[Math] Proof that a ring homomorphism is isomorphic if and only if there is another ring homomorphism such that there is an identity map on R and S

Question

A ring homomorphism $\phi:R \to S$ is an isomorphism if and only if there exists another ring homomorphism $\psi:S \to R$ such that $\psi\circ\phi:R\to R$ is the identity map on R and $\phi\circ\psi:S \to S$ is the identity map on S.

I'm struggling to find a proof to help my understanding, thank you.

Answer 1

$\impliedby$

$\phi$ is injective because it has kernel zero: if $x\in \ker(\phi)$, then $\phi(x)=0$ implies $x=\psi\phi(x)=0$

$\phi$ is surjective because, given $y\in S$, $\phi\psi(y)=y$, so $\psi(y)\in R$ maps to $y$.

$\implies$

In the other direction, you just check that $\phi^{-1}=\psi$.