3
$\begingroup$

This is Exercise II.4 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to Approach0, it is new to MSE.

The Details:

Adapted from p. 13, ibid. . . .

Definition 1: A functor $F: \mathbf{A}\to \mathbf{B}$ is an equivalence of categories if for any $\mathbf{A}$-objects $A, A'$, we have that

$$\begin{align} {\rm Hom}_{\mathbf{A}}(A, A')&\to{\rm Hom}_{\mathbf{B}}(FA, FA')\\ p&\mapsto F(p) \end{align}$$

is a bijection and, moreover, any object of $\mathbf{B}$ is isomorphic to an object in the image of $F$.

On p. 66, ibid. . . .

Definition 2: A sheaf of sets $F$ on a topological space $X$ is a functor $F:\mathcal{O}(X)^{{\rm op}}\to\mathbf{Sets}$ such that each open covering $U=\bigcup_iU_i, i\in I$, of open subsets of $U$ of $X$ yields an equaliser diagram

$$ FU\stackrel{e}{\dashrightarrow}\prod_{i\in I}FU_i\overset{p}{\underset{q}{\rightrightarrows}}\prod_{i,j\in I}(U_i\cap U_j),$$

where for $t\in FU,$ $e(t)=\{ t\rvert_{U_i}\mid i\in I\}$ and for a family $t_i\in FU_i$,

$$p\{ t_i\}=\{t_i\rvert_{(U_i\cap U_j)}\}\quad\text{ and }\quad q\{ t_i\}=\{t_j\rvert_{(U_i\cap U_j)}\}.$$

Here $\mathcal{O}(X)$ is the set of open sets of $X$.

The Question:

Prove that for a basis $\mathcal{B}$ of the topology on a space $X$, the restriction functor $\mathbf{r}:{\rm Sh}(X)\to{\rm Sh}(\mathcal{B})$ is an equivalence of categories.

[Hint: Define a quasi-inverse $\mathbf{s}:{\rm Sh}(\mathcal{B})\to{\rm Sh}(X)$ for $\mathbf{r}$ as follows. Given a sheaf $F$ on $\mathcal{B}$, and an open set $U\subset X$, consider the cover $\{B_i\mid i\in I\}$ of $U$ by all basic open sets $B_i\in\mathcal{B}$ which are contained in $U$. Define $\mathbf{s}(F)(U)$ by the equaliser

$$\mathbf{s}(F)(U)\to\prod_{i\in I}F(B_i)\rightrightarrows\prod_{i, j}F(B_i\cap B_j).]$$

Thoughts:

I need to show that, following Definition 1,

$$\begin{align} {\rm Hom}_{{\rm Sh}(X)}(V, V')&\to{\rm Hom}_{{\rm Sh}(\mathcal{B})}(\mathbf{r}V, \mathbf{r}V')\\ p&\mapsto \mathbf{r}(p) \end{align}$$ is a bijection for all $V, V'\in{\rm Ob}({\rm Sh}(X))$ and any ${\rm Sh}(\mathcal{B})$-object is isomorphic to an object in the image of $\mathbf{r}$.

Further Context:

Related questions of mine include:

I am teaching myself topos theory for fun. I have read Goldblatt's book, "Topoi [. . .]", although I did not fully understand it. For example,

Please help :)

$\endgroup$
4
  • $\begingroup$ I suppose revising the definition of ${\rm Sh}(X)$ would be useful . . . $\endgroup$
    – Shaun
    Feb 25 '20 at 19:11
  • 1
    $\begingroup$ You can find this towards the beginning of Eisenbud's The Geometry of Schemes I believe. $\endgroup$
    – jgon
    Feb 25 '20 at 19:49
  • 2
    $\begingroup$ I would suppose that what they expect is for you to show isomorphisms of functors $\mathbf{r} \circ \mathbf{s} \simeq \operatorname{id}$ and $\mathbf{s} \circ \mathbf{r} \simeq \operatorname{id}$. $\endgroup$ Feb 25 '20 at 22:09
  • 1
    $\begingroup$ Incidentally, I think what they might have meant in the last line is not $\prod_{i,j} F(B_i \cap B_j)$, but $\prod_{i,j,k \mid B_k \subseteq B_i \cap B_j} F(B_k)$. $\endgroup$ Feb 25 '20 at 22:25
1
$\begingroup$

First of all, when the hint speaks of a "quasi-inverse" it is referring to the following equivalent of the given definition: a functor $F : \mathbf{C} \to \mathbf{D}$ is an equivalence of categories if and only if there exists a functor $G : \mathbf{D} \to \mathbf{C}$ such that $F \circ G \simeq \operatorname{id}_{\mathbf{D}}$ and $G \circ F \simeq \operatorname{id}_{\mathbf{C}}$; and in this case, $G$ is called a quasi-inverse of $F$.

So, one way to follow the hint would be to explain how $\mathbf{s}$ becomes a functor (i.e. how it operates on morphisms, and show it preserves identities and compositions), and then establish isomorphisms $\mathbf{r} \circ \mathbf{s} \simeq \operatorname{id}$ and $\mathbf{s} \circ \mathbf{r} \simeq \operatorname{id}$.


