Show that if f is strictly increasing on A, then inverse f is strictly increasing on B

Question

Let A, B ⊆ R, and let f : A → B be a bijective function. Show that if $f$ is strictly increasing on A, then $f^{-1}$ is strictly increasing on B.

How would I write this proof? I think by contradiction but I don't know where to start.

Answer 1

Suppose $x<y$. We have to show that $f^{-1}(x) <f^{-1}(y)$. Let us prove this by contradiction.

Suppose $f^{-1}(x) \geq f^{-1}(y)$. Since $f$ is increasing this implies $f(f^{-1}(x))\geq f(f^{-1}(y))$ which means $x \geq y$, a contradiction.

Answer 2

Let $x,y\in B$ with $x<y$.

Now find $u,v\in A$ with $x=f(u)$ and $y=f(v)$ (possible because $f$ is surjective).

From $x\neq y$ it follows that $u\neq v$ (because $f$ is injective).

$v<u$ would lead to the false statement $y=f(v)<f(u)=x$ (since $f$ is strictly increasing).

We conclude that $u<v$.

Now realize that $u=f^{-1}(x)$ and $v=f^{-1}(y)$ so actually we proved that:$$x<y\implies f^{-1}(x)<f^{-1}(y)$$q.e.d.