a)\\\
\text{max}\\{v\in V\ |\ v\leq f\left((y\leq\mathcal{P}(a))\setminus(c\leq
I)\right)\\}&\text{otherwise }\end{cases}$

According to the above formula, if $h(y)<0$, then there is $x$ in
$\mathcal{P}(a)$ for which $y=c\wedge x$ and $f(x)<0$. Any common ancestor of
$c$ and $\text{supp}(f)$ is therefore also an ancestor of $y$. As this happens
for every $y$ for which $h(y)<0$, the support $\text{supp}(h)$ has an
ancestor, and hence $(c,h)$ belongs to the realisation $\mathcal{R}(I)$. Since
$h$ has values in $V,$ the element $(c,h)$ belongs to $\mathcal{R}_{I\leq
d}(I,V)$.

Consider the translation $T_{c\leq a}h$. Let $x$ be a parent of $a$ for which
$f(x)<0$. Consistency of $I$ and the fact $c$ and $\text{supp}(f)$ have a
common ancestor, imply that the product $c\wedge x$ exists and is either $c$
or is a parent of $c$. If $c\leq x$, then $T_{c\leq a}h(x)=-1\leq f(x)$, and,
if $c\wedge x$ is a parent of $c$, then $T_{c\leq c}h(x)=h(c\wedge x)\leq
f(x)$ by the formula defining $h$. These relations imply $(c,h)\leq(a,f)$, and
hence the set $\mathcal{R}_{I\leq d}(I,V)\leq(a,f)$ is non-empty.

We are going to prove $(c,h)=\bigvee\left(\mathcal{R}_{I\leq
d}(I,V))\leq(a,f)\right)$, which gives:

$\alpha^{!}(a,f)=(c,h)$

Let $(b,g)$ be in $\mathcal{R}_{I\leq d}(I,V)\leq(a,f)$. We need to show
$(b,g)\leq(c,h)$. The relation $b\leq c$ holds since $b$ is an ancestor of
$\\{d,\text{supp}(f),a\\}$. Consider $T_{b\leq c}g$. Let $y$ be a parent of
$c$ for which $h(y)<0$. Then there is $x$ in $\mathcal{P}(a)$ such that
$y=c\wedge x$ and $f(x)<0$. By the assumption $T_{b\leq c}g(y)\leq f(x)$.
Since the values of $T_{b\leq c}g$ are in $V$, then $T_{b\leq c}g(y)\leq
h(y)$, which gives $(b,g)\leq(c,h)$.

Consider $V=\\{-0.5\\}$ and $d=(2,2)$. Recall that the subposet inclusion
$\alpha\colon\mathcal{R}_{\mathbb{N}^{2}\leq
d}(\mathbb{N}^{2},V)\hookrightarrow\mathcal{R}(\mathbb{N}^{2})$ can be
identified with $0.5[2]^{2}\subset[0,\infty)^{2}$ (see 4.16). Figure 5
illustrates the effect of the transfer of $\alpha$ via this identification.

Figure 5. The intersections of the solid lines correspond to points in
$\mathbb{N}^{2}$. All the intersections of solid and dashed lines correspond
to points in
$\mathcal{R}_{\mathbb{N}^{2}\leq(2,2)}(\mathbb{N}^{2},\\{-0.5,0\\})$.

##  8\. Tame functors

######  8.1 Definition.

Let $\mathcal{C}$ be category and $J$ a poset. A functor $F\colon
J\to\mathcal{C}$ is called tame or discretisable if there is a finite poset
$I$ and functors $f\colon I\to J$ and $G\colon I\to\mathcal{C}$ for which $F$
is isomorphic to the left Kan extension $f^{k}G$ of $G$ along $f$ (see 2.11),
in which case $F$ is also said to be discretised by $f$, and $G$ is called a
discretisation of $F$.

The symbol $\text{Tame}(J,\mathcal{C})$ denotes the full subcategory of
$\text{Fun}(J,\mathcal{C})$ whose objects are tame functors.

If $J$ is a finite poset, then any functor $F\colon J\to\mathcal{C}$ is tame.
The notion of tameness is therefore restrictive and meaningful only in case
$J$ is infinite. To determine if $F\colon J\to\mathcal{C}$ is tame we need to
look for two things: a functor $f\colon I\to J$ that discretises $F$ and a
discretisation $G\colon I\to\mathcal{C}$ of $F$.

How can we search for a discretising $f$? In many situations it is enough to
consider only finite subposet inclusions $I\subset J$. This is the case if for
example $\mathcal{C}$ is closed under finite colimits. For example, consider
$\mathcal{C}$ the category of vector spaces over a given field. The closure of
$\mathcal{C}$ under finite colimits guarantees the existence of the left Kan
extension $f^{k}\colon\text{Fun}(I,\mathcal{C})\to\text{Fun}(J,\mathcal{C})$
for every functor $f\colon I\to J$ with finite $I$. This has the following
consequences:

######  8.2 Proposition.

Let $\mathcal{C}$ be a category closed under finite colimits and $J$ a poset.

  1. (1)

Consider the following commutative diagram of poset functors with $I_{0}$ and
$I_{1}$ finite:

${I_{0}}$${I_{1}}$${J}$$\scriptstyle{g}$$\scriptstyle{f_{0}}$$\scriptstyle{f_{1}}$

If $F\colon J\to\mathcal{C}$ is discretised by $f_{0}$, then it is also
discretised by $f_{1}$.

  2. (2)

If $F\colon J\to\mathcal{C}$ is discretised by $f\colon I\to J$, then it is
also discretised by every subposet $I^{\prime}\subset J$ for which
$f(I)\subset I^{\prime}$.

  3. (3)

Let $F_{1},\ldots,F_{n}\colon J\to\mathcal{C}$ be a finite sequence of tame
functors. Then there is a finite subposet $I\subset J$ that discretises
$F_{i}$ for all $i$.

  4. (4)

Let $L$ a finite category, and $\phi\colon L\to\text{\rm Fun}(J,\mathcal{C})$
a functor. Assume $\phi(l)$ is tame for every object $l$ in $L$. Then the
functor $\text{\rm colim}_{L}\phi$ is also tame.

###### Proof.

(1): As long as they exist, Kan extensions commute with compositions:
$f_{0}^{k}$ and $f_{1}^{k}g^{k}$ are naturally isomorphic. Thus if $F$ is
isomorphic to $f_{0}^{k}G$ (is discretised by $f_{0}$), then it is also
isomorphic to $f_{1}^{k}(g^{k}G)$.

(2): It is a direct consequence of (1).

(3): Let $I_{i}\subset J$ be a finite subposet discretising $F_{i}$. Then,
according to (2), $\cup_{i=1}^{n}I_{i}\subset I$ discretises $F_{i}$, for
every $i$.

(4): Use (3) to choose a finite subposet $f\colon I\subset J$ that discretises
$\phi(l)$ for every $l$ in $L$ (see 8.3.(3)). Since $f$ is an injection, the
natural transformation $f^{k}(\phi f)\to\phi$, adjoint to the identity
$\text{id}\colon\phi f\to\phi f$, is an isomorphism (see the end of 2.11).
Consequently $\text{\rm colim}_{L}\phi$ is isomorphic to $\text{\rm
colim}_{L}f^{k}(\phi f)$. Kan extensions commute with colimits and hence we
can conclude $\text{\rm colim}_{L}\phi$ is isomorphic to $f^{k}\text{\rm
colim}_{L}(\phi f)$ proving its tameness. ∎

According to Proposition 8.2, if $\mathcal{C}$ is closed under finite
colimits, the search for discretising functors can be restricted to finite
subposet inclusions $I\subset J$, in which case a discretisation is given by
restricting to $I$ (see 2.11). That is a considerable simplification. For
example, a tame functor indexed by a realisation $\mathcal{R}(I)$ with values
in a category closed under finite colimits can always be discretised by a
subposet of the form $\mathcal{R}_{D}(I,V)$, for some finite $D\subset I$ and
a finite $V\subset(-1,0)$ (see 4.16). However not all the categories
$\mathcal{C}$ for which we would like to have a simpler way of verifying
tameness are closed under finite colimits. For example, the category of
simplicial complexes fails to have this property. In such a case Corollary 7.8
can be used where the only assumption on $\mathcal{C}$ is that it has an
initial object $e$. It gives a characterisation of functors which are
discretised by a homomorphism $f\colon I\to J$ out of a finite upper
semilattice $I$. According to this corollary, a functor $F\colon
J\to\mathcal{C}$ is discretised by $f$ if and only if, for every $a$ in $J$,
if $(f\leq a)\not=\emptyset$, then $F(f(\bigvee_{I}(f\leq a))\leq a)$ is an
isomorphism, and if $(f\leq a)=\emptyset$, then $F(a)$ is isomorphic to $e$.
For an arbitrary $J$, not all tame functors are discretised by homomorphisms.
To make sure all tame functors are discretised by a homomorphism, $J$ itself
needs to be an upper semilattice. The rest of this section is devoted to
discussing tameness under this assumption on $J$ and when $\mathcal{C}$ is a
category with an initial object $e$. We start with a statement analogous to
Proposition 8.2:

######  8.3 Proposition.

Let $J$ be an upper semilattice and $\mathcal{C}$ a category with an initial
object.

