# Effective epimorphisms in $G$-set

$$\newcommand{\gset}{G\text{-}\mathsf{set}}$$ 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)

• The category of $G$-sets is a topos, so every epimorphism is effective. – Malice Vidrine Nov 15 '18 at 17:54
• Yes, you can obtain such a map: $v=v'\circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$. – Kevin Carlson Nov 15 '18 at 17:57

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.