In trying to find a proof for the proposition below i came up with the following solution, which i haven't found on the internet. Can anyone tell me, if this proof is correct?
Proposition: If $f$ is a strictly increasing function, then $f^{-1}$ is also strictly increasing.
Proof: Let $(a,f(a)))$ and $(b,f(b))$ be two arbitrarily pairs in $f$ with $a<b$. Since $f$ is strictly increasing this implies that $f(a)<f(b)$. Now, the pairs $(f(a),a))$ and $(f(b),b))$ are in $f^{-1}$ with $f(a)<f(b)$. To complete the proof we must verify, that this implies that $a<b$. But we know already that this is true by assumption. $qed$