# Bijective hom sets

Let $f: G \to H$ be a group homomorphism. Suppose that the induced map $F: \text{Hom}(H,H) \to \text{Hom}(G,H)$ defined by $F(\phi) \stackrel{\text{def}}{=} \phi \circ f$ is a bijection. Show that if $G$ is abelian, then so is $H$.

I'm wondering if there is a fancy categorical proof of this theorem.

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The proof I know you show f(G) is contained in Z(H), so that all inner automorphisms in H agree with the identity on f(G). Hence they are all the identity and H is abelian. But it looks like something nicer should exist. –  Joe Aug 15 '11 at 1:02
What I'm hoping (and I have no reason to) is that Yoneda lemma somehow applies, because I'd like a better understanding of it. –  Joe Aug 15 '11 at 1:03
It's quite easy to show that the condition on the hom-set map implies $f$ is an epimorphism. But at the moment I'm not seeing a good abstract-nonsense proof that epimorphisms transport abelian group structure... –  Zhen Lin Aug 15 '11 at 2:03
An epimorphism in the category of groups is necessarily surjective... –  Keenan Kidwell Aug 15 '11 at 3:19
@Zhen: there are tons of surjective functions from a group $G$ to a set $X$ which are not homomorphisms for any group structure on $X$... –  Mariano Suárez-Alvarez Oct 23 '11 at 23:03