Sheaves on the site of $\pi$-sets Let $\pi$ be a group, and let $\mathcal{C}$ be the site whose underlying category is that of $\pi$-sets (with $\pi$-linear maps as morphisms). The covers are jointly surjective families of such $\pi$-linear maps (ie, maps commuting with the $\pi$-action).
Let $\langle\pi\rangle$ denote the $\pi$-set whose underlying set is $\pi$, and where the $\pi$-action is given by left-multiplication. Thus, $\pi$ acts as $\pi$-linear automorphisms of $\langle\pi\rangle$ via right-multiplication.
Let $F$ be a sheaf of sets on $\mathcal{C}$. Thus, for any $a\in\pi$, we get an automorphism of $\langle\pi\rangle$, giving an automorphism $F(a)$ of $F(\langle\pi\rangle)$. Let $S$ be a set with transitive $\pi$-action, so that after picking a particular $s\in S$, we can define a surjective $\pi$-linear map $p : \langle\pi\rangle\rightarrow S$, which is defined by $p(a) = a.s$
The sheaf axiom tells us that we have an equalizer diagram:
$$F(S)\stackrel{i}{\hookrightarrow} F(\langle\pi\rangle)\stackrel{\longrightarrow}{\longrightarrow} F(\langle\pi\rangle\times_S\langle\pi\rangle)$$
Let $H\subseteq\pi$ denote the stabilizer of $s\in S$. Then, the image of $F(S)$ in the above diagram lies in the set $F(\langle\pi\rangle)^H$ of $H$-invariant elements of $F(\langle\pi\rangle)$. This follows from applying $F$ to the identity $p\circ h = p$ for any $h\in H$.
Question: Why is the image of $F(S)$ in $F(\langle\pi\rangle)$ the full $H$-invariant subset $F(\langle\pi\rangle)^H$?
I feel like I need to relate the property of an element of $F(\langle\pi\rangle)$ being $H$-invariant to it equalizing the two maps in the equalizer diagram, but I've thought about this for way too long and can't figure out how to do it.
Mumford asserts this is true on page 40 of Mumford's "Picard Groups of Moduli Problems", which can be found here: http://www.mathcs.emory.edu/~brussel/Scans/mumfordpicard.pdf
 A: I think you're right that the key is to show that all $H$-fixed elements of $F(\langle\pi\rangle)$ equalize the two maps from $F(\langle\pi\rangle)$ to $F(R)$, where I wrote $R$ to abbreviate $\langle\pi\rangle\times_S\langle\pi\rangle$.  For this purpose, let me first look more closely at $R$ and its two maps to $\langle\pi\rangle$.  Of course, being a pullback, $R$ consists of those pairs $(a,b)$ of elements of $\pi$ such that $p(a)=p(b)$, i.e., $as=bs$, i.e., $b=ah$ for some $h\in H$.  So I can regard $R$ as the disjoint union, over all $h\in H$, of the sets $R_h=\{(a,ah):a\in\pi\}$.  Notice that the $\pi$-set structure of $R$, i.e., the action of $\pi$ on $R$ by left multiplication, respects this decomposition, so $R$ is, as a $\pi$-set, the coproduct of the $R_h$'s, each of which is isomorphic to $\langle\pi\rangle$ via the isomorphism $\langle\pi\rangle\to R_h:a\mapsto(a,ah)$.  Identifying each $R_h$ with $\langle\pi\rangle$ this way, we have that $R$ is, as a $\pi$-set, the $H$-indexed coproduct of copies of $\langle\pi\rangle$.  Its two projections to $\langle\pi\rangle$ act, on the summand indexed by $h$, as the identity and (the right multiplication by) $h$.  Now because $F$ is a sheaf, $F(R)$ is the product of the $F(R_h)$'s.  The two maps from $F(\langle\pi\rangle)$ to $F(R)$ are, when regarded as $H$-indexed systems of maps to the factors $F(R_h)$, the identity maps and the maps $F(h)$.  So their equalizer is exactly the part of $F(\langle\pi\rangle)$ on which every $F(h)$ for $h\in H$ agrees with the identity.
