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!
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$\begingroup$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 $f^{-1}(f(a))=a$.
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