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?

Solution 1:

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 $\mathscr A(y_j)$.

Solution 2:

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 $$ f^{-1}(W_{y_j}) = f^{-1}( Y \backslash f(X \backslash U_{y_j})) \subset X \backslash f^{-1}(f(X \backslash U_{y_j})) \subset X \backslash (X \backslash U_{y_j}) = U_{y_j}$$ 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.