$\newcommand{\gset}{G\text{-}\mathsf{set}}$
Let $G$ be a group and $\gset$ the category of $G$-sets, whose morphisms are $G$-maps (i.e. set maps where the map commutes with $G$-action).
What are the effective epimorphisms in $\gset$? Recall that in a category $\mathsf{C}$ an epimorphism $f:A\to B$ is a morphisms such that:
$$\hom(B,Z)\hookrightarrow \hom(A,Z),$$
is injective for every $Z\in\text{ob}(\mathsf{C})$, and it's an effective epimorphism if:
$$\hom(B,Z)\to \hom(A,Z)\stackrel{\longrightarrow}\to \hom(A\times_BA,Z),$$
is exact, where the two arrows are $pr_1^*,pr_2^*$, and by exact I mean $\hom(B,Z)$ is the equalizer of this diagram.
In my previous question I make a claim what epimorphisms and the fibred products are. Apparently the effective epimorphisms in $\gset$ are simply the surjective maps, but this seems false. I could take $A\to \{1\}$ where $\{1\}$ is just some singleton, and $A\times_{\{1\}}A=A\prod A$ in set, with 'product action' $$\rho(g)(a,a')=(\rho(g)a,\rho(g)a').$$
Certainly I can't obtain map $v:A\to Z$ such that $v\circ pr_1 = v\circ pr_2$ from $v'\circ f$, where $v'$ is just a map $\{1\}\to Z$?
$$\require{AMScd}
\begin{CD}
A\times_{\{1\}}A@>>> Z;\\
@V{pr_1,pr_2}VV@V{id}VV \\
A @>{v}>> Z ;\\
@V{f}VV @V{id}VV\\
\{1\}@>{v'}>>Z
\end{CD}$$
(Sorry about the diagram, not sure how to make commuting triangles)
Best Answer
The category of $G$-sets is a topos, so every epimorphism is effective.
The fact that it is a topos is apparent if you consider $G$ as a category $\mathcal{C}$ where $\text{Mor}(\mathcal{C})=G$ and $\text{Ob}(\mathcal{C})=*$ (a point). Then the category of $G$-sets is just the presheaf category $\text{Sets}^{G^{\text{op}}}$. Indeed, such a presheaf $\mathcal{F}$ assigns to the single object a set $\mathcal{F}(*)$, and to each $g\in G$ a morphism $\mathcal{F}(g):\mathcal{F}(*)\to\mathcal{F}(*)$. I.e. such a presheaf is a set with a $G$-action. The morphisms are natural transformations of functors $\mathcal{F}\Rightarrow\mathcal{G}$, in which case we have a set map $\mathcal{F}(*)\to\mathcal{G}(*)$ commuting with the $G$-action.