  1. (1)

A functor $F\colon J\to\mathcal{C}$ is tame if and only if it is discretised
by a finite sublattice $I\subset J$ (see 5.7).

  2. (2)

Let $I_{0}\subset I_{1}\subset J$ be finite sublattices of $J$. If $F\colon
J\to\mathcal{C}$ is discretised by $I_{0}\subset J$, then it is also
discretised by $I_{1}\subset J$.

  3. (3)

Let $F_{1},\ldots,F_{n}\colon J\to\mathcal{C}$ be a finite sequence of tame
functors. Then there is a finite sublattice $I\subset J$ that discretises
$F_{i}$ for all $i$.

###### Proof.

(1): Let $F\colon J\to\mathcal{C}$ be tame. Chose a finite poset $I^{\prime}$
and functors $f\colon I^{\prime}\to J$ and $G\colon I^{\prime}\to\mathcal{C}$
for which there is an isomorphism $f^{k}G\to F$. Let $\phi\colon G\to Ff$ be
the natural transformation adjoint to this isomorphism. Since
$f(I^{\prime})\subset J$ is finite, the sublattice generated by this image
$I:=\langle f(I^{\prime})\rangle\subset J$ is also finite (see 5.7). Let us
denoted the inclusion $I^{\prime}\subset J$ by $g$ and by $h\colon I\to
I^{\prime}$ the functor that maps $x$ to $f(x)$. Since $g$ is a homomorphism
of finite type between upper semilattices, the left Kan extension
$g^{k}\colon\text{Fun}(I,\mathcal{C})\to\text{Fun}(J,\mathcal{C})$ exists (see
Section 7). We claim the natural transformation $g^{k}(Fg)\to F$, adjoint to
the identity $\text{id}\colon Fg\to Fg$, is an isomorphism. This would mean
$F$ is discretised by the finite sublattice $I\subset J$ proving the lemma.
According to Proposition 7.7.(1), we have $g^{k}(Fg)f=g^{k}(Fg)gh=Fgh=Ff$.
Thus the natural transformation $F\to g^{k}(Fg)$, whose adjoint is $\phi\colon
G\to Ff=g^{k}(Fg)f$, is the inverse to $g^{k}(Fg)\to F$.

(2): Exactly the same argument as in 8.2.(2) can be used to prove this
statement since the left Kan extensions along $I_{0}\subset I_{1}$,
$I_{0}\subset J$, and $I_{1}\subset J$ exist by 7.5.

(3): For every $i$ choose a finite sublattice $I_{i}\subset J$ discretising
$F_{i}$. Define $I:=\langle I_{1}\cup\cdots\cup I_{n}\rangle\subset Q$ (see
5.7). Since $I$ is finite, according to (2), $F_{i}$ is discretised by the
inclusion $I\subset J$ for every $i$. ∎

######  8.4.

For example, let $J$ be an upper semilattice of finite type. Then for every
finite subset $I\subset J$, there is $b$ (for example $\bigvee_{J}I$) for
which $I\subset(J\leq b)\subset J$. Consequently, according to Proposition
8.3, a functor $F\colon J\to\mathcal{C}$ is tame if and only if it is
discretised by the sublattice $(J\leq b)\subset J$, for some $b$ in $J$. Thus,
according to Corollary 7.8, $F$ is tame if and only if, there is $b$ in $J$
such that, for every $a$ in $J$, if $(J\leq b)\cap(J\leq a)\not=\emptyset$,
then $F\left((\bigvee(J\leq b)\cap(J\leq a))\leq a\right)$ is an isomorphism,
and if $(J\leq b)\cap(J\leq a)=\emptyset$, then $F(a)$ is isomorphic to $e$.

If $\mathcal{C}$ is closed under taking finite colimits and limits, then so is
the category of all functors $\text{Fun}(J,\mathcal{C})$ for every poset $J$.
In this case the category $\text{Tame}(J,\mathcal{C})$ is also closed under
finite colimits (see 8.2.(4)). In general, we do not know if
$\text{Tame}(J,\mathcal{C})$ is closed under finite limits. A sufficient
assumption to guarantee this is again $J$ being an upper semilattice.

######  8.5 Proposition.

Let $J$ be an upper semilattice, $\mathcal{C}$ a category closed under finite
limits, $L$ a finite category, and $\phi\colon L\to\text{\rm
Fun}(J,\mathcal{C})$ a functor. Assume $\phi(l)$ is tame for every object $l$
in $L$. Then the functor $\text{\rm lim}_{L}\phi$ is also tame.

###### Proof.

Let $f\colon I\subset J$ be a finite sublattice that discretises $\phi(l)$ for
every object $l$ in $L$ (see 8.3.(3)). Then the natural transformation
$f^{k}(\phi f)\to\phi$, adjoint to the identity $\text{id}\colon\phi f\to\phi
f$, is an isomorphism (see 7.8). Consequently $\text{\rm lim}_{L}\phi$ is
isomorphic to $\text{\rm lim}_{L}f^{k}(\phi f)$ and hence, by Proposition 7.7,
it is also isomorphic to $f^{k}(\text{\rm lim}_{L}\phi f)$. ∎

According to 8.2.(4) and 8.5, if $J$ is an upper semilattice and $\mathcal{C}$
is closed under finite limits and colimits, then so is the category of tame
functors $\text{Tame}(J,\mathcal{C})$. Furthermore the inclusion
$\text{Tame}(J,\mathcal{C})\subset\text{Fun}(J,\mathcal{C})$ preserves finite
limits and colimits.

##  9\. Homotopy theory of tame functors

Let $\mathcal{M}$ be a model category as defined in [14]. This means three
classes of morphisms in $\mathcal{M}$ are chosen: weak equivalences
($\xrightarrow{\sim}$), fibrations ($\twoheadrightarrow$), and cofibrations
($\hookrightarrow$), which are required to satisfy the following axioms:

  1. MC1

Finite limits and colimits exist in $\mathcal{M}$.

  2. MC2

If $f$ and $g$ are morphisms in $\mathcal{M}$ for which $gf$ is defined and if
two of the three morphisms $f$, $g$, $gf$ are weak equivalences, then so is
the third.

  3. MC3

The three classes of morphisms are preserved by retracts.

  4. MC4

Consider a commutative square in $\mathcal{M}$ consisting of the solid
morphisms:

${X}$${E}$${Y}$${B}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$

Then a morphism, depicted by the dotted arrow, making this diagram
commutative, exists under either of the following two assumptions: (i)
$\alpha$ is a cofibration and a weak equivalence and $\beta$ is a fibration,
or (ii) $\alpha$ is a cofibration and $\beta$ is a fibration and a weak
equivalence.

  5. MC5

Every morphism in $\mathcal{M}$ can be factored in two ways: (i)
$\beta\alpha$, where $\alpha$ is a cofibration and $\beta$ is a fibration and
a weak equivalence, and (ii) $\beta\alpha$, where $\alpha$ is a cofibration
and a weak equivalence and $\beta$ is a fibration.

In particular, the axiom MC1 guarantees the existence of the initial object
$\text{colim}_{\emptyset}F$, denoted by $e$, and of the terminal object
$\text{lim}_{\emptyset}F$, denoted by $\ast$. An object $X$ in $\mathcal{M}$
is called cofibrant if the morphism $e\to X$ is a cofibration. If the morphism
$X\to\ast$ is a fibration, then $X$ is called fibrant.

