Let $f:X\to Y$ and $g:Z\to Y$ be continuous maps of locally compact Hausdorff second-countable topological spaces. Assume that $f$ is proper (the preimage of compact set is compact). Is it necessarily true that $X\times_Y Z\to Z$ is a closed map (here $X\times_Y Z$ is the fiber product i.e. the pullback of f and g in TOP)? I know it is a standard result that $f$ is "universally closed" in this case (i.e. $X\times_{pt} Z\to Y\times_{pt} Z$ is always closed), but here I am working with a slightly different notion of "universally closed" and I'm not sure if the two are equivalent.
Proper implies “universally closed”
general-topology
Best Answer
You have closed immersions $i: X \times_Y Z \rightarrow X\times_{pt} Z$, $j: Z \rightarrow Z \times_{pt} Y$, and you know that $f’: X \times_{pt} Z \rightarrow Y \times_{pt} Z$ is closed.
Let $b: X \times_Y Z \rightarrow Z$, then $j \circ b=f’ \circ i$.
Let $F \subset X \times_Y Z$ be a closed subset. Then $f’(i(F))$ is a closed subset of $Z \times Y$, and it is $j(b(F))$. As $j$ is a closed immersion, $b(F)$ is closed in $Z$. QED.