Section 26: Problem 12 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

In this exercise we are asked to prove that $X$ is compact given that $f$ is continuous, surjective and closed, $Y$ is compact, and the preimages of all singletons are compact. We prove a little more: if $f$ is closed, and the preimage of each point is compact, then the preimage of any compact subset is compact.

When we can guarantee that the preimage of any compact set is compact? Let us try to collect minimum requirements. A constant function is an example that shows that we may need to require that the preimage of any point is compact. So, suppose it is. Let $B$ be a compact subset of $Y$ , and $A=f^{-1}(B)$ . We know nothing about $A$ : it need not even be closed (neither $Y$ is assumed to be Hausdorff, nor $f$ is assumed to be continuous, so far). Let $\{U_{\alpha}\}$ be an open covering of $A$ . $\cup_{\alpha}U_{\alpha}$ covers $\cup_{b\in B}f^{-1}(\{b\})$ . We know that for each $b$ there is a finite subcovering $\{U_{j}(b)\}$ of $f^{-1}(\{b\})$ . If we only could group all $b$ ’s in a finite number of groups $\cup_{i}V(b_{i})=B$ where $V(b_{i})$ contains $b_{i}$ and is such that $\{U_{j}(b_{i})\}$ covers not only $f^{-1}(\{b_{i}\})$ but $f^{-1}(V(b_{i}))$ as well, then we are done. Take any collection of neighborhoods over all $b\in B$ . Since $B$ is compact, there is a finite subcollection that covers $B$ . Therefore, we just need to find a neighborhood $V(b)$ for each $b$ such that $\{U_{j}(b)\}$ covers $f^{-1}(V(b))$ as well. In other words, what we need for $f$ is the property that if an open set $U$ contains $f^{-1}(\{b\})$ , then there is an open neighborhood $V$ of $b$ such that $f^{-1}(V)\subset U$ . Now, this property is equivalent to the requirement that any $b$ is an interior point of $Y-f(X-U)$ where $U$ is open and contains $f^{-1}(\{b\})$ . The requirement that $f$ is closed is obviously sufficient for this. (How did we use the continuity of $f$ ? I think we did not.)

If $f:X\to Y$ is closed, $y\in Y$ , and $U$ is an open subset of $X$ such that $f^{-1}(\{y\})\subset U$ , then there is an open neighborhood $V$ of $y$ such that $f^{-1}(V)\subset U$ . In other words, if $f$ is closed, then every open set containing the preimage of a point, contains the preimage of some neighborhood of the point.

Consider $V=Y-f(X-U)$ . $V$ is clearly open. Further, since $U$ contains all points mapped to $y$ , $y\notin f(X-U)$ , and $y\in V$ . Moreover, for every $z\in V$ , $z$ is not the image of any point in $X-U$ , hence, $f^{-1}(\{z\})\subset U$ .

From the lemma, as discussed above, we conclude the following.

If $f:X\to Y$ is closed, and for every $y\in Y$ , $f^{-1}(\{y\})$ is compact, then the preimage of every compact set is compact. If further $f(X)$ or $Y$ itself is compact, then $X$ is compact.

Let $B\subset Y$ be compact. Suppose $\{U_{\alpha}\}$ is an open cover of $f^{-1}(B)$ . For every $b\in B$ , $f^{-1}(\{b\})$ is compact, hence, there is a finite subcover $\{U_{\alpha_{j}^{b}}\}_{j=1}^{n_{b}}$ of $f^{-1}(\{b\})$ . Then, $U^{b}=\cup_{j=1}^{n_{b}}U_{\alpha_{j}^{b}}$ is an open set containing $f^{-1}(\{b\})$ , and, by lemma, there is an open neighborhood $V^{b}$ of $b$ such that $f^{-1}(V^{b})\subset U^{b}$ . $\{V^{b}\}$ covers $B$ , and there is a finite subcover $\{V^{b_{i}}\}_{i=1}^{n}$ . Therefore, $f^{-1}(B)\subset$ $f^{-1}(\cup_{i=1}^{n}V^{b_{i}})=$ $\cup_{i=1}^{n}f^{-1}(V^{b_{i}})\subset$ $\cup_{i=1}^{n}\cup_{j=1}^{n_{b_{i}}}U_{\alpha_{j}^{b_{i}}}$ , a finite subcover of $\{U_{\alpha}\}$ .

If $f:X\to Y$ is continuous and closed, $X$ is compact, and $Y$ is a $T_{1}$ -space, then the preimage of every compact set is compact.

For every $y\in Y$ , $\{y\}$ is closed, and $f^{-1}(\{y\})$ is closed, hence, compact. The result follows from the theorem above.

This corollary generalizes the following well-known fact: if $f:X\to Y$ is a continuous function from a compact space to a Hausdorff space, then it is closed, and the preimage of every compact set is compact. If we assume that $Y$ is just a $T_{1}$ -space (without additional assumption that $f$ is closed), then we cannot say that every compact subset of $Y$ is closed, and, hence, its preimage is closed and compact in $X$ . So, we need this additional assumption explicitly. Consider, for example, $f:[0,1]\to[0,1]_{fc}$ where the range is in the finite complement topology.

Subpages

Section 26: Problem 12 Solution

Section 26: Problem 12 Solution