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If $f_i : X_i \to Y_i$ with $i=1,2,\ldots,n$ are closed maps between topological space it is known that their product map

$$f : X_1 \times \cdots \times X_n \to Y_1 \times \cdots \times Y_n : (x_1, \ldots, x_n) \mapsto (f_1(x_1), \ldots, f_n(x_n))$$ doesn't need to be closed. However the question is: are there some nice conditions on $X_i$, $Y_i$ (like compactness, connection, Hausdorff...) such that $f$ will be closed? I have looked in "Bourbaki - General Topology Part 1,2" and I have found nothing about it. By the way I'm more interested to the case $X_1 = \cdots = X_n$, $Y_1 = \cdots = Y_n$.

Thank you for your help!

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According to Remark 7.3.1.15 in Higher Topos Theory, a sufficient conditition is that the fibers of $f_i$ are compact (not necessarily Hausdorff). In fact this is equivalent to $f_i$ being universally closed, and by abstract nonsense a finite product of universally closed maps is universally closed. –  Marc Hoyois Apr 10 '12 at 17:46
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There is a useful notion of a proper map $f: X \to Y$. The definition is that $f$ is proper if for all spaces $Z$ the map $f \times 1_Z: X \times Z \to Y \times Z$ is a closed map. This notion comes I think from Bourbaki. You can find it on p. 90 of my book "Topology and groupoids" (2006). The first result on this is that if $f$ is a closed map such that $f^{-1}(y)$ is compact for all $y \in Y$, then $f$ is proper.

The Exercises to that Section contain quite a few further results on proper maps.

You should quickly be able to prove in your question that if the $f_i$ are all proper, then so is $f$, and hence $f$ is closed.

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So, if $X_1, \ldots, X_n$ are compacts, $Y_1, \ldots, Y_n$ are Hausdorff, $f_1, \ldots, f_n$ are closed and continuous then (for any $y \in Y_i$, $\{y\}$ is closed, $f_i^{-1}(y)$ is closed, $f_i^{-1}(y)$ is compact) $f$ is closed. Right? This will be a nice condiction for my purpouse. –  Richard Bonne Apr 10 '12 at 18:41
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If each $X_i$ is compact and each $Y_i$ is Hausdorff then $X:=X_1 \times \cdots \times X_n$ is compact and $Y:=Y_1 \times \cdots \times Y_n$ is Hausdorff. So any map $f:X \to Y$ is closed, whether it is a product of (closed) maps or not.

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