The Stacks project

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