Why does this exact sequence of sheaves imply the maps are $G$-equivariant? I'm confused about something in Tamme's Introduction to Étale Cohomology (page 27). Let $G$ be a group and let $G$-$\mathsf{Set}$ denote the category of left $G$-sets with $G$-equivariant maps as morphisms. Let $T_G$ denote the canonical topology on this category. Then a family $\left\{\phi_i: U_i \to U\right\}$ is a family of universal effective epimorphisms if and only if $U = \bigcup_i \phi_i (U_i)$.
Tamme is showing that the category $G$-$\textsf{Set}$ is equivalent to the category of sheaves of sets on the site ($G$-$\textsf{Set}$, $T_G$). To a $G$-set $Z$ we associate the representable sheaf $h_Z = \text{Hom}_G (-, Z)$ and to a sheaf $F$ we associate the set $F(G)$ made into a left $G$-set by applying $F$ to the right-multiplication of $G$ on itself. It is easy to see that
$$ Z\mapsto h_Z \mapsto h_Z (G) = \text{Hom}_G (G, Z)$$
gives a natural isomorphism in one direction of the equivalence. The thing I'm struggling with is the other direction: showing that
$$ F \cong \text{Hom}_G (-, F(G))$$
naturally in $F$.
To set things up specifically for my problem, let $U$ be a $G$-set and let $u\in U$. Define $\phi_u : G\to U$ by $\phi_u (g) = g\cdot u$. Then $\phi_u$ is a $G$-map and $\left\{\phi_u : G\to U\right\}_{u\in U}$ is a covering family in $T_G$. The sheaf property of $F$ gives us an equaliser diagram
$$ F(U)\to\prod_{u\in U} F(G)\rightrightarrows \prod_{u, v \in U} F(G\times_U G)$$
Hence $F(U)$ can be identified with a subset of the set $\prod_{u\in U } F(G) \cong \left\{ \text{functions } U\to F(G)\right\}$.
I'm struggling to interpret how the exactness condition implies that each of these functions $U\to F(G)$ is actually a $G$-map i.e. $F(U) \cong \text{Hom}_G (U, F(G))$ (as is quickly mentioned in the book). I know that this is just a matter of spelling out what the maps to the term on the right actually are, but I've let myself get extremely confused. I'd really appreciate if anyone can explain how it works in more detail than the book. Thanks!

Edit: Zhen Lin's comment proves the equivalence with a far more general statement about sheaves on categories of sheaves; this example follows as a special case where the underlying site has one object. However I would still like to see an answer to this question because the thing that is actually confusing me is how one interprets the equaliser diagram to see that elements of $F(U)$ are naturally $G$-equivariant functions $U\to F(G)$. I am not sure if this is obvious and I am just looking at it in the wrong way or if it requires some work.
 A: First, let us introduce some notation :
Let $F$ be any presheaf of sets on $G$-Set. If $f:U\rightarrow V$ is any map of $G$-Sets, we write $f^*:F(V)\rightarrow F(U)$ for the restriction.
Let $R_g:G\rightarrow G$ be the right multiplication map $h\mapsto hg$. If $F$ is any presheaf on $G$-Set, then $F(G)$ is a $G$-set via $(g,x)\mapsto R_g^*(x)$.
Also, if $U$ is any $G$-set and $u\in U$, you have a map $\phi_u:G\rightarrow U$ such that $\phi_u(g)=g.u$. This defines a map 
$$\Psi:F(U)\rightarrow\prod_u F(G)=F(G)^U.$$
Concretely, if $\alpha\in F(U)$, then $\Psi(\alpha)$ is the map $u\mapsto \phi_u^*(\alpha)$.

Now, without any further hypotheses, I claim that $\Psi(\alpha)$ is already a $G$-equivariant map. Indeed, the structure of $G$-Set on $F(G)$, we have :
$$g.(\Psi(\alpha)(x))=R_g^*(\Psi(\alpha)(u))=R_g^*\phi_u^*(\alpha)=(\phi_u\circ R_g)^*(\alpha)=\phi_{gu}^*\alpha=\Phi(\alpha)(g.u)$$
since you can easily check that $\phi_u\circ R_g=\phi_{gu}$ as maps $G\rightarrow U$.