On the other hand, it is possible to proceed using the definition you stated. First, as a preliminary, I don't know if MacLane and Moerdik specified what exactly $\operatorname{Sh}(\mathcal{B})$ means; but the reasonable definition would be that it is the presheaves on the poset category of $\mathcal{B}$ such that whenever $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$ is a cover of $U \in \mathcal{B}$, we have an equalizer diagram $$F(U) \rightarrow \prod_{i\in I} F(V_i) \rightrightarrows \prod_{i, j \in I, W\in \mathcal{B}, W \subseteq V_i \cap V_j} F(W).$$

(The first step would be to see why $\mathbf{r}$ of a sheaf on $X$ would satisfy this condition; I will leave that as an exercise.)

So, first let us see that $\mathbf{r}$ is injective on morphisms; so, suppose that we have two morphisms $f, g : F \to G$ such that $f(V) = g(V)$ whenever $V \in \mathcal{B}$. Then for any open $U$ and $x \in F(U)$, there is a cover of $U$ by elements $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. Now, by the hypothesis, $$f(x) {\mid_{V_i}} = f(V_i)(x {\mid_{V_i}}) = g(V_i)(x {\mid_{V_i}}) = g(x) {\mid_{V_i}}$$ for each $i$; and by the injectivity part of the equalizer condition defining that $G$ is a sheaf, we conclude that $f(x) = g(x)$. Since this is true for any open $U$ and any $x \in F(U)$, then $f = g$.

Similarly, to see that $\mathbf{r}$ is surjective on morphisms, suppose we have $f : \mathbf{r}(F) \to \mathbf{r}(G)$. Then for any open $U \subseteq X$ and $x \in F(U)$, again choose a cover of $U$ by $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. (In fact, to forestall questions of the following construction being well-defined, let us use the canonical maximal cover of all elements of $\mathcal{B}$ contained in $U$.) Then for each $i \in I$, define $y_i := f(V_i)(x {\mid_{V_i}})$. Then for each $i,j$, we can find the canonical maximal cover of $V_i \cap V_j$ by $\{ W_k \mid k \in K_{i,j} \} \subseteq \mathcal{B}$. Now for each $k$, we have $$y_i {\mid_{W_k}} = f(V_i)(x {\mid_{V_i}}) {\mid_{W_k}} = f(W_k)((x {\mid_{V_i}}) {\mid_{W_k}}) = F(W_k)(x {\mid_{W_k}}) = y_j {\mid_{W_k}}.$$ Therefore, by the injectivity part of the sheaf condition on $G$, we have $y_i {\mid_{U_i \cap U_j}} = y_j {\mid_{U_i \cap U_j}}$. Then, by the exactness part of the sheaf condition on $G$, there exists a unique $y \in G(U)$ such that $y {\mid_{U_i}} = y_i$. We now define $f'(U)(x) := y$.

It remains to show that $f'$ defines a morphism of sheaves, and that $\mathbf{r}(f') = f$. (Hint for the morphism of sheaves part: given $U' \subseteq U$ and $x \in F(U)$, show that $(f'(U) {\mid_{U'}}) {\mid_{V_i}}$ is equal to $y_i$ when you put $x {\mid_{U'}}$ in place of $x$, and then apply the injectivity part of the sheaf condition on $G$.)

Now, to show that $\mathbf{r}$ is essentially surjective, suppose we have $F \in \operatorname{Sh}(\mathcal{B})$. Then for each open $U$, define $G(U)$ to be the equalizer in the diagram $$G(U) \rightarrow \prod_{V \in \mathcal{B}, V \subseteq U} F(V) \rightrightarrows \prod_{V, V', W \in \mathcal{B}, V \subseteq U, V' \subseteq U, W \subseteq V \cap V'} F(W).$$ The restriction maps of $G$ will then be constructed based on the universal property of equalizers. We now need to see that $G$ is a sheaf on $X$, and that $\mathbf{r}(G) \simeq F$. The latter follows fairly directly from the sheaf condition on $F$.

For the sheaf condition, suppose we have a cover $\{ U_i \mid i \in I \}$ of $U$ and sections $x_i \in G(U_i)$ such that $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ for each $i,j$. Then each $x_i$ can be decomposed into the compatible data of an element of $F(V)$ for every $V \in \mathcal{B}$, $V \subseteq U_i$ which we will call $x_i {\mid_V}$. But then, the union of the canonical covers of each $U_i$ will form a cover of $U$; and for each $W$ in this cover, we can choose $i$ such that $W \subseteq U_i$, and define $y_W := x_i {\mid_W}$. If we have two different indices $i,j$ such that $W \subseteq U_i$ and $W \subseteq U_j$, then from the condition $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ we get $x_i {\mid_W} = x_j {\mid_W}$, which makes this definition of $y_V$ well-defined. Once we verify the compatibility condition on $(y_W)$, we get a section $z_V \in F(V)$ from the definition of $F$ being a sheaf. It now remains to show that this family of $z_V$ satisfies the compatibility condition from the definition of $G$, and that the section $x \in G(U)$ we get in this way satisfies $x {\mid_{U_i}} = x_i$ for each $i$. It also remains to establish the uniqueness of $x$.


In the above, you can see that our construction in the "essential surjectivity" proof amounted to specifying the object part of a quasi-inverse $\mathbf{s}$, and our construction in the "surjectivity on morphisms" proof amounted to specifying the morphism part of $\mathbf{s}$. (Note that the definition of $\mathbf{s}$ as you wrote it does not necessarily make sense if $\mathcal{B}$ is not closed under intersections.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.