Assume $I$ is a finite poset. Then the following choices of weak equivalences,
fibrations, and cofibrations in $\text{Fun}(I,\mathcal{M})$ is a model
structure (see for example [23, 14]). A natural transformation $\varphi:F\to
G$ in $\text{Fun}(I,\mathcal{M})$ is:

  * •

a weak equivalence (resp. a fibration) if, for all $a$ in $I$, the morphism
$\varphi_{a}:F(a)\to G(a)$ is a weak equivalence (resp. a fibration) in
$\mathcal{M}$;

  * •

a cofibration if, for all $a$ in $I$, the morphism

$\operatorname{colim}\left(\operatorname{colim}_{I<a}G\leftarrow\operatorname{colim}_{I<a}F\rightarrow
F(a)\right)\to G(a),$

induced by the following commutative diagram, is a cofibration in
$\mathcal{M}$:

${\operatorname{colim}_{I<a}F}$${F(a)}$${\operatorname{colim}_{I<a}G}$${G(a)}$$\scriptstyle{\operatorname{colim}_{I<a}\varphi}$$\scriptstyle{\varphi_{a}}$

The described model structures on functors indexed by finite posets are
compatible in the following sense. Let $f\colon I_{0}\to I_{1}$ be a functor
between finite posets. If $\phi\colon F\to G$ is a cofibration (resp. a
cofibration and a weak equivalence) in $\text{\rm Fun}(I_{0},\mathcal{M})$,
then so is $f^{k}\phi\colon f^{k}F\to f^{k}G$ in $\text{\rm
Fun}(I_{1},\mathcal{M})$. This is a consequence of the universal property of
left Kan extensions and the axiom (MC4). However, left Kan extensions in
general fail to preserve weak equivalences and fibrations (compare with
Proposition 9.4).

To verify if a natural transformation in $\text{Fun}(I,\mathcal{M})$ is a
cofibration, we need to perform colimits over subposets $(I<a)\subset I$. In
general, such subposets can be large and constructing colimits over them may
require performing a lot of identifications. When $I$ is an upper semilattice
however, we can be more efficient. For $a$ in a finite upper semilattice $I$,
define a sublattice:

$I_{a}:=\\{\text{$\bigwedge$}S\ |\ S\subset\mathcal{P}(a)\text{ has an
ancestor}\\}\subset(I<a).$

For all $b$ in $I<a$, the set $b\leq\mathcal{P}(a)$ is non-empty and
$b\leq\bigwedge(b\leq\mathcal{P}(a))$. The element
$\bigwedge(b\leq\mathcal{P}(a))$ is therefore the initial object in $b\leq
I_{a}$ and consequently $b\leq I_{a}$ is contractible (see [5]). As this
happens for all $b$ in $I<a$, the inclusion $I_{a}\subset(I<a)$ is cofinal
(see [18, 5]), which proves:

######  9.1 Proposition.

Let $I$ be a finite upper semilattice and $a$ its element. For every functor
$F\colon(I<a)\to\mathcal{M}$, the morphism $\text{\rm
colim}_{I_{a}}F\to\text{\rm colim}_{I<a}F$ is an isomorphism and $\text{\rm
hocolim}_{I_{a}}F\to\text{\rm hocolim}_{I<a}F$ is a weak equivalence.

According to Proposition 9.1, when $I$ is a finite upper semilattice, to
verify if a natural transformation in $\text{Fun}(I,\mathcal{M})$ is a
cofibration, we need only to perform colimits over the subposets $I_{a}\subset
I$ for $a$ in $I$. How can such colimits be calculated? One way is to
consider, for $a$ in $I$, the subposet of the the discrete cube (see 2.1):

$C_{a}:=\\{S\subset\mathcal{P}(a)\ |\ \mathcal{P}(a)\setminus S\text{ has an
ancestor}\\}\subset 2^{\mathcal{P}(a)}$

and the functor $\bigwedge^{c}\colon C_{a}\to I_{a}$ mapping $S$ to
$\bigwedge(\mathcal{P}(a)\setminus S)$. Since, for every $b$ in $I_{a}$, the
subset $\mathcal{P}(a)\setminus(b\leq\mathcal{P}(a))\subset\mathcal{P}(a)$ is
the initial object in the poset $b\leq\bigwedge^{c}$ (see 2.7), this poset is
contractible. The functor $\bigwedge^{c}$ is therefore cofinal. Thus, for
every $F\colon I_{a}\to\mathcal{M}$, the morphism
$\text{colim}_{C_{a}}(F\bigwedge^{c})\to\text{colim}_{I_{a}}F$ is an
isomorphism and
$\text{hocolim}_{C_{a}}(F\bigwedge^{c})\to\text{hocolim}_{I_{a}}F$ is a weak
equivalence.

The assumption on the indexing poset being an upper semilattice is not only
helpful in verifying if a natural tranformation is a cofibration. It is also
crucial in proving that the following choices of weak equivalences,
fibrations, and cofibrations in $\text{Tame}(J,\mathcal{M})$ satisfy the
axioms of a model structure.

######  9.2 Definition.

Let $J$ be an upper semilattice and $\mathcal{M}$ a model category. A natural
transformation $\phi\colon F\to G$ in $\text{Tame}(J,\mathcal{M})$ is called:

  * •

a weak equivalence (resp. fibration) if, for all $a$ in $J$, the morphism
$\phi_{a}\colon F(a)\to G(a)$ is a weak equivalence (resp. fibration) in
$\mathcal{M}$;

  * •

a cofibration if there is a finite subposet inclusion $f\colon I\subset J$
discretising $F$ and $G$, and for which $\phi^{f}\colon Ff\to Gf$ is a
cofibration in $\text{Fun}(I,\mathcal{M})$.

######  9.3 Theorem.

Let $J$ be an upper semilattice and $\mathcal{M}$ a model category. The
choices described in Definition 9.2 satisfy the axioms of a model structure on
$\text{\rm Tame}(J,\mathcal{M})$.

Before we prove Theorem 9.3, we first show:

######  9.4 Proposition.

Let $\mathcal{M}$ be a model category and $f\colon I\to J$ a homomorphism of
finite type between upper semilattices. If a natural transformation
$\phi\colon F\to G$ in $\text{\rm Fun}(I,\mathcal{M})$ is a weak equivalence
(resp. fibration), then so is its left Kan extension $f^{k}\phi\colon
f^{k}F\to f^{k}G$ in $\text{\rm Fun}(J,\mathcal{M})$.

###### Proof.

Let $\phi\colon F\to G$ in $\text{\rm Fun}(I,\mathcal{M})$ be a weak
equivalence (resp. fibration). Since $f$ is a finite-type homomorphism between
upper semilattices, the morphism $(f^{k}\phi)_{a}\colon f^{k}F(a)\to
f^{k}G(a)$ is isomorphic to either $\text{id}\colon e\to e$, if ($f\leq
a)=\emptyset$, or to $\phi_{\bigvee_{I}(f\leq a)}\colon F(\bigvee_{I}(f\leq
a))\to G(\bigvee_{I}(f\leq a))$, if $(f\leq a)\not=\emptyset$ (see 7.6). As
these morphisms are weak equivalences (resp. fibrations), for all $a$, then so
is the left Kan extension $f^{k}\phi$. ∎

###### Proof of Theorem 9.3.

Requirement MC1 follows from Proposition 8.5. MC2 and MC3 are clear since
$\mathcal{M}$ satisfies them.

MC4: Let $\alpha\colon F\hookrightarrow G$ be a cofibration in $\text{\rm
Tame}(J,\mathcal{M})$. Choose a finite sublattice $I\subset J$ and functor
$f:I\to J$ that discretises both $F$ and $G$ and for which $\alpha^{f}\colon
Ff\to Gf$ is a cofibration in $\text{\rm Fun}(I,\mathcal{M})$. Assume $\alpha$
is part of a commutative square in $\text{\rm Tame}(J,\mathcal{M})$ depicted
in Figure 6 on the left, where $\beta$ is a fibration. By applying $(-)^{f}$
to this square we get a commutative square in $\text{\rm Fun}(I,\mathcal{M})$,
depicted by the solid arrows square on the right in Figure 6, where as
indicted the natural transformations are a cofibration and a fibration in
$\text{\rm Fun}(I,\mathcal{M})$. If in addition either $\alpha$ or $\beta$ is
a weak equivalence, then the lift, depicted by the dotted arrow in the right
of Figure 6, exists since $\text{\rm Fun}(I,\mathcal{M})$ is a model category.

${F}$${E}$${G}$${B}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$
${Ff}$${Ef}$${Gf}$${Bf}$$\scriptstyle{\alpha^{f}}$$\scriptstyle{\beta^{f}}$$\scriptstyle{s}$
Figure 6.

By applying the left Kan extension $f^{k}$ to the square on the right in
Figure 6 and comparing the result to the original square on the left in Figure
6 we can form the commutative $3$-dimensional cube in Figure 7 in $\text{\rm
Tame}(J,\mathcal{M})$.

${F}$${E}$${f^{k}(Ff)}$${f^{k}(Ef)}$${G}$${B}$${f^{k}(Gf)}$${f^{k}(Bf)}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$$\scriptstyle{(f^{k}\alpha)^{f}}$$\scriptstyle{f^{k}s}$$\scriptstyle{(f^{k}\beta)^{f}}$
Figure 7.

