# Prob 12, Sec 26 in Munkres' TOPOLOGY, 2nd ed: How to show that the domain of a perfect map is compact if its range is compact?

Let $X$ and $Y$ be topological spaces such that $Y$ is compact, and let $f \colon X \to Y$ be a closed, surjective, and continuous map such that, for each $y \in Y$, the inverse image $f^{-1} ( \ \{y \} \ )$ is compact.

Then how to show that $X$ is compact also?

My effort:

Let $\mathscr{A}$ be an open covering of $X$. Then, for each $y \in Y$, the collection $\mathscr{A}$ is a covering of $f^{-1} ( \ \{y\} \ )$ by sets open in $X$, and since $f^{-1} ( \ \{y\} \ )$ is compact, some finite subcollection of $\mathscr{A}$, say $\mathscr{A}(y)$, also covers $f^{-1} ( \ \{y\} \ )$.

How to proceed from here?

#### Solutions Collecting From Web of "Prob 12, Sec 26 in Munkres' TOPOLOGY, 2nd ed: How to show that the domain of a perfect map is compact if its range is compact?"

Continuing from your effort: Let $C(y)$ be the closed set of $X$ that is not covered by $\mathscr A(y)$. We know that $f(C(y))\subseteq Y$ is closed, which means that the complement $U(y)\subseteq Y$ of that again is open.

$\mathscr A(y)$ cover the fiber of $y$, so no point of $f^{-1}(\{y\})$ is in $C(y)$, and therefore $y \in U(y)$, so the $U(y)$ cover $Y$. Since $Y$ is compact, it can be covered by finitely many such $U(y)$, let’s say $U(y_i), 1 \leq i \leq n$ for some $n$. I claim that the corresponding $\mathscr A(y_i)$ for all $i$ cover $X$.

To prove it, take an $x_0 \in X$. The point $f(x_0)$ is in some $U(y_j)$. That means that $f(x_0) \notin f(C(y_j))$, which again means that $x_0 \notin C(y_j)$, which again means that $x_0$ is inside one of the open sets in $\mathscr A(y_j)$.

Let $\{ U_i \}_{i \in I}$ be an open cover of $X$. Then this is also an open cover of $f^{-1}(\{y\})$ for each $y \in Y$. Since $f^{-1}(\{y\})$ is compact, there exists a finite subset $I_y \subset I$, such that
$$f^{-1}(\{y\}) \subset \bigcup_{i \in I_y} U_i =: U_y \;$$
Since $U_y$ is open, $X \backslash U_y$ is closed subset of $X$, and since $f$ is a closed map, $f(X \backslash U_y)$ is a closed subset of $Y$. Note that $y \not\in f(X \backslash U_y)$. We define $W_y := Y \backslash f(X \backslash U_y)$. Now we see that $W_y$ is open in $Y$ and $y \in W_y$, so $W_y$ is an open neighbourhood of $y$. This means that $\{ W_y \}_{y \in Y}$ is an open cover of $Y$, and since $Y$ is compact, there exists a finite subset $\{ y_1, \ldots, y_m \} \subset Y$, such that
$$Y = \bigcup_{j=1}^m W_{y_j} \; .$$
We note that
$$f^{-1}(W_{y_j}) = f^{-1}( Y \backslash f(X \backslash U_{y_j})) = X \backslash f^{-1}(f(X \backslash U_{y_j})) \subset X \backslash (X \backslash U_{y_j}) = U_{y_j}$$
for each $j \in \{1, \ldots, m\}$, and from that it follows that
$$X = f^{-1}(Y) = \bigcup_{j=1}^m f^{-1}(W_{y_j}) \subset \bigcup_{j=1}^m U_{y_j} = \bigcup_{j=1}^m \bigcup_{i \in I_{y_j}} U_i \; ,$$
so we have found a finite subcover of $\{ U_i \}_{i \in I}$, which means, that $X$ is compact.

Please check all these steps carefully, I’m not completely sure, if everything is working.

Let $\{U_\alpha\}_{\alpha\in \Gamma}$ be an open cover of $X$. Then for each $y\in Y$, $f^{-1}(y)\subset \cup_{\alpha \in \Gamma}U_{\alpha}$. Since $f^{-1}(y)$ is compact, for each $y\in Y$, there exists a finite subcollection $\{U^y_{j}\}_{j=1}^n$ of $\{U_\alpha\}_{\alpha\in \Gamma}$ such that $f^{-1}(y)\subset\cup_{j=1}^nU^y_{j}$. So, $\cap_{j=1}^n(X-U^y_j)\subset X-f^{-1}(y)$. Notice that the left hand side of the previous expression is closed, so applying $f$ to both sides, the image remains closed. So using the fact that $f$ is surjective, we get, $f(\cap_{j=1}^n(X-U^y_j))\subset Y-y$. Therefore, $y\in Y-f(\cap_{j=1}^n(X-U^y_j))$, and $Y-f(\cap_{j=1}^n(X-U^y_j))$ is open. Now, taking union over $y\in Y$, we get, $Y=\cup_{y\in Y}(Y-f(\cap_{j=1}^n(X-U^y_j)))$. Now, since $Y$ is compact, there exists a finite collection of $y\in Y$ such that $Y=\cup_{k=1}^l(Y-f(\cap_{j=1}^n(X-U^{y_k}_j)))$. This implies
\begin{align} X = f^{-1}(Y) &=f^{-1}(\cup_{k=1}^l(Y-f(\cap_{j=1}^n(X-U^{y_k}_j)))) \\ &= \cup_{k=1}^l(f^{-1}(Y-f(\cap_{j=1}^n(X-U^{y_k}_j)))) \\ &= \cup_{k=1}^l(X-f^{-1}(f(\cap_{j=1}^n(X-U^{y_k}_j)))) \\ & \subset \cup_{k=1}^l(X-\cap_{j=1}^n(X-U^{y_k}_j)) \\ &=\cup_{k=1}^l\cup_{j=1}^nU^{y_k}_j. \end{align}
So, we’ve found a finite subcover, so $X$ is compact.