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Let $Sets$ be the category of finite sets and all maps. I have come accross several example of functors $F : Sets^{op} \to Sets$ which satisfy the condition below:

-- There exists an integer $k$ such that, whenever $X$ is a finite set, with subsets $U_1, U_2, U_3...$ whose union is $X$ and with $|U_i| \ge k$, then we have the usual glueing condition: given elements $a_i \in F(U_i)$ with $a_i|_{U_i\cap U_j} = a_j|_{U_i\cap U_j}$, there exists a unique $a \in F(X)$ with $a |_{U_i} = a_i$.

(I hope the meaning of the "restriction" notation is obvious.)

So far I have taken this, informally, to mean that $F$ "is determined by what it does to sets of size $\le k$" (whenever $|X| > k$, we can recover $F(X)$ from all the sets $F(U)$ where $U$ runs through the subsets of $X$ of size $k$). However, it would be instructive to have a more conceptual understanding of what's going on.

Obviously the condition is almost the sheaf condition, but of course, the "subsets of size $\ge k$" do not form a Grothendieck topology...

I know of one way to rephrase this: given a contravariant functor $G$ from $C_k$, the category of sets of size $\le k$, to $Sets$, you can extend it to a functor $R(G)$ on $Sets^{op}$ in such a way that $R$ is a right adjoint to the "truncation" functor, which takes $F$ to its restriction $F_{\le k}$ to $C_k^{op}$. And my condition is that $F$ is isomorphic to $R(F_{\le k})$ (I think this is called the $k$-th "coskeleton" of $F$). However, I don't really know what to make of this.

Does anyone know of a good conceptual framework for this situation ? Perhaps in terms of sheaves and Grothendieck topologies?

Edit : I like the answer by T; Goodwillie below.

If there is such a framework, then what can I learn about $F$, concretely?

Edit : let me more specific. For a given $F$, I'm trying to look at the minimal $k$ such that the above holds, and I'm hoping that $k$ is a good measure of the "complexity" of $F$, with a low value of $k$ meaning that $F$ is "easier" to understand. For example you get $k=1$ for $F(X) = Hom(X, S)$ for a fixed $S$; I think the converse holds, with $S=F(\{x\})$, which is defined up to a canonical bijection.

So I would like to

(1) Substantiate the hope that $k$ is a good quantity to look at. Any interpretation à la $n$-lab is good for this, and by now I'm convinced that it's reasonable. Extra comments are welcome of course.

(2) compute $k$ concretely in some examples. Which is why any observations about the consequences of the above condition may help put some bounds on $k$.

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    $\begingroup$ Just a quick note: what about a "coverage" instead of a Grothendieck topology? I don't have the liberty to check right now. $\endgroup$
    – David Roberts
    Commented Apr 17, 2023 at 10:39
  • $\begingroup$ The concept of rank seems related. $\endgroup$
    – varkor
    Commented Apr 17, 2023 at 10:54

3 Answers 3

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Consider the Grothendieck topology in which a cover of $X$ is a way of expressing $X$ as a union of sets $U$ with $|U|\le k$. It seems to me that you are talking about sheaves for this topology.

This looks wrong (because I have "$\le$" where you had "$\ge$"), but I think it's right.

EDIT: This was nonsense. I should have said:

For each $k\ge 0$ there is a topology on the category of (all) sets, in which a cover of $X$ by subsets $U$ means that every subset of $X$ having at most $k$ elements belongs to some $U$. Your presheaves are the sheaves for this topology.

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  • $\begingroup$ ah, very nice! Definitely an alternative to the "right adjoing to truncation" point of view. Let me edit the question to emphasize the second part now. $\endgroup$
    – Pierre
    Commented Apr 17, 2023 at 13:12
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    $\begingroup$ Well, it was the "right adjoint to truncation" (a.k.a. right Kan extension) POV that got me pointed in the right direction. $\endgroup$ Commented Apr 17, 2023 at 14:55
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More generally, given a fully faithful functor $i: C \hookrightarrow D$, there is a Grothendendieck topology on $D$ such that the category of sheaves identifies with $Psh(C)$. That topology is given by the families $d_i \to d$ such that any $c \to d$ for $c \in C$ factors as $c \to d_i$ for some $i$.

The case you are considering is $D = Set$ and $C$ the full subcategory of sets or cardinality $\leqslant k$, or equivalently the full subcategory on the object $\{1,\dots,k\}$. Which give back the topology given in Tom Goodwillie answer's.

I'm going to assume that $C$ and $D$ are small - which isn't quite what you have in your situation. The case of non-small categories is handle by moving to a larger universe so that they are small and then checking at the end that the functor $i_*$ in the notation below preserves the presheaves with value in small sets (which I don't think is true in general, but often true, and definitely true when $C$ is small as $i_*$ can be defined as a $C$ indexed end.)

One can give a direct proof by checking that this defines a topology on $D$ and then applying Grothendieck's comparison lemma to check that sheaves for this topology identifies with presheaves on $C$, but I'd rather give the following proof which I find more interesting:

So, the restriction functor $i^* : Psh(D) \to Psh(C)$ has a left adjoint $i_!$ and a right adjoint $i_*$.

The left adjoint $i_!$ is defined as a coend

$$ (i_! \mathcal{F}) (d) = \int^{c \in C} \mathcal{F}(c) \times Hom_D(d,c) $$

In particular $i^*i_! \mathcal{F}$ is:

$$ (i_! \mathcal{F}) (c) = \int^{c' \in C} \mathcal{F}(c') \times Hom_D(c,c') = \int^{c' \in C} \mathcal{F}(c') \times Hom_C(c,c') = \mathcal{F}(c) $$

Where I've used that the inclusion is fully faithful and one of the coend version of the Yoneda lemma.