Since the natural transformation $f^{k}(Gf)\to G$ is an isomorphism, a desired
lift exists in the right square of Figure 6 under the additional assumption
that either $\alpha$ or $\beta$ is a weak equivalence.

MC5: Let $\phi\colon F\to G$ be a natural transformation in $\text{\rm
Tame}(J,\mathcal{M})$. Choose a finite sublattice $f\colon I\subset J$ that
discretises both $F$ and $G$. Factor the natural transformation
$\phi^{f}\colon Ff\to Gf$ in $\text{\rm Fun}(I,\mathcal{M})$ as
$\phi^{f}=\beta\alpha$ where $\alpha\colon Ff\to H$ is a cofibration,
$\beta\colon H\to Gf$ is a fibration, and either $\alpha$ or $\beta$ is a weak
equivalence. By applying the left Kan extension $f^{k}$ to these
factorisations we obtain a commutative diagram in $\text{\rm
Tame}(J,\mathcal{M})$ where the vertical natural transformations are
isomorphisms:

${f^{k}(Ff)}$${f^{k}H}$${f^{k}(Gf)}$${F}$${G}$$\scriptstyle{f^{k}\alpha}$$\scriptstyle{f^{k}\beta}$$\scriptstyle{\phi}$

Since $(f^{k}\alpha)^{f}$ is isomorphic to $\alpha$, it is a cofibration in
$\text{\rm Fun}(I,\mathcal{M})$ and consequently $f^{k}\alpha$ is a
cofibration in $\text{\rm Tame}(J,\mathcal{M})$. According to Proposition 9.4,
$f^{k}\beta$ is a fibration in $\text{\rm Tame}(J,\mathcal{M})$. The same
proposition assures also that if $\alpha$ (resp. $\beta$) is a weak
equivalence, then so is $f^{k}\alpha$ (resp. $f^{k}\beta$). This gives the
desired factorisation of $\phi$. ∎

##  10\. Betti diagrams of vector spaces valued functors.

In this section we are going to discuss a standard strategy of retrieving
Betti diagrams of functors indexed by posets with values in the category
$\text{vect}_{K}$ of finite dimensional $K$-vector spaces, where $K$ is a
chosen field.

If $J$ is a finite poset, then the language of model categories can be used
for phrasing homological properties of functors in
$\text{Fun}(J,\text{vect}_{K})$. This is because, in this case, the category
$\text{ch}(\text{Fun}(J,\text{vect}_{K}))$ of non-negative chain complexes of
such functors, which can be identified with
$\text{Fun}(J,\text{ch}(\text{vect}_{K}))$, has a natural model structure
where weak equivalences are given by the homology isomorphisms. See Section 9
for a recollection of how such a model structure can be obtained. It is
however unlikely that the category $\text{Fun}(J,\text{ch}(\text{vect}_{K}))$
has a natural model structure with same weak equivalences for a general
infinite $J$. The reason is the restriction to functors with values in finite
dimensional $K$ vector spaces which might prevent the existence of the
required factorisations. This restriction however is important as our primary
interest is in circumstances when Betti diagrams can be defined and
calculated. For that the finite dimensionality assumption is essential.

In Section 9 (see 9.3), it was also explained how the model structures on
$\text{Fun}(I,\text{ch}(\text{vect}_{K}))$, for finite $I$, can be extended to
$\text{Tame}(J,\text{ch}(\text{vect}_{K}))$ where $J$ is an arbitrary upper
semilattice. However, if $J$ is not finite and not an upper semilattice, then
$\text{Tame}(J,\text{ch}(\text{vect}_{K}))$ may even fail to be closed under
finite limits, preventing for example the existence of resolutions.

This means that for $J$ which is not finite, care needs to be exercised in
order to be able to consider resolutions and Betti diagrams of functors
indexed by $J$. In this section we recall necessary foundations for defining
Betti diagrams, and also present a strategy based on Koszul complexes to
calculate them.

######  10.1Freeness.

Let $J$ be a poset with the poset relation denoted by $\leq$. Consider the
poset $(J,=)$ with the trivial poset relation on the set $J$, where two
elements are related if and only if they are equal. The identity function
$\iota\colon(J,=)\to J$, mapping $a$ to $\iota(a)=a$, is a functor. A functor
$V\colon(J,=)\to\text{vect}_{K}$ is just a sequence $\\{V_{a}\\}_{a\in J}$ of
finite dimensional $K$-vector spaces. The set $\\{a\in J\ |\ V_{a}\not=0\\}$
is called the support of $V$ and is denoted by $\text{supp}(V)$.

A functor $F\colon J\to\text{vect}_{K}$ is called free finitely generated  if
it is isomorphic to the left Kan extension along $\iota\colon(J,=)\to J$ of
some $V=\\{V_{a}\\}_{a\in J}$ whose support is finite. Since all free functors
considered in this article are finitely generated, we use the term free,
without mentioning finite generation, to describe such functors. The name free
is justified by the universal property of the left Kan extension, which gives
a linear isomorphism between $\text{Nat}_{J}(\iota^{k}V,H)$ and $\prod_{a\in
J}\text{Hom}(V_{a},H(a))$, for every functor $H\colon J\to\text{vect}_{K}$.
For example, for $b$ in $J$ and a finite dimensional vector space $U$,
consider the simple functor $U[b]\colon J\to\text{vect}_{K}$:

$U[b](a):=\begin{cases}0&\text{ if }a\not=b\\\ U&\text{ if }a=b\end{cases}$

Then the vector spaces $\text{Nat}_{J}(\iota^{k}V,U[b])$ and
$\text{Hom}(V_{b},U)$ are isomorphic. By varing $b$ in $J$ and taking $U$ to
be $K$, we can conclude that free functors $\iota^{k}V$ and $\iota^{k}W$ are
isomorphic if and only if $V_{a}$ and $W_{a}$ are isomorphic for all $a$ in
$J$. Thus, a free functor $F$ determines a unique sequence $\beta F=\\{(\beta
F)_{a}\\}_{a\in J}$ of vector spaces whose support is finite, called the Betti
diagram of $F$, for which $F$ is isomorphic to the left Kan extension
$\iota^{k}(\beta F)$. If $\text{supp}(\beta F)=\\{a\\}$, then $F$ is called
homogeneous and is also denoted by the symbol $F(a)[a,-)$. The restriction of
$F(a)[a,-)$ to the subposet $(a\leq J)\subset J$ is isomorphic to the constant
functor with value $F(a)$. Its restriction to $\\{x\in J\ |\ a\not\leq x\\}$
is isomorphic to the constant functor with value $0$. If $F$ is free, then it
is isomorphic to the direct sum $\oplus_{a\in J}(\beta F)_{a}[a,-)$ and,
consequently, $F$ is isomorphic to the left Kan extension of $\\{(\beta
F)_{a}\\}_{a\in\text{supp}(\beta F)}$ along $(\text{supp}(\beta
F),=)\hookrightarrow J$. Thus, every free functor $F$ is tame and is
discretised by the subposet inclusion $\text{supp}(\beta F)\subset J$.
Moreover every collection $V=\\{V_{a}\\}_{a\in J}$ with finite support is the
Betti diagram of a free functor.

If $G\colon I\to\text{vect}_{K}$ is free, then, for every functor $f\colon
I\to J$, the left Kan extension $f^{k}G\colon J\to\text{vect}_{K}$ is also
free and $(\beta f^{k}G)_{a}=0$ if $f^{-1}(a)$ is empty, and $(\beta
f^{k}G)_{a}$ is isomorphic to $\oplus_{b\in f^{-1}(a)}(\beta G)_{b}$, if
$f^{-1}(a)$ is non-empty.

######  10.2Resolutions.

Let $J$ be a poset and $F\colon J\to\text{vect}_{K}$ a functor. An exact
sequence $P_{n-1}\to\cdots\to P_{0}\to F\to 0$ in
$\text{Fun}(J,\text{vect}_{K})$, with $P_{i}$ free for all $i$, is called an
$n$-resolution of $F$. A $1$-resolution $P_{0}\to F\to 0$ is also called a
cover of $F$. A $2$-resolution $P_{1}\to P_{0}\to F\to 0$ is also called a
presentation of $F$. An infinite exact sequence $\cdots\to P_{1}\to P_{0}\to
F\to 0$, with $P_{i}$ free for all $i$, is called an $\infty$-resolution of
$F$. An $n$-resolution of $F$ is also denoted, as a map of chain complexes, by
$P\to F$, where $F$ is a chain complex concentrated in degree $0$, and either
$P=(P_{n-1}\to\cdots\to P_{0})$ or $P=(\cdots\to P_{1}\to P_{0})$, depending
if $n$ is finite.