You have an diagram :
$$F(U)\overset{\Psi}\rightarrow\prod_u F(G)\rightrightarrows\prod_{u,v} F(G\times_U G)$$
By definition, if $F$ is separated, then the first map is a mono.
Let us shows that, if $F$ is a sheaf, then the equalizer of the two maps are exactly the set of maps $U\rightarrow F(G)$ which are $G$-equivariant. This will show that $\Psi$ maps isomorphically $F(U)$ to the set of $G$-equivariant maps $U\rightarrow F(G)$.
Write $d^1,d^2$ for these two maps. Concretely, if $r=(r_u)_{u\in U}$ is an element of $\prod_{u\in U}F(G)$, then $d^1(r),d^2(r)$ are families $$d^1(r)_{u,v},d^2(r)_{u,v}\in F(G\times_U G)=F(\{(g,h)\in G\times G, g.u=h.v\})$$
Now $d^1(r)_{u,v}=p_1^*(r_u)$ and $d^2(r)_{u,v}=p_2^*(r_v)$.
So assume first that $d^1(r)=d^2(r)$ and let us show that $r=(r_u)$ is $G$-equivariant (this is not needed to prove the equivalence of categories, but for sake of completeness...). We have $p_1^*(r_u)=p_2^*(r_v)$ for all $u,v$ and we want to prove that $r_{f.u}=R_f^*r_u$ for all $u\in U$ and all $f\in G$. So fix $u\in U, f\in G$ and let $v=f.u$. Let
$$ i_f:G\rightarrow G\times_U G=\{(g,h), g.u=hf.u \} $$
be the map $g\mapsto (gf,g)$. Clearly $p_2i_f=\operatorname{id}$ and $p_1i_f=R_f$. Now compute :
$$ r_{f.u}=r_v=i_f^*p_2^*(r_v)=i_f^*p_1^*(r_u)=R_f^*r_u$$
Hence $r$ is $G$-equivariant as claimed (we didn't use the fact that $F$ was a sheaf).
Conversely and finally, assume that $r=(r_u)_{u\in U}$ is $G$-equivariant ($R_g^*r_u=r_{g.u}$). We want to show that for all $u,v$, $p_1^*r_u=p_2^*r_v\in F(\{(g,h)\in G\times G, g.u=h.v\})$. Let $E_{u,v}=\{f\in G, f.u=v\}$ and consider the map
$$\begin{array}{rcl}G\times E_{u,v}&\rightarrow& G\times_U G\\
(g,f)&\mapsto& (gf,g)\end{array}$$
In fact, consider $G\times E_{u,v}$ as the disjoint union $\coprod_{f\in E_{u,v}} G$. So that this map is actually a map of $G$-sets, and is component-wise given by the maps $i_f$ considered previously.
This map is obviously an isomorphism with inverse $(g,h)\mapsto (h, h^{-1}g)$. Because $F$ is a sheaf, it is in particular additive (meaning $F(A\sqcup B)=F(A)\times F(B)$), this implies that
$$ F(G\times_U G)\simeq F(G\times E_{u,v})\simeq\prod_{f\in E_{u,v}}F(G)$$
So by construction an element $x\in F(G\times_U G)$ maps to the family $(i_f^*x)_{f\in E_{u,v}}$. Because this is an isomorphism, to prove $p_1^*r_u=p_2^*r_v$, it is enough to prove this equality after applying $i_f^*$ for all $f\in E_{u,v}$, in other words for all $f\in G$ such that $f.u=v$.
Now compute :
$$i_f^*p_1^*r_u=R_f^*r_u=r_{f.u}=r_v=i_f^*p_2^* r_v$$
This concludes the proof. 
