Let $f: A\to B$ and that $f$ is a bijection. Show that the inverse of $f$ is bijective.
Surjectivity: Since $f^{-1} : B\to A$, I need to show that $\operatorname{range}(f^{-1})=A$. But since $f^{-1}$ is the inverse of $f$, and we know that $\operatorname{domain}(f)=\operatorname{range}(f^{-1})=A$, this proves that $f^{-1}$ is surjective.
Injectivity: I need to show that for all $a\in A$ there is at most one $b\in B$ with $f^{-1}(b)=a$. But we know that $f$ is a function, i.e. for all $a\in A$ there is exactly one (at least one and never more than one) $b\in B$ with $f(a)=b$. 'Exactly one $b\in B$' obviously complies with the condition 'at most one $b\in B$'.
Since $f^{-1}$ is the inverse of $f$, $f^{-1}(b)=a$. So combining the two, we get for all $a\in A$ there is exactly one (at least one and never more than one) $b\in B$ with $f^{-1}(b)=a$.
I think my surjective proof looks ok; but my injective proof does look rather dodgy - especially how I combined '$f^{-1}(b)=a$' with 'exactly one $b\in B$' to satisfy the surjectivity condition. Could someone verify if my proof is ok or not please? Thank you so much!