A functor $F\colon J\to\text{vect}_{K}$ is called $n$-resolvable if it has an
$n$-resolution. Functors which are $1$-resolvable are also called finitely
generated. Functors which are $2$-resolvable are also called finitely
presented. If the indexing poset $J$ is finite, then all functors are
$\infty$-resolvable. This is because all such functors are finitely generated,
and hence by taking covers of successive kernels, an $\infty$-resolution can
be constructed.

If $J$ is infinite, then not all $n$-resolvable functors have to be
$(n+1)$-resolvable. For example, let $J=[0,2)\coprod\\{a,b\\}\coprod(2,3]$,
with $a$ and $b$ incomparable and $x<a<y$ and $x<b<y$, for $x$ in $[0,2)$ and
$y$ in $(2,3]$. Consider the subposet $I:=\\{1,a,b\\}\subset J$ and a functor
$G\colon I\to\text{\rm Vect}_{K}$ where both $G(1<a)$ and $G(1<b)$ are given
by the function $K\to 0$. Let $F\colon J\to\text{\rm Vect}_{K}$ be the left
Kan extension of $G$ along $I\subset J$. Then $F$ is finitely presented
($2$-resolvable), however it is not $3$-resolvable.

Here is a characterisation of finitely generated and presented functors.

######  10.3 Proposition.

Let $J$ be a poset and $F\colon J\to\text{\rm vect}_{K}$ a functor.

  1. (1)

$F$ is finitely generated if and only if there is a finite poset $I$ and a
functor $f\colon I\to J$ for which the natural transformation $\mu\colon
f^{k}(Ff)\to F$, adjoint to the identity $\text{\rm id}\colon Ff\to Ff$, is
surjective.

  2. (2)

$F$ is finitely presented if and only if it is tame, i.e., if and only if
there is a finite poset $I$ and a functor $f\colon I\to J$ for which the
natural transformation $\mu\colon f^{k}(Ff)\to F$, adjoint to the identity
$\text{\rm id}\colon Ff\to Ff$, is an isomorphism.

###### Proof.

(1): Let $\pi\colon P_{0}\to F$ be a cover. Since $P_{0}$ is tame, there is a
functor $f\colon I\to J$, with finite $I$, discretising $P_{0}$ (for example
$I=\text{supp}(\beta P_{0})\subset J$). The commutativity of the following
square and the fact that left Kan extensions preserve surjections imply the
surjectivity of $\mu$:

${f^{k}(P_{0}f)}$${P_{0}}$${f^{k}(Ff)}$${F}$$\scriptstyle{f^{k}(\pi^{f})}$$\scriptstyle{\pi}$$\scriptstyle{\mu}$

Let $I$ be a finite poset, $f\colon I\to J$ a functor for which $\mu\colon
f^{k}(Ff)\to F$ is surjective, and $\pi\colon P_{0}\to Ff$ a cover. The
surjectivity of $\mu$ implies the surjectivity of the following composition,
which is then a cover of $F$:

${f^{k}(P_{0})}$${f^{k}(Ff)}$${F}$$\scriptstyle{f^{k}\pi}$$\scriptstyle{\mu}$

(2): If $P_{1}\to P_{0}\to F\to 0$ is a $2$-resolution, then
$\text{colim}(0\leftarrow P_{1}\to P_{0})$ is isomorphic to $F$. Since
tameness is preserved by finite colimits (see 8.2.(4)), $F$ is tame.

Assume $F$ is discretised by a subposet inclusion $f\colon I\subset J$ with
finite $I$. Consider a $2$-resolution $P_{1}\to P_{0}\to Ff\to 0$, which
exists since $I$ is finite. As before, $f^{k}(Ff)$, and hence $F$, is
isomorphic to $\text{colim}(0\leftarrow f^{k}P_{1}\to f^{k}P_{0})$. Since
$f^{k}P_{1}$ and $f^{k}P_{0}$ are free, $F$ is finitely presented. ∎

Proposition 10.3 characterises $n$-resolvable functors for $n\leq 2$. We do
not have a similar characterisation for $n>2$, only a partial result:

######  10.4 Proposition.

Let $J$ be a poset. Assume $F\colon J\to\text{\rm vect}_{K}$ is discretised by
a functor $f\colon I\to J$ out of a finite poset $I$ and for which
$f^{k}\colon\text{\rm Fun}(I,\text{\rm vect}_{K})\to\text{\rm Fun}(J,\text{\rm
vect}_{K})$ is exact. Then $F$ is $\infty$-resolvable.

###### Proof.

Let $G\colon I\to\text{\rm vect}_{K}$ be a functor for which $f^{k}G$ is
isomorphic to $F$. Choose an $\infty$-resolution $\pi\colon P\to G$. Exactness
of $f^{k}$ means that the left Kan extension $f^{k}\pi\colon f^{k}P\to f^{k}G$
is an $\infty$-resolution $f^{k}G$. ∎

Here is a condition guaranteeing exactness of the left Kan extension:

######  10.5 Lemma.

Assume $f\colon I\to J$ is a functor of posets with finite $I$ satisfying the
following property: for every $a$ in $J$, every pair of elements in $f\leq a$
that have an ancestor also has a descendent (such posets are called weakly
directed in [3]). Then $f^{k}\colon\text{\rm Fun}(I,\text{\rm
vect}_{K})\to\text{\rm Fun}(J,\text{\rm vect}_{K})$ is exact.

###### Proof.

We show $\text{colim}_{f\leq a}\colon\text{\rm Fun}(f\leq a,\text{\rm
vect}_{K})\to\text{\rm vect}_{K}$ is exact for all $a$ in $J$. Let
$M\subset(f\leq a)$ consists of all the maximal elements. The assumption on
$f$ implies $M\subset(f\leq a)$ is cofinal and hence $\text{colim}_{f\leq a}F$
is isomorphic to $\bigoplus_{x\in M}F(x)$. The lemma follows from the
exactness of direct sums. ∎

######  10.6 Corollary.

Let $J$ be a poset.

  1. (1)

Every functor $F\colon J\to\text{\rm vect}_{K}$ discretised by a homomorphism
$f\colon I\to J$ from a finite upper semilattice $I$ is $\infty$-resolvable.

  2. (2)

If $J$ is an upper semilattice, then a functor $F\colon J\to\text{\rm
vect}_{K}$ is tame if and only if it is $\infty$-resolvable.

###### Proof.

In both cases (1) and (2) the poset $f\leq a$, for all $a$ in $J$, has a
terminal object given by its coproduct in $I$. Thus the assumption of Lemma
10.5 is satisfied and hence the conclusion of 10.4 holds. ∎

######  10.7Minimality and Betti diagrams.

An $n$-resolution $\pi\colon P\to F$ is called minimal if every chain map
$\phi\colon P\to P,$ for which the following triangle commutes, is an
isomorphism:

${P}$${P}$${F}$$\scriptstyle{\phi}$

If $P\to F$ and $Q\to F$ are minimal $n$-resolutions of $F$, then $P$ and $Q$
are isomorphic. Thus, if $P\to F$ is a minimal $n$-resolution, then the
isomorphism type of $P_{i}$, for $i<n$, is uniquely determined by the
isomorphism type of $F$, and its Betti diagram $\beta P_{i}$ is called the
$i$-th Betti diagram of $F$, and is denoted by $\beta^{i}F$. For example if
$F$ is free, then $\text{id}\colon F\to F$ is a minimal $\infty$-resolution of
$F$ and, hence, for every $a$ in $J$, the vector space $(\beta^{0}F)_{a}$ is
isomorphic to $(\beta F)_{a}$, and $(\beta^{i}F)_{a}=0$ for $i>0$.

A minimal $1$-resolution $P_{0}\to F$ is also called a minimal cover of $F$.
An $n$-resolution $P\to F$ is minimal if and only if, for every $i<n$, the
following natural transformations are minimal covers: $P_{0}\to F$,
$P_{1}\to\text{ker}(P_{0}\to F)$, …, $P_{i}\to\text{ker}(P_{i-1}\to P_{i-2})$.
Thus, if $P_{0}\to F$ is a minimal cover of an $n$-resolvable functor, then,
for $1\leq i<n$, $\beta^{i}F$ is isomorphic to $\beta^{i-1}\text{ker}(P_{0}\to
F)$. Also, if $P\to F$ is a minimal $n$-resolution, then $\beta^{0}F$ is
isomorphic to $\beta P_{0}$, $\beta^{1}F$ is isomorphic to
$\beta^{0}\text{ker}(P_{0}\to F)$, and for $i>1$, $\beta^{i}F$ is isomorphic
to $\beta^{0}\text{ker}(P_{i-1}\to P_{i-2})$.

Our strategy for constructing minimal resolutions of $n$-resolvable functors
indexed by an arbitrary poset is to reduce this problem to the case when the
indexing poset is finite.

