Let $f:M\to Y$ be a continuous proper bijective map from a metrizable space $M$ onto a $T\_1$-space $Y$. The properness of $f$ means that for every compact subspace $K\subseteq Y$ the preimage $f^{-1}[K]$ is a compact subset of $M$.


Let $g:H\to Y$ be any continuous map from a compact Hausdorff space $H$ into $Y$.



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> **Question.** Is the function $f^{-1}\circ g:H\to M$ continuous?
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**Remark 1.** The map $f^{-1}\circ g:H\to M$ is sequentially continuous.


*Proof.* Assuming that the map $h:=f^{-1}\circ g:H\to M$ is not sequentially continuous, we can find a sequence $(x\_n)\_{n\in\omega}$ in $H$ that converges to some point $x\in H$ but no subsequence of the sequence $(h(x\_n))\_{n\in\omega}$ converges to $h(x)$. Then the set $\{n\in\omega:h(x\_n)=h(x)\}$ is finite and we can assume that $h(x\_n)\ne h(x)$ and hence $g(x\_n)\ne g(x)$ for all $n\in\omega$. Since $Y$ is a $T\_1$-space, we can replace $(x\_n)\_{n\in\omega}$ by a suitable subsequence and assume that $g(x\_n)\ne g(x\_m)$ for any distinct numbers $n,m$. Then the sequence $(h(x\_n))\_{n\in\omega}$ consists of pairwise distinct points. Since the space $M$ is metrizable, we can replace $(x\_n)\_{n\in\omega}$ by a suitable subsequence and assume that $\{h(x\_n):n\in\omega\}$ is a discrete subspace of $M$.


By the continuity of $g$, the set $K=\{g(x)\}\cup\{g(x\_n)\}\_{n\in\omega}$ is compact in $Y$ and by the properness of $f$, $f^{-1}[K]=\{h(x)\}\cup\{h(x\_n)\}\_{n\in\omega}$ is a compact subset of $M$. Then some subsequence $(h(x\_{n\_k}))\_{k\in\omega}$ converges to some point $z$ of the compact set $f^{-1}[K]$. The choice of the sequence $(x\_n)\_{n\in\omega}$ ensures that $z\ne h(z)$. Then $z=h(x\_m)$ for some $m\in\omega$, which also is not possible as the space $\{h(x\_n):n\in\omega\}$ is discrete. $\square$


**Remark 2.** Requirement on $Y$ to be a $T\_1$-space is essential as shown by the following


**Example.** Let $M$ be the doubleton $\{0,1\}$ with discrete topology and $Y$ be the doubletin $\{0,1\}$ with the $T\_0$-topology $\{\emptyset,\{1\},\{0,1\}\}$. It is clear that the identity map $i:M\to Y$ is continuous and proper. Consider the compact metrizable subspace $H=\{0\}\cup\{\frac1n:n\in\mathbb N\}$ of the real line and the continuous function $g:H\to Y$ defined by $g(0)=0$ and $g(\frac1n)=1$ for all $n\in\mathbb N$. It is easy to see that the function $i^{-1}\circ g:H\to M$ is discontinuous.



The continuity of certain maps on compact Hausdorff spaces
The answer to this question is affirmative.



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> **Proposition.** Let $p:X\to Y$ be a proper bijective map from a Hausdorff topological space $X$ onto a $T\_1$-space $Y$. Then for every continuous map $f:K\to Y$ from a compact Hausdorff space $K$, the map $p^{-1}\circ f:K\to X$ is continuous.
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*Proof.* By the $T\_1$-property of $Y$ and the continuity of $f$, for every $x\in X$ the set $f^{-1}(p(x))$ is closed in $K$. By the compactness of $K$, for every closed set $F\subseteq K$ the image $f[F]$ is compact in $Y$ and by the properness of $p$, the preimage $p^{-1}[f[F]]$ is compact in $X$ and closed in $X$, by the Hausdorff property of $X$. This means that the map $p^{-1}\circ f:K\to X$ is closed and has compact preimages of points. The following lemma implies that $p^{-1}\circ f$ is continuous. $\quad\square$



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> **Lemma.** A function $f:X\to Y$ from a normal topological space $X$ to a topological space $Y$ is continuous if $f$ is closed, $f[X]$ is compact and for every $y\in Y$, the preimage $f^{-1}(y)$ is a closed subset of $X$.
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*Proof.* To see that the function $f:X\to Y$ is continuous, take any point $x\in X$ and any neighborhood $O\_y$ of its image $y=f(x)$ in $Y$.
Let $\mathcal F$ be the family of all closed sets $F\subseteq X$ containing the set $f^{-1}(f(x))$ in its interior in $X$. Since the map $f$ is closed, for every $F\in \mathcal F$ its image $f[F]$ is a closed subset of $Y$. We claim that $\bigcap\_{F\in\mathcal F}f[F]=\{y\}$. Indeed, given any point $z\ne y$, we obtain that $f^{-1}(z)$ is a closed subset of $X$, disjoint with the closed set $f^{-1}(y)$. By the normality of the space $X$, there exists a set $F\in\mathcal F$ such that $F\cap f^{-1}(z)=\emptyset$ and hence $z\notin f[F]$. Since the space $f[X]$ is compact and $\bigcap\_{F\in\mathcal F}f[F]=\{y\}\subseteq O\_y$, we can apply Corollary 3.1.5. from Engelking's "General Topology", and conclude that $f[F]\subseteq O\_y$ for some $F\in\mathcal F$. Then $F$ is a neighborhood of $x$ with $f[F]\subseteq O\_y$, witnessing that the function $f$ is continuous. $\quad\square$


