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I never used the fact the $X$ is compact in my proof below. Which makes me worry if my proof is complete.

Let $X$ and $Y$ be metric spaces, $X$ compact, and $T: X \to Y$ bijective and continuous. To show that $T^{-1}$ is continuous, we will proceed by contradiction. That is, suppose that $T^{-1}$ is discontinuous. Since $T$ is bijective, then every element $y \in Y$ is uniquely determined by an element $x \in X$, \emph{viz.}, $Tx = y$ or equivalently $x = T^{-1}y$. By continuity of $T$, if $(x_n)$ is a convergent sequence in $X$ with limit point $x_0 \in X$, then $Tx_n = y_n \to Tx_0 = y_0$. However, this would imply that if $y_n \to y_0$, then $T^{-1}y_n = x_n \to T^{-1}y_0$. A contradiction, since $T^{-1}$ was assumed to be discontinuous.

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marked as duplicate by J.-E. Pin, José Carlos Santos proof-verification Oct 3 at 10:01

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    $\begingroup$ Do you really want to put the tag of functional analysis?? Is there some linear map between linear spaces somewhere? $\endgroup$ – Praphulla Koushik Sep 25 at 18:47
  • $\begingroup$ I put it since it came from my Functional Analysis book by Kreyszig. I can add a linear algebra tag. $\endgroup$ – Taylor McMillan Sep 25 at 18:52
  • $\begingroup$ I don’t think that’s how it works.,, linear algebra is irrelevant as there are no linear map here.. $\endgroup$ – Praphulla Koushik Sep 25 at 18:53
  • $\begingroup$ That makes sense. $\endgroup$ – Taylor McMillan Sep 25 at 18:55
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You have already assumed $(T^{-1}(y_n))=(x_n)\rightarrow x_0=T^{-1}(y_0)$. That’s why it looks that $T^{-1}$ is continuous.

What you have to do is, take an arbitrary sequence $y_n\rightarrow y$ and prove that $T^{-1}(y_n)\rightarrow T^{-1}(y)$.

How to use compactness of $X$? Remember that, as $X$ is compact, every sequence has convergent subsequence. From $y_n\rightarrow y$ you can get a sequence $(x_n)$ in $ X$ with $f(x_n)=y_n$ (because of surjective property). This has a convergent subsequence, can you finish from here?

Another hint : In case you can prove $(x_n)$ is Cauchy, then you are done. A Cauchy sequence is convergent if it has a convergent subsequence. By above observation, it is clear that it has convergent subsequence. Showing that Cauchy, and using the injective property, you will see that $(x_n)\rightarrow x$. Thus $T^{-1}$ is continuous.

As $T$ is continuous, $x_{n_k}\rightarrow x_0$ which imply $T(x_{n_k})\rightarrow T(x_0)$; as $T(x_{n_k})=y_{n_k}\rightarrow y$, uniqueness of limits says that $T(x_0)=y_0$. This says there exists a subsequence of $(x_n)$ that converge to inverse image of $y_0$. How do you see whole sequence converge?

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  • $\begingroup$ I have attempted this. Let $y_n \to y_0 \in Y$. We want to show that $T^{-1}(y_n) \to T^{-1}(y_0)$. By the subjectiveness of $T$ we can pick a sequence $x_n \in X$ with $Tx_n = y_n$. By compactness of $X$, there exists a convergent subsequence $x_{n_k} \to x_0$. By continuity of $T$, we then have that $Tx_{n_k} = y_{n_k} \to Tx_0 = y_0$. I am just not sure if I can then just take the inverse of everything and then say that I am done. $\endgroup$ – Taylor McMillan Sep 25 at 18:40
  • $\begingroup$ Where did you fail in your attempt? How do you know $T(x_0)=y_0$?. $\endgroup$ – Praphulla Koushik Sep 25 at 18:40
  • $\begingroup$ Awww. Continuity only tells me that $Tx_{n_k} \to Tx_0$. $\endgroup$ – Taylor McMillan Sep 25 at 18:50
  • $\begingroup$ Yes.. do you see if it is Cauchy?? $\endgroup$ – Praphulla Koushik Sep 25 at 18:50
  • $\begingroup$ Trying to work through that now. $\endgroup$ – Taylor McMillan Sep 25 at 18:51
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Your proof is not clear. Yes, if $x_n\to x_0$ then $T(x_n)\to T(x_0)=y_0$. But it is not necessary an "if and only if" relation. Anyway, you must use the fact that $X$ is compact because otherwise the statement is false. (take the identity map from the discrete metric space in $\mathbb{R}$ to the standard metric space).

Here is a proof. It is enough to show that if $F\subseteq X$ is closed then $(T^{-1})^{-1}(F)$ is a closed set in $Y$. Since $X$ is compact and $F$ is a closed subset we know that $F$ is compact as well. Continuous functions preserve compact sets, hence $T(F)$ is compact in $Y$. But $Y$ is a Hausdorff space (since it is a metric space), so it follows that $T(F)$ is closed in $Y$. This means $(T^{-1})^{-1}(F)$ is closed.

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Your proof is surely incorrect, as you need compactness. Consider the identity function from $\Bbb R$ with the discrete topology to $\Bbb R$ with the standard topology.

You just need to prove that $f$ is open. Let $U\subset X$ be open. Then $U^c$ is compact. Then $f(U^c)\subset Y$ is compact. Since $Y$ is a metric space, it's Hausdorff and $f(U^c)$ is closed. Then $f(U)=f(U^c)^c$ is open.

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$T$ is closed because $C \subseteq X$ closed implies $C$ compact, so $T[C]$ is compact by continuity of $T$ so $T[C]$ is closed (as $Y$ is metric).

And a closed continuous bijection is a homeomorphism.

Or if you want to go the sequence route: suppose $y_n \to y$ in $Y$.

Let $x_n,x$ be the unique points in $X$ such that $T(x_n)=y_n$ and $T(x)=y$.

As $X$ is compact there is $x_0 \in X$ and a subsequence $x_{n_k}$ such that $x_{n_k} \to x_0$, and then continuity of $T$ implies $$y_{n_k}=T(x_{n_k}) \to T(x_0)$$ and as $y_{n_k} \to y$ (as subsequences have the same limit as the total sequence) and as limits are unique we know that $T(x_0)=y=T(x)$ so $x=x_0$. So $T^{-1}(y_n) \to T^{-1}(y)$ and $T^{-1}$ is continuous.

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