######  10.8 Proposition.

Let $J$ be a poset and $F\colon J\to\text{\rm vect}_{K}$ a functor. Assume
$P\to F$ is an $n$-resolution which is discretised by a subposet inclusion
$f\colon I\subset J$ with finite $I$. If $Q\to Ff$ is a minimal $n$-resolution
of $Ff$, then its adjoint $f^{k}Q\to F$ is a minimal resolution of $F$ and,
for $i<n$ and $a$ in $J$:

$(\beta^{i}F)_{a}\text{ is isomorphic to
}\begin{cases}(\beta^{i}(Ff))_{a}=(\beta Q_{i})_{a}&\text{ if }a\in I\\\
0&\text{ if }a\not\in I\end{cases}$

###### Proof.

By restricting the $n$-resolution $P\to F$ of $F$ along $f$, we obtain a
resolution $Pf\to Ff$ of $Ff$. Let $Q\to Ff$ be a minimal $n$-resolution. Then
there are chain maps $\phi\colon Q\to Pf$ and $\psi\colon Pf\to Q$ for which
the composition $\psi\phi\colon Q\to Q$ is an isomorphism and the diagram on
the left commutes:

${Q}$${Pf}$${Q}$${Ff}$$\scriptstyle{\phi}$$\scriptstyle{\psi}$
${f^{k}Q}$${f^{k}(Pf)}$${f^{k}Q}$${F}$$\scriptstyle{\phi}$$\scriptstyle{\psi}$

By taking the adjoints of the vertical maps in the left diagram, we obtain a
commutative diagram on the right. Since $f^{k}(Pf)\to F$ is an $n$-resolution,
then so is its retract $f^{k}Q\to F$. Its minimality follows from the
minimality of $Q\to Ff$ and the fact that
$f^{k}\colon\text{Nat}_{I}(Q_{i},Q_{i})\to\text{Nat}_{J}(f^{k}Q_{i},f^{k}Q_{i})$
is a bijection for every $i$ (see 2.12). ∎

According to Proposition 10.8, to construct a minimal $n$-resolution of a
functor $F\colon J\to\text{\rm vect}_{K}$, the first step is to find a finite
subposet inclusion $f\colon I\subset J$ for which there is an $n$-resolution
$P\to F$ with $P$ being discretised by $f$ (the natural transformation
$f^{k}(Pf)\to P$ is an isomorphism). For $n=1$, according to the proof of
10.3.(1), such a subposet inclusion is given by any $f\colon I\subset J$ for
which $f^{k}(Ff)\to F$ is a surjection. For $n=2$, according to the proof of
10.3.(2), such a subposet inclusion is given by any $f\colon I\subset J$ that
discretises $F$. We do not have a similar statement for $n>2$.

The second step is to construct a minimal $n$-resolution $Q\to Ff$ of the
restriction $Ff$. The adjoint of this minimal resolution $f^{k}Q\to F$ is then
the desired minimal $n$-resolution of $F$. This process reduces finding a
minimal $n$-resolution of $F$ to finding a minimal $n$-resolution of $Ff$
which is a functor indexed by a finite poset. Constructing minimal resolutions
of functors indexed by finite posets is standard and involves radicals (see
for example [2, 9]).

######  10.9Radical.

Let $I$ be a finite poset. The radical of $G\colon I\to\text{\rm vect}_{K}$ is
a subfunctor $\text{rad}(G)\subset G$ given by
$\text{rad}(G)(a)=\sum_{s\in(I<a)}\text{im}(G(s<a))$ for $a$ in $I$. The
quotient functor $G/\text{rad}(G)$ is semisimple as it is isomorphic to a
direct sum $\oplus_{a\in I}U_{a}[a]$ of simple functors (see 10.1), where
$U_{a}:=(G/\text{rad}(G))(a)$. For example, for a free functor $G=\oplus_{a\in
I}(\beta G)_{a}[a,-)$, the quotient $G/\text{rad}(G)$ is isomorphic to
$\oplus_{a\in I}(\beta G)_{a}[a]$.

A key property of the quotienting by the radical, when the indexing poset is
finite, is the surjectivity detection: a natural transformation $H\to G$ is
surjective if and only if its composition with the quotient $G\to
G/\text{rad}(G)$ is surjective. The surjectivity detection may fail for
infinite posets.

The surjectivity detection can be used to construct minimal covers. Consider
the quotient $G/\text{rad}(G)$ of $G\colon I\to\text{\rm vect}_{K}$. Set
$P_{0}:=\bigoplus_{a\in I}U_{a}[a,-)$, where $U_{a}=(G/\text{rad}(G))(a)$.
Note that there is an isomorphism $P_{0}/\text{rad}(P_{0})\to
G/\text{rad}(G)$. Let $\pi\colon P_{0}\to G$ be any natural transformation
fitting into the following commutative square, where the bottom horizontal
arrow represent the chosen isomorphism:

${P_{0}}$${G}$${P_{0}/\text{rad}(P_{0})}$${G/\text{rad}(G)}$$\scriptstyle{\pi}$

Such $\pi$ exists since $P_{0}$ is free. The composition $P_{0}\to
G/\text{rad}(G)$ is surjective, and hence so is $\pi$. The same argument gives
the surjectivity of every $\phi\colon P_{0}\to P_{0}$ for which $\pi\phi=\pi$.
Since the values of $P_{0}$ are finite dimensional, every such $\phi$ is
therefore an isomorphism, and hence $\pi\colon P_{0}\to G$ is a minimal cover.
This shows:

######  10.10 Proposition.

Let $I$ be a finite poset and $G\colon I\to\text{\rm vect}_{K}$ a functor. A
natural transformation $P_{0}\to G$ is a minimal cover if and only if $P_{0}$
is free and the induced natural transformation $P_{0}/\text{\rm rad}(P_{0})\to
G/\text{\rm rad}(G)$ is an isomorphism. Moreover $(\beta^{0}G)_{a}$ is
isomorphic to $(G/\text{\rm rad}(G))(a)$ for all $a$.

Since all the functors indexed by a finite poset $I$ are finitely generated,
by taking minimal covers of successive kernels, every functor indexed by $I$
admits a minimal $\infty$-resolution. This inductive step-wise construction of
a minimal $\infty$-resolution can be used for a step wise inductive procedure
of calculating the Betti diagrams. Can the Betti diagrams be retrieved in one
step without the need of an inductive procedure? For this purpose Koszul
complexes are standardly used.

######  10.11.

Let $I$ be a finite poset and $a$ its element. Choose a linear ordering
$\prec$ on the set of parents $\mathcal{P}(a)$ of $a$. For every functor
$G\colon I\to\text{vect}_{K}$, we define a non-negative chain complex denoted
by $\mathcal{K}_{a}G$ and called the Koszul complex of $G$ at $a$. Let $k$ be
a natural number. Define:

$(\mathcal{K}_{a}G)_{k}:=\begin{cases}G(a)&\text{ if }k=0\\\
\displaystyle\bigoplus_{\begin{subarray}{c}S\subset\mathcal{P}(a),\ |S|=k\\\
S\text{ has an
ancestor}\end{subarray}}\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(I\leq
s)\end{subarray}}G&\text{ if }k>0\end{cases}$

For example $(\mathcal{K}_{a}G)_{1}=\bigoplus_{s\in\mathcal{P}(a)}G(s)$ and
$(\mathcal{K}_{a}G)_{k}=0$ if $k>\text{par-dim}_{I}(a)$ (see 3.5).

Define $\partial\colon(\mathcal{K}_{a}G)_{k+1}\to(\mathcal{K}_{a}G)_{k}$ as
follows:

  * •

If $k=0$, then
$\partial\colon(\mathcal{K}_{a}G)_{1}\to(\mathcal{K}_{a}G)_{0}=G(a)$ is the
linear function which on the summand $G(s)$ in $(\mathcal{K}_{a}G)_{1}$,
indexed by $s$ in $\mathcal{P}(a)$, is given by $G(s<a)$.

