$\newcommand{\gset}{G\text{-}\mathsf{set}}$
Let $G$ be any group and $\gset$ be the category of $G$-sets, with morphisms being $G$-maps. That is an object of $\gset$ is a pair $(X,\rho)$ where $\rho:G\to\text{aut}(X)$ is some group action of $G$ on $X$, and $$f:(X,\rho)\to (Y,\tau),\text{ satisfies } f(\rho(g)x)=\tau(g)f(x).$$
Am I correct that the fibred product $(X,\rho)\times_{f,(W,\tau),g}(Y,\chi)$ is given by $(Z,\rho\oplus \chi)$ where
$$Z=\{(x,y)\in X\times Y\mid f(x)=g(y),$$
with the standard projections, and where by $\rho\oplus \chi$ I mean that:
$$(\rho\oplus \chi)(g)(x,y)=(\rho(g)x,\chi(g)y),$$
which makes the projection maps $G$-maps.
(And where by $f,g$ I mean that we are taking the fibred product with respect to the maps $f:(X,\rho)\to (W,\tau)$ and $g:(Y,\chi)\to (W,\tau)$.)
Further, are epimorphisms just set surjections that are $G$-maps?
Best Answer
That is correct. Conveniently, the forgetful functor $U:\boldsymbol{G}\mathbf{-set}\to\mathbf{Set}$ is monadic, so it creates limits; the upshot of which is that all limit constructions in $\boldsymbol{G}\mathbf{-set}$ have, as their underlying structure, the limit of the image of the same diagram under $U$ in $\mathbf{Set}$. That doesn't tell you everything about the limit in $\boldsymbol{G}\mathbf{-set}$, but it's a good sanity check. That the set you describe has the requisite universal property takes very little checking.
And yes, epimorphisms are surjective $G$-maps. You can tell this is the case because the category of $G$-sets is equivalent to (if not defined as) the functor category $\mathbf{Set}^G$, so $G$-maps are essentially just natural transformations, and a natural transformation of set-valued functors is epimorphic exactly when all of its components are epimorphic.