Epimorphisms are surjective in the category of $G$-sets? Let $G$ be a group. If you take the category of $G$-sets with $G$-equivariant maps as morphism, then the epimorphisms are surjective?
 A: Here are some general facts. Let $F : C \to D$ be a functor.

Exercise 1: If $F$ preserves pushouts, then it preserves epimorphisms. Dually, if $F$ preserves pullbacks, then it preserves monomorphisms.
Exercise 2: The forgetful functor from $G$-sets to sets has both a right and a left adjoint, so preserves both limits and colimits. By Exercise 1, this means it preserves both epimorphisms and monomorphisms.

So epimorphisms are surjective and monomorphisms are injective. But there's more! Recall that a functor $F$ is faithful if, whenever $f, g : c \to d$ are parallel morphisms, if $F(f) = F(g)$ then $f = g$.

Exercise 3: Faithful functors reflect monomorphisms and epimorphisms: if $F(f)$ is a monomorphism, then $f$ is a monomorphism, and the same for epimorphisms.

So in $G$-sets the epimorphisms are precisely the surjections and the monomorphisms are precisely the injections.
A: Suppose a map $f: X \to Y$ of $G$-sets is not surjective.  Let $Z = \{a,b\}$ have trivial $G$ action.  Define two maps $g,h: Y \to Z$ where $g$ sends everything to $a$, and $h$ sends everything in the image of $f$ to $a$ and everything else to $b$.  Clearly $g \circ f = h \circ f$, but $g\ne h$ since $f$ is not surjective. Therefore $f$ is not an epimorphism.
