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Jun 17, 2022 at 7:00 history edited David White
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Feb 28, 2014 at 19:17 answer added David White timeline score: 11
Feb 28, 2014 at 12:47 comment added Anonymous That answers my question. Thanks for reference.
Feb 28, 2014 at 3:50 comment added Ricardo Andrade If one assumes that both inclusions are closed, then this is true in the usual category of topological spaces. It follows from the statement of theorem 6 in Arne Strøm's article "Note on cofibrations II" (Mathematica Scandinavica, volume 22, 1968, pages 130-142, available at mscand.dk/article/view/10877), which states that the inclusion of the subspace $(A\times Y) \cup (X\times B)$ in $X\times Y$ is a cofibration. Since both maps $A\to X$ and $B\to Y$ are closed, the pushout $(A\times Y)\coprod_{A\times B} (X\times B)$ is necessarily a subspace of $X\times Y$.
Feb 27, 2014 at 22:55 history edited Anonymous CC BY-SA 3.0
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Feb 27, 2014 at 22:14 review First posts
Feb 27, 2014 at 22:29
Feb 27, 2014 at 21:56 history asked Anonymous CC BY-SA 3.0