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

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


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 :)

  • $\begingroup$ 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

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.)


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