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!
Best Answer
Your proof is logically correct (except you may want to say the "at least one and never more than one" comes from the surjectivity of $f$) but as you said it is dodgy, really you just needed two lines:
(1) $f^{-1}(x)=f^{-1}(y)\implies f(f^{-1}(x))=f(f^{-1}(y))\implies x=y$.
(2) Let $a\in A$, then $f^{-1}(f(a))=a$.