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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.
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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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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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