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Consider a topological space $(X, \mathcal{T})$. Suppose $X$ is compact and $(Y, \mathcal{T}_Y)$ is Hausdorff. Let $\Phi: X \times Y \rightarrow Y$ be the projection map. We show that $\Phi$ is a closed mapping. Suppose $A \subset X \times Y$ is closed. Consider an arbitrary $y \notin \Phi(A)$ i.e $(x,y) \notin {A}$ for all $x \in X$. It follows that $X \times {y} \subset (X \times Y) - A$. As $X$ is compact, by the Tube Lemma there exists an open set $V$ of $y$ such that $X \times V \subset (X \times Y) - A$. This implies that $V \subset Y- \Phi(A)$. Since $Y- \Phi(A)$ is open, $\Phi(A)$ is closed.

Is this correct?

Is the Hausdorfness of Y relevant?

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Fatsho
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  • Please do not delete questions with good answers. This is not only vandalism of the site, but also impolite to those who have spent time answering your question. – robjohn Feb 24 '13 at 08:50
  • See also" http://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map/ – Martin Sleziak Mar 08 '14 at 06:49

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Just a couple of comments:

  • I know that some mathematicians write $X\times y$, but it’s very sloppy; $X\times\{y\}$ is much better.

  • Where you have ‘an open set $V$ of $y$ such that’, you really mean ‘an open nbhd $V$ of $y$ such that’.

Otherwise it’s correct. And since you did not use Hausdorffness of $Y$ at any point, that means that Hausdorffness of $Y$ is not needed.

Brian M. Scott
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