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

  • 2
    $\begingroup$ The category of $G$-sets is a topos, so every epimorphism is effective. $\endgroup$ – Malice Vidrine Nov 15 '18 at 17:54
  • $\begingroup$ 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 $Z$. $\endgroup$ – 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.


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