Section 31: Problem 6 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

Let $p:X\to Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$ . [Hint: If $U$ is an open set containing $p^{-1}(\{y\})$ , show there is a neighborhood $W$ of $y$ such that $p^{-1}(W)\subset U$ .]

$Y$ is clearly a $T_{1}-$ space. Now, take a closed set $B\subset Y$ and an open set $V\supset B$ . $A=p^{-1}(B)\subset U=p^{-1}(V)$ is a closed set in an open set. Find an open neighborhood $W$ of $A$ such that $\overline{W}\subset U$ . $p(X-W)$ is closed, and $B\subset Y-p(X-W)\subset p(W)\subset V$ , because $p^{-1}(B)\cap X-W=\emptyset$ and $p(X-W)\cup p(W)=Y$ ( $p$ is surjective). So, $B\subset W'\subset V$ , where $W'=Y-p(X-W)$ is open. Moreover, $p(\overline{W})\subset V$ is closed and contains $W'$ . Therefore, $\overline{W'}\subset V$ .

For an alternative solution, see §73, Lemma 73.3.

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Section 31: Problem 6 Solution

Section 31: Problem 6 Solution