# Projection map being a closed map

Let $\pi: X \times Y \to X$ be a projection map where $Y$ is compact. Prove that $\pi$ is a closed map.

• First i would like to see a proof of this claim.

• I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold.

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 @Subramani: Welcome to the site. Have a nice time over here. – anonymous Feb 19 '11 at 4:28

There is a standard example for why some hypothesis on $Y$ is necessary: let $X=Y=\mathbb R$, and consider the closed subset $F=\{(x,y)\in \mathbb R\times\mathbb R:xy=1\}\subset\mathbb R\times\mathbb R$. What is its projection to the first factor?

In fact, one can prove that a space $Y$ is compact iff for all spaces $X$ the projection $X\times Y\to X$ is closed. So while compactness is not necessary (I think...) for the closedness of the projection for one $X$, it is necessary if you want all such projections to be closed.

As for the proof you want in the first bullet point... this is a standard exercise in topology: what have you tried?

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And to see that compactness isn't necessary for closedness of the projection for one $X$, let $X$ be discrete with 0 or more points. – Jonas Meyer Feb 18 '11 at 19:31
Heh. The empty space makes for a great example :) – Mariano Suárez-Alvarez Feb 18 '11 at 19:34

Suppose $Z \subset X \times Y$ is closed, and suppose $x_0 \in X \setminus \pi[Z]$. For any $y \in Y, (x_0, y) \notin Z$, and as $Z$ is closed we find a basic open subset $U(y) \times V(y)$ of $X \times Y$ that contains $(x_0, y)$ and misses $Z$. The $V(y)$ cover $Y$, so finitely many of them cover $Y$ by compactness, say $V(y_1),\ldots,V(y_n)$ do. Now define $U = \cap_{i=1}^{n} U(y_i)$, and note that $U$ is an open neighbourhood of $x_0$ that misses $\pi[Z]$. So $\pi[Z]$ is closed.

To see that the closed projection property implies compactness (sketch): suppose $X$ has the closed projection property along $X$, and let $\cal{F}$ be a filter on $X$. Define a space $Y$ that is as a set $X \cup \{\ast\}, \ast \notin X$, where $X$ has the discrete topology and a neighbourhood of $\ast$ is of the form $A \cup \{\ast\}$ with $A \in \cal{F}$. Then $D = \{(x,x): x \in X\}$ is a subset $X \times Y$ and closedness of the projection $p: X \times Y \rightarrow Y$ implies that some point $(x,\ast)$ is in its closure, and this $x$ is an adherence point of the filter.

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