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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

4 Answers 4

up vote 17 down vote accepted

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
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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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I'll add a proof using nets. I think that nets are often useful, since we have good intuition about sequences in metric spaces and many things work very similarly for nets in general topological spaces. (For example, we know that a metric space is compact if and only if every sequence has a convergent subsequence. If we work with topological spaces, we have a similar characterization with nets: A topological space is compact if and only if every net has a convergent subnet.)

Proof. Let $C$ be a closed subset of $X\times Y$. We want to show that $\pi[C]$ is a closed subset of $X$.

Let $(x_d)_{d\in D}$ be a net in $X$ such that each $x_d$ belongs to $\pi[C]$ and $x=\lim_{d\in D} x_d$. We want to show that $x\in\pi[C]$.

Since $x_d\in\pi[C]$, we can choose (for each $d\in D$) a point $y_d\in Y$ such that $(x_d,y_d)\in C$. Now $(y_d)_{d\in D}$ has a convergent subnet $(y_e)_{e\in E}$. (This follows from compactness of $Y$.) This means that there is an $y\in Y$ such that $y=\lim_{e\in E} y_e$.

Now we have $\lim_{e\in E} x_e = x$ and $\lim_{e\in E} y_e = y$, which implies that $\lim_{e\in E} (x_e,y_e)=(x,y)$ and $(x,y)\in C$. Therefore $x\in\pi[C]$. $\hspace{2cm}\square$

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This proof is adapted from this lecture note. It is interesting to figure out that this statement is actually a reformulation of The Tube Lemma.

Let $C$ be a closed subset of $X \times Y$, we want to show that $\pi_{1}(C) \subset X$ is closed. To this end, we take any point $x \notin \pi_1(C)$ and show that there exists a neighborhood of $x$ which is disjoint from $\pi_1(C)$.

Since $x \notin \pi_1(C)$, the slice $\{ x \} \times Y$ is disjoint from $C$. Because $Y$ is compact, by The Tube Lemma (replace open with closed and contain with disjoint, respectively), there is a neighborhood $W$ of $x$ such that the whole tube $W \times Y$ is disjoint from $C$. Therefore, $W$ is the neighborhood of $x$ which is disjoint from $\pi_1(C)$, as desired.

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