# Proving the inverse of a bijection is bijective

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!

• In stead of this I would recommend to prove the more structural statement: "$f:A\to B$ is a bijection if and only if it has an inverse". An inverse is a map $g:B\to A$ that satisfies $f\circ g=1_B$ and $g\circ f=1_A$. So it is immediate that the inverse of $f$ has an inverse too, hence is bijective. – drhab Nov 6 '15 at 9:55

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 $f^{-1}(f(a))=a$.
• Do you know about the concept of contrapositive? $b\neq b \implies f^{-1}(b)\neq f^{-1}(b)$ is logically equivalent to $f^{-1}(b)= f^{-1}(b)\implies b=b$ – cr001 Nov 6 '15 at 9:50
• There is never a need to prove $b\neq b \implies f^{-1}(b)\neq f^{-1}(b)$ because $b\neq b$ is never true in the first place. What we want to prove is $a\neq b \implies f^{-1}(a)\neq f^{-1}(b)$ for any $a,b$ – cr001 Nov 6 '15 at 13:03