The inversion map is a homomorphism precisely on abelian groups

Let $G$ be a group. Prove that the map $\varphi : G \rightarrow G$ given by $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian.

Solution:
($\Rightarrow$) Suppose $G$ is abelian. Then $$\varphi(ab) = (ab)^{-1} = b^{-1} a^{-1} = a^{-1} b^{-1} = \varphi(a) \varphi(b),$$ so that $\varphi$ is a homomorphism.
($\Leftarrow$) Suppose $\varphi$ is a homomorphism, and let $a,b \in G$. Then $$ab = (b^{-1} a^{-1})^{-1} = \varphi(b^{-1} a^{-1}) = \varphi(b^{-1}) \varphi(a^{-1}) = (b^{-1})^{-1} (a^{-1})^{-1} = ba,$$ so that $G$ is abelian.