  * •

Let $k>0$. For $k\geq j\geq 0$, let
$\partial_{j}\colon(\mathcal{K}_{a}G)_{k+1}\to(\mathcal{K}_{a}G)_{k}$ be the
linear function mapping the summand
$\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(I\leq s)\end{subarray}}G$ in
$(\mathcal{K}_{a}G)_{k+1}$, indexed by $S=\\{s_{0}\prec\cdots\prec
s_{k}\\}\subset\mathcal{P}(a)$, to the summand
$\text{colim}_{\begin{subarray}{c}\cap_{s\in S\setminus\\{s_{j}\\}}(I\leq
s)\end{subarray}}G$ in $(\mathcal{K}_{a}G)_{k}$, indexed by
$S\setminus\\{s_{j}\\}\subset\mathcal{P}(a)$, via the function of the colimits
$\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(I\leq
s)\end{subarray}}G\to\text{colim}_{\begin{subarray}{c}\cap_{s\in
S\setminus\\{s_{j}\\}}(I\leq s)\end{subarray}}G$ induced by the poset
inclusion $\cap_{s\in S}(I\leq s)\subset\cap_{s\in
S\setminus\\{s_{j}\\}}(I\leq s)$. Define
$\partial\colon(\mathcal{K}_{a}G)_{k+1}\to(\mathcal{K}_{a}G)_{k}$ to be the
alternating sum $\partial=\sum_{j=0}^{k}(-1)^{j}\partial_{j}$.

The linear functions $\partial$ form a chain complex as it is standard to
verify that composition of two consecutive such functions is the $0$ function.

For a natural transformation $\phi\colon F\to G$, define:

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{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}:=\begin{cases}\phi_{a}&\text{
if }k=0\\\ \displaystyle\bigoplus_{\begin{subarray}{c}S\subset\mathcal{P}(a),\
|S|=k\\\ S\text{ has an
ancestor}\end{subarray}}\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(I\leq
s)\end{subarray}}\phi&\text{ if }k>0\end{cases}$

These linear functions, for all $k$, form a chain map denoted by
$\mathcal{K}_{a}\phi\colon\mathcal{K}_{a}F\to\mathcal{K}_{a}G$. The
association $\phi\mapsto\mathcal{K}_{a}\phi$ is a functor.

The image of the differential
$\partial\colon(\mathcal{K}_{a}G)_{1}\to(\mathcal{K}_{a}G)_{1}=G(a)$ coincides
with $\text{rad}(G)(a)=\sum_{s\in\mathcal{P}(a)}\text{im}(G(s<a))$, and
consequently, according to 10.10, the vector spaces $H_{0}(\mathcal{K}_{a}G)$,
$(G/\text{rad}(G))(a)$, and $(\beta^{0}G)_{a}$ are isomorphic.

######  10.12.

Let an element $a$ in a finite poset $I$ have the following property: every
subset $S\subset\mathcal{P}(a)$ which has an ancestor has the product
$\bigwedge S$ in $I$. For example, if $I$ is an upper semilattice, then all
its elements satisfy this property. Under this assumption, for every subset
$S\subset\mathcal{P}(a)$ that has an ancestor, the product $\bigwedge S$ is
the terminal object in the category $\cap_{s\in S}(I\leq s)$ and consequently,
for $k>0$,

$(\mathcal{K}_{a}G)_{k}=\displaystyle\bigoplus_{\begin{subarray}{c}S\subset\mathcal{P}(a),\
|S|=k\\\ S\text{ has an ancestor}\end{subarray}}G(\bigwedge S)$

######  10.13.

Let $I$ be a finite poset and $a$ its element. Since colimits commute with
direct sums, so does the functor $\mathcal{K}_{a}$, i.e., the natural
transformation
$\mathcal{K}_{a}F\oplus\mathcal{K}_{a}G\to\mathcal{K}_{a}(F\oplus G)$ is an
isomorphism.

As the colimit operation is right exact, then so is our Koszul complex
construction: if $0\to F\to G\to H\to 0$ is an exact sequence in
$\text{Fun}(I,\text{vect}_{K})$, then
$\mathcal{K}_{a}F\to\mathcal{K}_{a}G\to\mathcal{K}_{a}H\to 0$ is an exact
sequence of chain complexes. In general colimits do not preserve
monomorphisms, and hence one does not expect the Koszul complex construction
to preserve monomorphisms in general either. Let $n$ be an extended positive
natural number (containing $\infty$). An element $a$ in $I$ is called Koszul
$n$-exact if, for every exact sequence of functors $0\to F\to G\to H\to 0$ and
$k<n$, the following sequence of vector spaces is exact:

$0\to(\mathcal{K}_{a}F)_{k}\to(\mathcal{K}_{a}G)_{k}\to(\mathcal{K}_{a}H)_{k}\to
0$

For example all elements in $I$ turn out to be Koszul $2$-exact.

######  10.14 Proposition.

Let $I$ be a finite poset.

  1. (1)

Then every element $a$ in $I$ is Koszul $2$-exact.

  2. (2)

Assume an element $a$ in $I$ has the following property: every subset
$S\subset\mathcal{P}(a)$ which has an ancestor has the product $\bigwedge S$
in $I$. Then $a$ is Koszul $\infty$-exact.

  3. (3)

If $I$ is an upper semilattice, then all its elements are Koszul
$\infty$-exact.

###### Proof.

Statement (1) is a consequence of the fact that direct sums preserves
exactness. Under the assumption of statement (2), for a functor $F$ and $k>0$,
the vector space $(\mathcal{K}_{a}F)_{k}$ can be also described as (see
10.12):

$\displaystyle\bigoplus_{\begin{subarray}{c}S\subset\mathcal{P}(a),\ |S|=k\\\
S\text{ has an ancestor}\end{subarray}}F(\bigwedge S)$

In this case, the statement also follows from the exactness of direct sums.
Finally statement (3) is a particular case of (2). ∎

Proposition 10.14 translates into the homology exact sequence:

######  10.15 Corollary.

Let $I$ be a finite poset and $0\to F\to G\to H\to 0$ be an exact sequence in
$\text{\rm Fun}(I,\text{\rm vect}_{K})$.

  1. (1)

For every element $a$ in $I$, there is an exact sequence of vector spaces:

${H_{2}(\mathcal{K}_{a}G)}$${H_{2}(\mathcal{K}_{a}H)}$${H_{1}(\mathcal{K}_{a}F)}$${H_{1}(\mathcal{K}_{a}G)}$${H_{1}(\mathcal{K}_{a}H)}$${H_{0}(\mathcal{K}_{a}F)}$${H_{0}(\mathcal{K}_{a}G)}$${H_{0}(\mathcal{K}_{a}H)}$${0}$

  2. (2)

Assume an element $a$ in $I$ has the following property: every subset
$S\subset\mathcal{P}(a)$ which has an ancestor has the product $\bigwedge S$
in $I$. Then there is an exact sequence of vector spaces:

${\cdots}$${H_{2}(\mathcal{K}_{a}F)}$${H_{2}(\mathcal{K}_{a}G)}$${H_{2}(\mathcal{K}_{a}H)}$${H_{1}(\mathcal{K}_{a}F)}$${H_{1}(\mathcal{K}_{a}G)}$${H_{1}(\mathcal{K}_{a}H)}$${H_{0}(\mathcal{K}_{a}F)}$${H_{0}(\mathcal{K}_{a}G)}$${H_{0}(\mathcal{K}_{a}H)}$${0}$

######  10.16.

Consider a homogeneous free functor $F=V[b,-)\colon I\to\text{vect}_{K}$,
where $I$ is a finite poset. We claim that $\mathcal{K}_{a}F$ has the
following homology:

$H_{i}(\mathcal{K}_{a}F)\text{ is isomorphic to }\begin{cases}V&\text{ if
}i=0\text{ and }b=a\\\ 0&\text{ otherwise }\end{cases}$

To prove the claim, recall $F(x)=0$ if $b\not\leq x$, and $F$ restricted to
$(b\leq I)\subset I$ is isomorphic to the constant functor with value $V$.
Thus, if $b\not\leq a$, then $\mathcal{K}_{a}F=0$, and the claim holds. If
$b=a$, then $(\mathcal{K}_{a}F)_{0}=F(a)=V$ and $(\mathcal{K}_{a}F)_{i}=0$ for
$i>0$, and again the claim holds. Assume $b<a$. Then
$(\mathcal{K}_{a}F)_{0}=F(a)$, which is isomorphic to $V$. Moreover, for a
subset that has an ancestor $S\subset\mathcal{P}(a)$ with $|S|>0$, the colimit
$\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(I\leq s)\end{subarray}}F$ is
isomorphic to $\text{colim}_{\begin{subarray}{c}\cap_{s\in S}(b\leq I\leq
s)\end{subarray}}F$, which is either isomorphic to $V$, in the case
$S\subset(b\leq\mathcal{P}(a))$, or is $0$ otherwise. Consequently, the
complex $\mathcal{K}_{a}F$ is isomorphic to $L\otimes V$, where $L$ is the
augmented chain complex of the standard $|b\leq\mathcal{P}(a)|$-dimensional
simplex whose homology is trivial in all degrees:

$L:=\left(\cdots\to\bigoplus_{\begin{subarray}{c}S\subset(b\leq\mathcal{P}(a))\\\
|S|=2\end{subarray}}K\to\bigoplus_{\begin{subarray}{c}S\subset(b\leq\mathcal{P}(a))\\\
|S|=1\end{subarray}}K\to K\right)$

Since the Koszul complex commutes with direct sums, if $F$ is free, isomorphic
to $\oplus_{b\in I}(\beta F)_{b}[b,-)$, then:

$H_{i}(\mathcal{K}_{a}F)\text{ is isomorphic to }\begin{cases}(\beta
F)_{a}&\text{ if }i=0\\\ 0&\text{ if }i>0\end{cases}$

We are now ready to state the key fact connecting the homology of the Koszul
complexes of a functor with its Betti diagrams.

