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.
Best Answer
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.