Let $X$ and $Y$ be metric spaces, $X$ compact, and $T: X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism.

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

$T$ 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.

Answer 4

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.