######  10.17 Theorem.

Let $I$ be a finite poset and $G\colon I\to\text{\rm vect}_{K}$ a functor.

  1. (1)

For every $a$ in $I$ and $i=0,1,2$, the vector spaces $(\beta^{i}G)_{a}$ and
$H_{i}(\mathcal{K}_{a}G)$ are isomorphic.

  2. (2)

Assume an element $a$ in $I$ has the following property: every subset
$S\subset\mathcal{P}(a)$ which has an ancestor has the product $\bigwedge S$
in $I$. Then, for every $i$, the vector spaces $(\beta^{i}G)_{a}$ and
$H_{i}(\mathcal{K}_{a}G)$ are isomorphic.

###### Proof.

The proof relies on Corollary 10.15. Since the arguments for the statements
(1) and (2) are analogous, we show only (1).

The case $i=0$ follows from Proposition 10.10 and the fact that
$H_{0}(\mathcal{K}_{a}G)$ is isomorphic to $(G/\text{rad}(G))(a)$.

Consider an exact sequence $0\to S_{1}\to P_{0}\xrightarrow{\pi}G\to 0$ where
$\pi$ is a minimal cover. It leads to an exact sequence of homologies (see
10.15):

${H_{2}(\mathcal{K}_{a}P_{0})}$${H_{2}(\mathcal{K}_{a}G)}$${H_{1}(\mathcal{K}_{a}S_{1})}$${H_{1}(\mathcal{K}_{a}P_{0})}$${H_{1}(\mathcal{K}_{a}G)}$${H_{0}(\mathcal{K}_{a}S_{1})}$${H_{0}(\mathcal{K}_{a}P_{0})}$${H_{0}(\mathcal{K}_{a}G)}$${0}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{H_{0}(\mathcal{K}_{a}\pi)}$

Minimality of $\pi$ is equivalent to $H_{0}(\mathcal{K}_{a}\pi)$ being an
isomorphism. Since $P_{0}$ is free,
$H_{1}(\mathcal{K}_{a}P_{0})=H_{2}(\mathcal{K}_{a}P_{0})=0$. These two
observations imply $H_{1}(\mathcal{K}_{a}G)$ is isomorphic to
$H_{0}(\mathcal{K}_{a}S_{1})$, and $H_{2}(\mathcal{K}_{a}G)$ is isomorphic to
$H_{1}(\mathcal{K}_{a}S_{1})$. By the already proven case $i=0$,
$H_{0}(\mathcal{K}_{a}S_{1})$ is isomorphic to $(\beta^{0}S_{1})_{a}$ which is
isomorphic to $(\beta^{1}G)_{a}$. This gives the case $i=1$. Applying this
case to $S_{1}$, we get that $H_{1}(\mathcal{K}_{a}S_{1})$ is isomorphic to
$(\beta^{1}S_{1})_{a}$, which is isomorphic to $(\beta^{2}G)_{a}$, and the
case $i=2$ also holds. ∎

Here are some consequences of the presented statements, which are proved by
the same strategy: first dicretise and then use the Koszul complex
construction.

######  10.18 Corollary.

Let $J$ be a poset and $F\colon J\to\text{\rm vect}_{K}$ a functor.

  1. (1)

Assume $F$ is $n$-resolvable and $f\colon I\subset J$ is a subposet inclusion
with finite $I$ discretising an $n$-resolution of $F$. Then $\text{\rm
supp}(\beta^{i}F)\subset I$ for all $0\leq i<n$. Moreover, for $0\leq
i<\text{\rm min}(3,n)$ and $a$ in $I$, $(\beta^{i}F)_{a}$ is isomorphic to
$H_{i}(\mathcal{K}_{a}(Ff))$.

  2. (2)

Assume $F$ is discretised by a subposet inclusion $f\colon I\subset J$ which
is a homomorphism out of a finite upper semilattice $I$. Then $F$ is
$\infty$-resolvable and $\text{\rm supp}(\beta^{i}F)\subset I$ for all $i\geq
0$. Moreover, for $i\geq 0$ and $a$ in $I$, $(\beta^{i}F)_{a}$ is isomorphic
to $H_{i}(\mathcal{K}_{a}(Ff))$.

  3. (3)

Assume $J$ is an upper semilattice and $F$ a tame functor. Then $F$ is
$\infty$-resolvable and, for $a$ in $J$ with $\text{\rm par-dim}_{J}(a)<i$,
$(\beta^{i}F)_{a}=0$.

###### Proof.

Statement 1 is a consequence of Proposition 10.8 and Theorem 10.17. Statement
2 is a consequence of Corollary 10.6.(1), and again Proposition 10.8 and
Theorem 10.17. Statement 3 follows from statement 2 and Proposition 3.9 since
tame functors indexed by an upper semilattice can be discretised by a finite
sublattice (see 8.3.(1)). ∎

We finish this long article with our key theorem describing how to determine
Betti diagrams of functors indexed by realisations of finite type posets that
admit discretisable resolutions. Our main result is that for such functors the
Koszul complex can also be used for this purpose, similarly to functors
indexed by upper semilattices (see 10.18). The key reason for this is the
following fact: functors indexed by realisations have natural grid-like
discretisations that can be refined in a way that every set of parents of an
element having an ancestor also has a product in the refinement.

######  10.19 Theorem.

Let $I$ be a finite type poset and $F\colon\mathcal{R}(I)\to\text{\rm
vect}_{K}$ an $n$-resolvable functor. Assume $d$ is an element in $I$ and $V$
is a finite subset of $(-1,0)$ for which $\text{\rm
supp}(\beta^{j}F)\subset\mathcal{R}_{I\leq
d}(I,V)\stackrel{{\scriptstyle\alpha}}{{\hookrightarrow}}\mathcal{R}(I)$ for
all $j\leq i$.

  1. (1)

Let $0\leq i<\text{\rm min}(3,n)$ and $(a,f)$ be in $\text{\rm
supp}(\beta^{i}F)$. Then $(\beta^{i}F)_{(a,f)}$ is isomorphic to
$H_{i}(\mathcal{K}_{(a,f)}(F\alpha))$.

  2. (2)

Let $i<n$ and $(a,f)$ be in $\text{\rm supp}(\beta^{i}F)$ for which there is
$\varepsilon$ in $V$ such that $f(x)>\varepsilon$ for all $x$ in
$\mathcal{P}(a)$. Then $(\beta^{i}F)_{(a,f)}$ is isomorphic to
$H_{i}(\mathcal{K}_{(a,f)}(F\alpha))$.

  3. (3)

If $\text{\rm par-dim}_{\mathcal{R}(I)}(a,f)<i<n$, then
$(\beta^{i}F)_{(a,f)}=0$.

###### Proof.

1: The assumption implies that a minimal $\text{min}(3,n)$-resolution of $F$
is discretised by $\mathcal{R}_{I\leq d}(I,V)\subset\mathcal{R}(I)$. This
statement is then a particular case of Corollary 10.18.(1).

2: The assumption implies that a minimal $i+1$ resolution of $F$ is
discretised by $\mathcal{R}_{I\leq d}(I,V)\subset\mathcal{R}(I)$. Moreover the
product of every set of parents of $(a,f)$ in $\mathcal{R}_{I\leq d}(I,V)$
exists. This statement is then a particular case of Theorem 10.17.(2).

3: Since $\text{\rm par-dim}_{\mathcal{R}_{I\leq d}(I,V)}(a,f)\leq\text{\rm
par-dim}_{\mathcal{R}(I)}(a,f)$ for $(a,f)$ in $\mathcal{R}_{I\leq d}(I,V)$,
this statement follows from 2. ∎

### Acknowledgments.

A. Jin and F. Tombari were supported by the Wallenberg AI, Autonomous System
and Software Program (WASP) funded by Knut and Alice Wallenberg Foundation. W.
Chachólski was partially supported by VR, the Wallenberg AI, Autonomous System
and Software Program (WASP) funded by Knut and Alice Wallenberg Foundation,
and MultipleMS funded by the European Union under the Horizon 2020 program,
grant agreement 733161, and dBRAIN collaborative project at digital futures at
KTH.

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