TITLE: An arrow is monic in the category of G-Sets if and only if its monic the category of sets
QUESTION [6 upvotes]: Let $G$ be a group, regarded as category with one object $*$ in which each arrow is invertible. Then the category of $G$-Sets is just the category of functors from $G$ to $\mathbf{Set}$. Now I've read that an arrow $\varphi: \tau \to \tau'$ (where $\tau$ and $\tau'$ are functors from $G$ to $\mathbf{Set}$) in the category of $G$-Sets is monic if and only if $\varphi_*$ is monic in $\mathbf{Set}$. I'm having trouble seeing why the only if part of this statement is true.

REPLY [2 votes]: Thanks to beroal for his answer. May I add the following perspective:
To prove that in a concrete category monic implies injective, one can sometimes use an identification between 'elements' and 'arrows', because 'monic' basically means injective on arrows.

In $\sf{Set}$, $\sf{Top}$: $Hom(\{\star\},X)\cong X$ via $f\mapsto f(\star)$.
In $\sf{Gr}$ and $\sf{Ab}$: $Hom(\mathbb{Z},G)\cong G$ via $f\mapsto f(1)$.
In $\sf{Ring}$: $Hom(\mathbb{Z}[x],R)\cong R$ via $f\mapsto f(x)$.

In other words, the forgetful functor is represented by $\{\star\},\mathbb{Z},\mathbb{Z}[x]$ (the free object on the singleton set) respectively.
The same works in $G$-$\sf{Set}$ by taking $G$ as 'Cayley' $G$-set: 
$Hom(G,X)\cong X$ via $f\mapsto f(e)$.