That is the unit $\mathcal{F} \to i^* i_! \mathcal{F}$ is an equivalence. By some abstract non-sense it follows that $i^* i_* \mathcal{F} \to \mathcal{F}$ is also an equivalence (thinking about it - one can probably get that from the exact same argument as above using the end description of $i_*$, but for some reasons I'm much more used to $i_!$ and coend than to $i_*$ and ends).

Now, because $i^*$ also preserves all limits this identifies $Psh(C)$ with a left exact localization of $Psh(D)$ and all such localization are given by Grothendieck topology.

That topology is the one such that the cover are the family of maps $d_i \to d$ which became covering family when applying $i^*$, that is, such that for each object $c \in C$, and each map $c \to d$ there exists a lift $c \to d_i$ for some $i$.

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It sounds to me you seek a solution to the following problem:

Given a suitable subcategory of $\textbf{Set}$, find a Grothendieck topology on $\textbf{Set}$ so that sheaves on $\textbf{Set}$ are equivalent to presheaves on the subcategory.

Fortunately, this can be done in great generality.

Theorem. Let $\mathcal{D}$ be a locally small category and let $\mathcal{C}$ be an essentially small full subcategory of $\mathcal{D}$. Then there is a unique Grothendieck topology on $\mathcal{D}$ with the following property:

  • Given a presheaf $F : \mathcal{C}^\textrm{op} \to \textbf{Set}$, there is a unique (up to isomorphism) sheaf $\tilde{F} : \mathcal{D}^\textrm{op} \to \textbf{Set}$ whose restriction to $\mathcal{C}$ is $F$.

Example. Let $\mathcal{D} = \textbf{Set}$ and let $\mathcal{C}$ be the full subcategory of sets of cardinality $\le k$, where $k \ge 1$. Then the hypotheses of the theorem are satisfied. The Grothendieck topology so obtained can be explicitly described as follows: a family of maps to $D$ is covering if and only if, for every map $f : C \to D$ where $C$ has cardinality $\le k$, $f$ factors through some member of the family. Put it even more simply, a family of maps to $D$ is covering if and only if every subset of $D$ of cardinality $\le k$ is contained in the image of some member of the family – which agrees with the answer of Tom Goodwillie here.

Proof of theorem. Consider the restriction functor $j^* : [\mathcal{D}^\textrm{op}, \textbf{Set}] \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$. Clearly, it preserves finite limits. Let $j_* : [\mathcal{C}^\textrm{op}, \textbf{Set}] \to [\mathcal{D}^\textrm{op}, \textbf{Set}]$ be right Kan extension along the inclusion, explicitly: $$(j_* F) (D) = \textrm{Hom} (j^* h_D, F)$$ Then $j_* : [\mathcal{C}^\textrm{op}, \textbf{Set}] \to [\mathcal{D}^\textrm{op}, \textbf{Set}]$ is the right adjoint of $j^*$. Given $F : \mathcal{C}^\textrm{op} \to \textbf{Set}$ and $C'$ in $\mathcal{C}$, we have $$(j_* F) (C') = \textrm{Hom} (j^* h_{C'}, F) = \textrm{Hom} (h_{C'}, F) \cong F (C')$$ by the Yoneda lemma. Hence, $j^* j_* F \cong F$, so – modulo some checking that this actually is the counit isomorphism – $j_* : [\mathcal{C}^\textrm{op}, \textbf{Set}] \to [\mathcal{D}^\textrm{op}, \textbf{Set}]$ is fully faithful. Thus – modulo size issues – the hypothesis means $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ is equivalent to a subtopos of $[\mathcal{D}^\textrm{op}, \textbf{Set}]$, with $j_*$ being identified with the inclusion. But subtoposes of a presheaf topos exactly correspond to Grothendieck topologies, so we are done (modulo size issues).

Explicitly – and this is how we check that there are no size issues – given a sieve $U$ on an object $D$ in $\mathcal{D}$, consider it as a subpresheaf of the representable presheaf $h_D$ and define it to be $J$-covering if $j^*$ sends the inclusion $U \hookrightarrow h_D$ to an isomorphism $j^* U \to j^* h_D$. Very explicitly, that means $U$ is a $J$-covering sieve on $D$ if and only if, for every object $C$ in $\mathcal{C}$ and every morphism $f : C \to D$ in $\mathcal{D}$, $f \in U (C)$. It is not hard to verify that $J$ so defined is a Grothendieck topology on $\mathcal{D}$. Given a presheaf $F : \mathcal{C}^\textrm{op} \to \textbf{Set}$ and a $J$-covering sieve $U$ on $D$, we have $$\textrm{Hom} (U, j_* F) \cong \textrm{Hom} (j^* U, F) \cong \textrm{Hom} (j^* h_D, F) \cong \textrm{Hom} (h_D, j_* F)$$ so $j_* F$ is a $J$-sheaf on $\mathcal{D}$, and $j^* j_* F \cong F$ as we previously noted. Thus, $J$ is the required Grothendieck topology. ◼

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  • $\begingroup$ that's nice, thank you. And likewise for the answer by Simon Henry below. $\endgroup$
    – Pierre
    Commented Apr 17, 2023 at 14:06

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