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Let $(X,d_X)$ and $(Y,d_Y)$ be compact metric spaces and $f:X\to Y$ be a continuous bijection. Prove that $f^{-1}:Y\to X$ is also continuous.

Since $(Y, d_Y)$ is compact, it is totally bounded. Thus, for every $\delta>0, \exists y_1, \dots, y_n\in X$ such that $X\subset \bigcup\limits_{k=1}^n \underbrace{\overline{B_\delta(y_k)}}_\text{closed ball about $x_k$}$. So let $\delta > 0$.

Since $(X, d_X)$ is totally bounded, $\forall\epsilon>0, \exists x_1, \dots, x_m\in X$ such that $X\subset \bigcup\limits_{k=1}^m {\overline{B_\epsilon(x_k)}}$. So let $\epsilon > 0$.

Let $y_0\in Y$, then $y_0$ is in some $\delta$-ball $B_\delta(y_i)$ for $1\le i \le n$. Since $f$ is continuous, $f^{-1}$ pulls back closed sets to closed sets, so that $f^{-1}\left(\overline{B_\delta(y_i)}\cap Y\right)\subset \bigcup\limits_{i=1}^m \overline{B_{\epsilon}(x_i)}\cap X$. Hence $f^{-1}$ is continuous.

Would appreciate if someone could look at my proof and let me know whether or not it is correct.

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  • $\begingroup$ $f$ continuous means $f$ pulls back open sets to open sets. So you start with a $y_0 \in Y$, and given is some $\varepsilon>0$ and you want some $\delta > 0$ such that $f^{-1}[B(y, \delta)]\ \subseteq B(f^{-1}(y_0), \varepsilon)$. You only know the left hand side is open, but produce no recipe to make $\delta$ such that the inclusion holds. $\endgroup$ – Henno Brandsma Oct 27 '17 at 6:08
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    $\begingroup$ Way easier is to use $f$ is a closed map (as it sends closed=compact subsets of $X$ to compact=closed subsets of $Y$). $\endgroup$ – Henno Brandsma Oct 27 '17 at 6:09
  • $\begingroup$ @HennoBrandsma Can you please take a look now? I've edited it. $\endgroup$ – sequence Oct 27 '17 at 7:05
  • $\begingroup$ You don't show at all it does hold. How do you make $\delta$ from $\varepsilon$? Whether you use closures of balls or open balls makes no difference. $\endgroup$ – Henno Brandsma Oct 27 '17 at 7:05
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    $\begingroup$ No, it doesn't follow from total boundedness alone. You would need completeness too. If you don't use the closed map argument you might go with sequences (or better nets, if you know about those). $\endgroup$ – Henno Brandsma Oct 27 '17 at 7:27
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The easiest way is to use that $f$ is a closed map: if $C \subseteq X$ is closed, then $C$ is compact as a closed subset of the compact $X$. $f$ continuous then implies that $f[C]$ is compact as well, and in a metric space we have that compact subsets are closed, so $f[C]$ is closed. Finally note that $f[C] = (f^{-1})^{-1}[C]$ so $f^{-1}$ pulls back closed sets to closed sets, so is continuous.

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  • $\begingroup$ Can you please clarify how does it follow from $f[C] = (f^{-1})^{-1}[C]$ that $f^{-1}$ pulls back closed sets to closed sets? $\endgroup$ – sequence Oct 27 '17 at 7:45
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    $\begingroup$ @sequence. For clarification call the inverse function of $f$ just $g: Y \to X$. We want to prove $g$ continuous so for any closed $C \subseteq X$ we want $g^{-1}[C]$ closed. But $g^{-1}[C] =f[C]$ by simple set theory. $\endgroup$ – Henno Brandsma Oct 27 '17 at 7:48

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