Let $f:R \rightarrow S$ be an isomorphism of rings.
Let $g:S \rightarrow R$ be the inverse function of $f$.
Show that $g$ is also an isomorphism.
I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.
because $f$ is a bijection, there are $r,r' \in R$ such that $f(r)=s$ and $f(r')=s'$. So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.