Inverse of a one-to-one and continuous function is continuous Show that if $f$ is continuous on an interval $[a,b]$ and one-to-one, then $f^{-1}$ is also continuous.
This is from Stephen Abbott's Understanding Analysis, Exercise 4.5.8.
I've read multiple proofs online but I can't seem to understand any.
What I've done so far:
$f$ is continuous, so it preserves connectedness and compactness which implies $f([a,b]) = [\alpha, \beta]$, for some $\alpha, \beta \in \mathbb{R}$.
$f$ is monotone, so we can assume WLOG that $f$ is strictly increasing and so $f^{-1}$ is strictly increasing. We know $f^{-1}$ exists since $f$ is one-to-one.
I've also done a proof which does not require $f$ to be continuous and I've read that continuity of $f$ is not required. Why is this the case? I'm looking for a proof which uses the continuity of $f$.
One more thing I tried:
Lets say we have $(y_n) \to c$, for some $c \in [\alpha, \beta]$, where $(y_n) \subset [\alpha, \beta]$. Lets assume $f^{-1} (y_n)$ does converge and to some $f^{-1} (d) \in [a,b]$, where $f^{-1} (d) \neq f^{-1} (c)$. By continuity of $f$, we have $(y_n) \to d$, where $d \neq c$, by injectiveness. This is a contradiction to our hypothesis. The issue with this proof is that we must first show that $f^{-1} (y_n)$ does indeed converge.
Thank you.
 A: Closed interval $[a, b]$ is compact, so every sequence has a converging subsequence. In our case the limit point is the same for all such subsequences and equal to $f^{-1}(c)$ (just like in your reasoning).
Now it's easy to prove (f.i. by contradiction) that thus the whole sequence converges to this limit.
A: I learned undergraduate analysis from Abbott as well, so I think the following argument should be something you can follow given you are in chapter 4 of Abbott. It just uses the fact that compact is the same as closed and bounded and the fact that continuous is equivalent to sequentially continuous. I cannot offer a remedy to your proof, but I can offer a proof that you might be able to understand (you mention you can't seem to understand any proofs that you have read) given your position in Abbott's book.
To show that $f^{-1}$ is continuous, we need to show that for any convergent sequence $(y_n)_{n\in\mathbb{N}}$ (converging to, say, $y$) in the domain of $f^{-1}$, we have that $f^{-1}(y_n)$ converges to $f^{-1}(y)$. This is by the sequential characterization of continuity that Abbott gives.
Since $f$ is continuous and its domain is compact, we have that the image of $f$ is also compact, hence closed. So, we have that $y = f(x)$ for some $x$, since $y$ is a limit point of the range of $f$.
Now, note that $(y_n)_{n\in\mathbb{N}}$ can be written as $\big(f(x_n)\big)_{n\in\mathbb{N}}$ for some $x_n$ in the domain of $f$. So we are trying to show that $x_n$ converges to $x$.
So, we have a sequence $(x_n)_{n\in\mathbb{N}}$ in $[a, b]$, which is a compact set. Thus, there is a convergent subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)_{n\in\mathbb{N}}$, who converges to, say, $x'$.
Now, by the continuity of $f$, we have that $f(x_{n_k})$ converges to $f(x')$. However, we have that $\big(f(x_{n_k})\big)_{k\in\mathbb{N}}$ is a subsequence of the convergent sequence $\big(f(x_n)\big)_{n\in\mathbb{N}}$, hence has the same limit. That is, we have that $f(x') = f(x)$. Lastly, we use injectivity of $f$ to conclude the result.
I would suggest reconciling your proof with this proof. Since your proof idea is to use injectivity to conclude strictly increasing, you could probably modify the above proof to say that $f(x') = f(x)$ immediately from the fact that $f$ is strictly increasing. Anyway, it would be a good exercise to try and modify the above proof to fit with your idea.
