Proposition 7.9.1. The functors $\mathcal{F} \mapsto \mathcal{F}({}_ GG)$ and $S \mapsto \mathcal{F}_ S$ define quasi-inverse equivalences between $\mathop{\mathit{Sh}}\nolimits (\mathcal{T}_ G)$ and $G\textit{-Sets}$.

## 7.9 The example of G-sets

As an example, consider the site $\mathcal{T}_ G$ of Example 7.6.5. We will describe the category of sheaves on $\mathcal{T}_ G$. The answer will turn out to be independent of the choices made in defining $\mathcal{T}_ G$. In fact, during the proof we will need only the following properties of the site $\mathcal{T}_ G$:

$\mathcal{T}_ G$ is a full subcategory of $G\textit{-Sets}$,

$\mathcal{T}_ G$ contains the $G$-set ${}_ GG$,

$\mathcal{T}_ G$ has fibre products and they are the same as in $G\textit{-Sets}$,

given $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T}_ G)$ and a $G$-invariant subset $O \subset U$, there exists an object of $\mathcal{T}_ G$ isomorphic to $O$, and

any surjective family of maps $\{ U_ i \to U\} _{i \in I}$, with $U, U_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T}_ G)$ is combinatorially equivalent to a covering of $\mathcal{T}_ G$.

These properties hold by Sets, Lemmas 3.10.2 and 3.11.1.

Remark that the map

is an isomorphism of groups. The inverse map sends $g \in G$ to the map $R_ g : s \mapsto sg$ (i.e. right multiplication). Note that $R_{g_1g_2} = R_{g_2} \circ R_{g_1}$ so the opposite is necessary.

This implies that for every presheaf $\mathcal{F}$ on $\mathcal{T}_ G$ the value $\mathcal{F}({}_ GG)$ inherits the structure of a $G$-set as follows: $g \cdot s$ for $g \in G$ and $s \in \mathcal{F}({}_ GG)$ defined by $\mathcal{F}(R_ g)(s)$. This is a left action because

Here we've used that $\mathcal{F}$ is contravariant. Note that if $\mathcal{F} \to \mathcal{G}$ is a morphism of presheaves of sets on $\mathcal{T}_ G$ then we get a map $\mathcal{F}({}_ GG) \to \mathcal{G}({}_ GG)$ which is compatible with the $G$-actions we have just defined. All in all we have constructed a functor

We leave it to the reader to verify that this construction has the pleasing property that the representable presheaf $h_ U$ is mapped to something canonically isomorphic to $U$. In a formula $h_ U({}_ GG) = \mathop{\mathrm{Hom}}\nolimits _ G({}_ GG, U) \cong U$.

Suppose that $S$ is a $G$-set. We define a presheaf $\mathcal{F}_ S$ by the formula^{1}

This is clearly a presheaf. On the other hand, suppose that $\{ U_ i \to U\} _{i\in I}$ is a covering in $\mathcal{T}_ G$. This implies that $\coprod _ i U_ i \to U$ is surjective. Thus it is clear that the map

is injective. And, given a family of $G$-equivariant maps $s_ i : U_ i \to S$, such that all the diagrams

commute, there is a unique $G$-equivariant map $s : U \to S$ such that $s_ i$ is the composition $U_ i \to U \to S$. Namely, we just define $s(u) = s_ i(u_ i)$ where $i\in I$ is any index such that there exists some $u_ i \in U_ i$ mapping to $u$ under the map $U_ i \to U$. The commutativity of the diagrams above implies exactly that this construction is well defined. All in all we have constructed a functor

We now have the following diagram of categories and functors

It is immediate from the definitions that $\mathcal{F}_ S({}_ GG) = \mathop{\mathrm{Mor}}\nolimits _ G({}_ GG, S) = S$, the last equality by evaluation at $1$. This almost proves the following.

**Proof.**
We have already seen that composing the functors one way around is isomorphic to the identity functor. In the other direction, for any sheaf $\mathcal{H}$ there is a natural map of sheaves

Namely, for any object $U$ of $\mathcal{T}_ G$ we let $can_ U$ be the map

Here $\alpha _ u : {}_ GG \to U$ is the map $\alpha _ u(g) = gu$ and $\alpha _ u^* : \mathcal{H}(U) \to \mathcal{H}({}_ GG)$ is the pullback map. A trivial but confusing verification shows that this is indeed a map of presheaves. We have to show that $can$ is an isomorphism. We do this by showing $can_ U$ is an isomorphism for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T}_ G)$. We leave the (important but easy) case that $U = {}_ GG$ to the reader. A general object $U$ of $\mathcal{T}_ G$ is a disjoint union of $G$-orbits: $U = \coprod _{i\in I} O_ i$. The family of maps $\{ O_ i \to U\} _{i \in I}$ is tautologically equivalent to a covering in $\mathcal{T}_ G$ (by the properties of $\mathcal{T}_ G$ listed at the beginning of this section). Hence by Lemma 7.8.4 the sheaf $\mathcal{H}$ satisfies the sheaf property with respect to $\{ O_ i \to U\} _{i \in I}$. The sheaf property for this covering implies $\mathcal{H}(U) = \prod _ i \mathcal{H}(O_ i)$. Hence it suffices to show that $can_ U$ is an isomorphism when $U$ consists of a single $G$-orbit. Let $u \in U$ and let $H \subset G$ be its stabilizer. Clearly, $\mathop{\mathrm{Mor}}\nolimits _ G(U, \mathcal{H}({}_ GG)) = \mathcal{H}({}_ GG)^ H$ equals the subset of $H$-invariant elements. On the other hand consider the covering $\{ {}_ GG \to U\} $ given by $g \mapsto gu$ (again it is just combinatorially equivalent to some covering of $\mathcal{T}_ G$, and again this doesn't matter). Note that the fibre product $({}_ GG)\times _ U ({}_ GG)$ is equal to $\{ (g, gh), g\in G, h\in H\} \cong \coprod _{h\in H} {}_ GG$. Hence the sheaf property for this covering reads as

The two maps $\text{pr}_ i^*$ into the factor $\mathcal{H}({}_ GG)$ differ by multiplication by $h$. Now the result follows from this and the fact that $can$ is an isomorphism for $U = {}_ GG$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)