
## 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 $\mathcal{T}_ G$:

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

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

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

4. 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

5. 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

$\mathop{\mathrm{Hom}}\nolimits _ G({}_ GG, {}_ GG) \longrightarrow G^{opp}, \varphi \longmapsto \varphi (1)$

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

$(g_1g_2) \cdot s = \mathcal{F}(R_{g_1g_2})(s) = \mathcal{F}(R_{g_2}\circ R_{g_1})(s) = \mathcal{F}(R_{g_1})( \mathcal{F}(R_{g_2})(s)) = g_1 \cdot (g_2 \cdot s).$

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

$\textit{PSh}(\mathcal{T}_ G) \longrightarrow G\textit{-Sets}, \quad \mathcal{F} \longmapsto \mathcal{F}({}_ GG).$

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 formula1

$\mathcal{F}_ S(U) = \mathop{Mor}\nolimits _{G\textit{-Sets}}(U, S).$

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

$\mathcal{F}_ S(U) = \mathop{Mor}\nolimits _{G\textit{-Sets}}(U, S) \longrightarrow \prod \mathcal{F}_ S(U_ i) = \prod \mathop{Mor}\nolimits _{G\textit{-Sets}}(U_ i, S)$

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

$\xymatrix{ U_ i \times _ U U_ j \ar[d] \ar[r] & U_ j \ar[d]^{s_ j} \\ U_ i \ar[r]^{s_ i} & S }$

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

$G\textit{-Sets} \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{T}_ G), \quad S \longmapsto \mathcal{F}_ S .$

We now have the following diagram of categories and functors

$\xymatrix{ \textit{PSh}(\mathcal{T}_ G) \ar[rr]^{\mathcal{F} \mapsto \mathcal{F}({}_ GG)} & & G\textit{-Sets} \ar[ld]_{S \mapsto \mathcal{F}_ S} \\ & \mathop{\mathit{Sh}}\nolimits (\mathcal{T}_ G) \ar[lu] & }$

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

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 $G\textit{-Sets}$.

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

$can : \mathcal{H} \longrightarrow \mathcal{F}_{\mathcal{H}({}_ GG)}.$

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

$\begin{matrix} \mathcal{H}(U) & \longrightarrow & \mathcal{F}_{\mathcal{H}({}_ GG)}(U) = \mathop{Mor}\nolimits _ G(U, \mathcal{H}({}_ GG)) \\ s & \longmapsto & (u \mapsto \alpha _ u^*s). \end{matrix}$

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, $\mathop{Mor}\nolimits _ G(U, \mathcal{H}({}_ GG)) = \mathcal{H}({}_ GG)^ H$ 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

$\xymatrix{ \mathcal{H}(U) \ar[r] & \mathcal{H}({}_ GG) \ar@<1ex>[r]^-{\text{pr}_0^*} \ar@<-1ex>[r]_-{\text{pr}_1^*} & \prod _{h \in H} \mathcal{H}({}_ GG). }$

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 $U = {}_ GG$. $\square$

[1] It may appear this is the representable presheaf defined by $S$. This may not be the case because $S$ may not be an object of $\mathcal{T}_ G$ which was chosen to be a sufficiently large set of $G$-sets.

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