Is the inverse of a bijective monotone function also monotone?

Question

If $f$ is bijective and monotonic function is $f^{-1}$ monotonic?

Here is my attempt at solving the question but I'm unsure wether it's the right way to proceed or not.

Mathematical translation of $f$ is bijective and monotonic:

$\forall x, f(x)\le f(x+1)$ or $f(x)\ge f(x+1)$

We have $f^{-1}(x)=x$ therefore and $x\le x+1$ therefore $f^{-1}$ is monotonic.

Answer 1

Hint #1: Let $f: D \rightarrow C$ be a bijective function with domain $D$ and codomain $C$. Then $f$ is monotonic implies that there exist $x, y \in D$ such that $$x \leq y \implies f(x) \leq f(y)$$ (if $f$ is monotonic increasing), or $$x \geq y \implies f(x) \geq f(y)$$ (if $f$ is monotonic decreasing).

Hint #2: If $f: D \rightarrow C$ is bijective, then $f^{-1}: C \rightarrow D$ is also bijective.

Hint #3: The compositions $$f(f^{-1}(x)) = x$$ and $$f^{-1}(f(x)) = x$$ hold.

Can you take it from here?

Answer 2

Let $U$ and $V$ be subsets of $\mathbb{R}$. Without loss of generality suppose that $f:U\rightarrow V$ is strictly increasing.

Suppose that $f^{-1}$ is not increasing, ie. there are $y_1<y_2$ in $V$ such that $f^{-1}(y_1)\ge f^{-1}(y_2)$. Let $x_1=f^{-1}(y_1)$ and $x_2=f^{-1}(y_2)$. Since $f$ is bijection we have$$f(x_1)=f(f^{-1}(y_1))=y_1<y_2=f(f^{-1}(y_2))=f(x_2)$$ which is contradiction because $x_1\ge x_2$.