# For basis $\mathcal{B}$ of the topology on $X$, the restriction functor $\mathbf{r}:{\rm Sh}(X)\to{\rm Sh}(\mathcal{B})$ is an equivalence.

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,

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