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This is a question from munkres section 26 problem 7. Show if Y is compact, then the projection $\pi_1:X \times Y \rightarrow X$ is a closed map. My question is why this is not trivial. Essentially we want that if $C_X \times C_Y$ is a closed subset of $X \times Y$ then, $\pi_1(C_X \times C_Y)=C_X$ is closed. But if $C_X \times C_Y$ is closed, then is it not the case that both $C_X $ and $ C_Y$ must be closed? Therefore $C_X$ must be closed. Thus $\pi_1$ is a closed map. Why do we even need $Y$ compact?

Akt904
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    Why do you think it's enough to consider closed sets of the form $C_X \times C_Y$? – Chris Eagle Feb 25 '13 at 07:35
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    The set ${\langle x,y\rangle:x^2+y^2\le 1}$ is a closed subset of $\Bbb R\times\Bbb R$ that is not of the form $A\times B$ for any sets $A,B\subseteq\Bbb R$. – Brian M. Scott Feb 25 '13 at 07:38
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    The projection of ${(x,y)\mid xy=1}$ from ${\Bbb R}^2$ into $\Bbb R$ is not closed. So, you definitely need $Y$ compact. – David Moews Feb 25 '13 at 07:40
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    I refer you to the answers on this question: http://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map – Potato Feb 25 '13 at 07:41

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The closed sets of $X \times Y$ in the product topology are complements of open sets. We know that the open sets are given by unions of sets of the form $A \times B$ where $A$, and $B$ are open sets in respectively $X$ and $Y$. Hence the closed sets are given by intersections of complements of sets of this form. However $(X \times Y) - A \times B $ is not $(X - A) \times (Y - B)$, since the complement is defined to be all elements in $X \times Y$ that are not in $A \times B$, we must have $$X \times Y - (A \times B) = (X - A) \times Y \cup X \times (Y - B) $$Hence the closed sets are intersections of sets of this form, however intersections aren't necessarily preserved under images. Hence you can't just consider sets of the form $C_X \times C_Y$. I think I have done this exercise in Munkres book some time ago, and if I am not mistaken, you can use the tube lemma to prove this.

Oliver E. Anderson
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