If $f$ is an increasing function, then $f^{-1}$ is increasing - Proof Critique

Question

I wrote a proof for this claim and wanted to get feedback on whether it was the correct approach. I've seen another proof of this claim that used contradiction, but since I'm not the biggest fan of using contradiction first that idea never came to me initially. Anyways my proof:

Claim: If $f$ is an increasing function, then $f^{-1}$ is increasing.

Proof: Choose $y_{1}, y_{2}$ in the domain of $f^{-1}$. Suppose $y_{1} < y_{2}$ and define $y_{1} = f(x_{1})$ and $y_{2} = f(x_{2})$.

Since $f$ is an increasing function, we know that for any $x_{1} < x_{2}$ that $$f(x_{1}) < f(x_{2}) \\ \Rightarrow\ y_{1} = f(x_{1}) < y_{2} = f(x_{2}) $$

Since we know the inverse for $f$ exists this then means $$ f^{-1}(y_{1}) = f^{-1}(f(x_{1})) < f^{-1}(y_{2}) = f^{-1}(f(x_{2})) \\ x_{1} < x_{2}$$

Therefore $f^{-1}$ is increasing.

Answer 1

One thing to add. You only have $x<y \implies f(x) < f(y)$. You need the other direction too, unless ofcourse you defined it as equivalent in the first palce. I would do it here with contradiction. saying if $f(x) < f(y)$ and $x>y$ we have a contradiction.

After that you can substitute like you did.

Answer 2

Your proof has errors. It can be rectified without using contradiction like this.

In the 3rd line: Since $f$ is increasing, so for any $x_1<x_2$ $\ldots$ (The bold part is the error)

You should write like this:

Since $y_1=f(x_1),y_2=f(x_2)$ with $y_1<y_2$ we should have $x_1<x_2$ since $f$ is increasing.

Now you can apply the fact that $x_1=f^{-1}(y_1), x_2=f^{-1}(y_2)$ to conclude that $f^{-1}(y_1)=x_1<x_2=f^{-1}(y_2)$