bijective continuous map and the preimage of a convergent sequence

Question

At first my question

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Let $X,Y$ are metric spaces with $X$ compact also let $T : X \to Y$ be a bijective continuous map and $(y_n)_{n=1}^{\infty}$ a sequence of $Y$. Is it true the following fact? $$(y_n) \text{ convergent } \implies (T^{-1} y_n) \text{ Cauchy }$$

Or the more general? $$(y_n) \text{ Cauchy } \implies (T^{-1} y_n) \text{ Cauchy }$$

Is the assumption that $X$ is a compact space necessary?

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And now the story behind

I came across the following problem about compact metric spaces

Let $X,Y$ are metric spaces and moreover $X$ is compact, also let $T : X \to Y$ be a bijective continuous map.

Prove that $T^{-1}$ is continuous

I am aware of the proof by contradiction. However a tried to make a direct proof but I got stuck.

my attempt

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The map $T^{-1}$ is continuous if for every convergent sequence $(y_n) \subseteq Y$, with $\lim y_n=y$, it follows that $\lim T^{-1}y_n=T^{-1}y$

Let $(y_n)$ be a sequence of $Y$ such that $\lim y_n=y$

The fact that $X$ is compact and $T$ continuous implies that there is a subsequence, $(T^{-1} y_{k_n})$ , of $(T^{-1}y_n)$ such that $$\lim T^{-1} y_{k_n}=T^{-1} y$$

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This is where a got stuck.

If a prove that $(T^{-1}y_n)$ is a Cauchy sequence.Then the fact that it has a convergent subsequence implies that $(T^{-1}y_n)$ and moreover $\lim T^{-1} y_{k_n}=\lim T^{-1} y$

Thank you for your time

Answer 1

The answer to both questions is "yes", and the assumption that $X$ is compact is necessary for both. Foobaz John has already provided an elegant proof for the first question, but here is another: since $X$ is compact, it suffices to show that $T^{-1}y$ is the only limit point of $\{T^{-1}y_n\}$ where $y=\lim_{n\to\infty}y_n$. But if $T^{-1}y_{n_k}\to x_0$, by continuity of $T$ we have $y_{n_k}\to Tx_0$, so $y=Tx_0$, i.e. $x_0=T^{-1}y$. This completes the proof.

For the second question: since $X$ is compact and $Y=T(X)$, with $T$ continuous, we know $Y$ is also compact. In particular, $Y$ is complete, so $\{y_n\}$ Cauchy implies $\{y_n\}$ convergent. Now we can just defer to the first question.

To see compactness is necessary as an assumption, consider $X=[0,1)$, $Y=\{z\in\mathbb C:|z|=1\}$ and $T(x)=e^{2\pi i x}$. Then $T$ is continuous and bijective, but $T^{-1}$ is not continuous. Moreover, even if we knew $T^{-1}$ was continuous, this would not necessarily mean $\{y_n\}$ Cauchy implies $\{T^{-1}y_n\}$ Cauchy; for that we would need uniform continuity (which is automatic in compact spaces, providing another proof for the second question).

Answer 2

It suffices to prove that the map $T$ is closed in order to show that $T$ is a homeomoprhism. To this end, let $E$ be a closed subset of $X$. Since $X$ is compact, it follows that $E$ is compact. Since $T$ is continuous, the continuous image of a compact set is compact and hence $T(E)$ is compact. But compact sets are closed (since $Y$ is a metric space); hence, $T(E)$ is closed as desired.

For your first question, since $T:X\to Y$ is a homeomorphism (as we have shown), it follows that $y_n\to y$ implies that $T^{-1}(y_n)\to T^{-1}(y)$ so that $T^{-1}(y_n)$ is Cauchy (as convergent sequences are clearly cauchy). The second question is similar since $T(X)$ is compact and hence complete. Thus if $\{y_n\}$ is Cauchy in $Y$, then it converges and we can use the answer of the previous question. The assumption that $X$ is compact is necessary as there exist continuous bijective maps which are not homeomorphisms (for example, the map $f:[0,1)\to S^1$ given by $f(x)=(\cos2\pi x,\sin 2\pi x)$.