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. $qed$
Best Answer
With the given answer and tips i make a second try. Hope this get's better. I included the existence of $f^{-1}$ in the proposition, since i didn‘t want to proof it here.
Proposition: If $f$ is a strictly increasing function and it’s inverse $f^{-1}$ exists, then $f^{−1}$ is also strictly increasing.
Proof: Let $f:X\to Y$ be a strictly increasing function with inverse $f^{-1}$. Then for all $x_1,x_2\in X$ we have unique $y_1=f(x_1),y_2=f(x_2) \in Y$ with $x_1<x_2 \implies f(x_1)<f(x_2)\iff y_1<y_2$. We now have to show that $y_1<y_2\implies f^{-1}(y_1)<f^{-1}(y_2)$. But since, per definition of the inverse function, $f^{-1}(y_1)=x_1$ and $f^{-1}(y_2)=x_2$. So we already know this is true. $\blacksquare$
Any tips or possible improvements welcome.