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?

Question

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: (A Mere Write-Up / Presentation Of The Proof Based On The Answers Below)

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 $f^{-1} ( \ \{y\} \ )$. That is, for each $y \in Y$, there exists a finite subcollection $\mathscr{A}_y$ of $\mathscr{A}$ such that $$ f^{-1} ( \ \{ y \} \ ) \subset \bigcup_{A \in \mathscr{A}_y } A. \tag{0} $$ For each $y \in Y$, let us put $$ U_y \colon= \bigcup_{A \in \mathscr{A}_y } A, \qquad \mbox{ and } \qquad C_y \colon= X \setminus U_y. \tag{A} $$ Then $U_y$ is an open set in $X$ and $C_y$ is a closed set; moreover we also have $$ f^{-1} ( \ \{ y \} \ ) \subset U_y, \tag{1} $$ and so $$ f^{-1} ( \ \{ y \} \ ) \cap C_y = \emptyset. \tag{2} $$ Now as $f \colon X \to Y$ is a closed map and as $C_y$ is a closed set in $X$, so the image set $f \left( C_y \right)$ is a closed set in $Y$. Let us put $$ W_y \colon= Y \setminus f \left( C_y \right). \tag{B} $$ Then $W_y$ is an open set in $Y$.

Now if $x \in f^{-1}\left( W_y \right)$, then by definition $x \in X$ is such that $f(x) \in W_y$, which implies that $f(x) \not\in f \left(C_y \right)$, [Refer to (B) above.] and so $x \not\in C_y$, which in turn implies that $x \in U_y$. [Refer to (A) above.] Therefore we have $$ f^{-1} \left( W_y \right) \subset U_y. \tag{3} $$

As $f$ is surjective, so, for each $y \in Y$, the set $f^{-}( \ \{ y \} \ )$ is non-empty, that is, there exists at least one $x_y \in X$ for which $$ y = f \left( x_y \right). \tag{4} $$ Any such $x_y \in f^{-1}( \ \{ y \} \ )$ of course, and so any such $x_y \not\in C_y$ because of (2) above.

So if $y \in f \left( C_y \right)$, then $y = f(v)$ for some element $v \in C_y$, which would imply that $v \in f^{-1}( \ \{ y \} \ ) \cap C_y$, which contradicts (2) above. Thus $y \not\in f \left( C_y \right)$, and therefore $ y \in W_y$. [Refer to (B) above.]

Thus we have shown that, for each $y \in Y$, there exists an open set $U_y$ in $X$ and there exists an open set $W_y$ in $Y$ such that $$ y \in W_y \ \qquad \mbox{ and } \qquad f^{-1} \left( W_y \right) \subset U_y; \tag{5} $$ [Refer to (3) above.] moreover the set $U_y$ is the union of some finite subcollection $\mathscr{A}_y$ of the open covering $\mathscr{A}$ of $X$. [Refer to (0) above.]

In this way, we obtain an open covering of $Y$, which is as follows: $$ \left\{ \ W_y \ \colon \ y \in Y \ \right\}. $$ And, since $Y$ is compact, this open covering has a finite subcollection that also covers $Y$; let $$ \left\{ \ W_{y_1}, \ldots, W_{y_n} \ \right\} $$ be this finite subcollection. Then we have $$ Y = \bigcup_{j=1}^n W_{y_j}. \tag{6} $$

Now we show that $$ X = \bigcup_{j=1}^n U_{y_j}. \tag{7} $$ Let $x$ be an arbitrary point of $X$. Let $y \colon= f(x)$. Then $y \in Y$, and so by (6) above we have $y \in W_{y_k}$ for at least one $k = 1, \ldots, n$; that is, $f(x) \in W_{y_k}$; this implies that $x \in f^{-1} \left( W_{y_k} \right)$ and hence $x \in U_{y_k}$, because of (3) above; therefore $$ x \in \bigcup_{j=1}^n U_{y_j}. $$ Thus one of the inclusions required for (7) above to hold holds; the other inclusion follows from the fact that, by our construction of these sets, each set $U_{y_j}$, for $j = 1, \ldots, n$, being a union of subsets of $X$, is itself a subset of $X$. [Refer to (0) and (1) above.]

Finally, as $X$ is a union of the collection $\left\{ \ U_{y_1}, \ldots, U_{y_n} \ \right\}$ consisting of finitely many sets and as each set $U_{y_j}$, for $j = 1, \ldots, n$, in this collection is itself a union of a finite subcollection $\mathscr{A}_{y_j}$ of our original open covering $\mathscr{A}$ of $X$ [Refer to (A) above.], so we can conclude that $X$ is eventually the union of the finite subcollection $\mathscr{A}_{y_1} \cup \ldots \cup \mathscr{A}_{y_n}$ of $\mathscr{A}$.

Since $\mathscr{A}$ is an arbitrary open covering of $X$, we can conclude that $X$ is compact.

Is my write-up correct? If so, then is it clear and easy enough to understand? If not, then where are the problems?

Answer 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 $\mathscr A(y_j)$.

Answer 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.

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

Answer 3

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.