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.

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

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

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.

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. The fact that I know almost nothing about $F$ doesn't really help.

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

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.

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