Group homomorphism/Inverse to inverse/Fact/Proof

Proof

To prove the first statement, consider

${}\varphi (e_{G})=\varphi (e_{G}e_{G})=\varphi (e_{G})\varphi (e_{G})\,.$

Multiplication with ${}\varphi (e_{G})^{-1}$ yields ${}e_{H}=\varphi (e_{G})$.
To prove the second claim, we use

${}\varphi {\left(g^{-1}\right)}\varphi (g)=\varphi {\left(g^{-1}g\right)}=\varphi (e_{G})=e_{H}\,.$

This means that ${}\varphi {\left(g^{-1}\right)}$ has the property that characterizes the inverse element of ${}\varphi (g)$. Since the inverse element in a group is, due to fact, uniquely determined, we must have ${}\varphi {\left(g^{-1}\right)}=(\varphi (g))^{-1}$.

To fact