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In the topology book I'm reading I found the following statement:

The "smash product" (of two pointed spaces) is defined as $X \bigwedge Y=X \times Y/(X \times \lbrace*\rbrace \bigcup Y \times \lbrace * \rbrace)$. If $X$ and $Y$ are compact, then $X \wedge Y$ is the one point compactification of $(X\setminus \lbrace*\rbrace) \times (Y \setminus \lbrace * \rbrace)$

(If I could prove this statement, it's easy to see that $S^p \bigwedge S^q \approx S^{p+q}$)

I don't know how to prove it and I hope, that someone can help.

Martin Sleziak
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Rungo
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  • this explaned in Hatchers book on Algebraic topology where the author defines the smash prodct – Norbert Oct 28 '13 at 19:39
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    I'd like to know, are the base points assumed to be closed, and does compact include Hausdorff in the book you are reading? – Stefan Hamcke Oct 28 '13 at 21:44
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    I'm reading Bredon's book "topology and geometry". On p. 199 you can find the problem. But there he didn't say that the spaces are Hausdorff. But Bredon defined the one point compactification only for locally compact Hausdorff spaces. So I think X and Y should be Hausdorff (and therefore the base points are closed) – Rungo Oct 28 '13 at 22:10
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    Today I was in the library and found Hatchers book. But I could only find an explanation for the following statement (on p.10): $S^p \bigwedge S^q \approx S^{p+q} $ but I'm much more interested in the other statement. Or did he explain it on another page? – Rungo Oct 30 '13 at 20:03
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    This may now be found spelled out as Prop. 1.6 in Tyrone Cutler's note "The Category of Pointed Topological Spaces" https://ncatlab.org/nlab/show/one-point+compactification+intertwines+Cartesian+product+with+smash+product#Cutler20 – Urs Schreiber Dec 14 '20 at 09:54

1 Answers1

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Let us assume $X$ and $Y$ are Hausdorff (otherwise, I'm not sure what "the one point compactification" is supposed to mean). Then $X\wedge Y$ is compact Hausdorff, being the quotient of the compact Hausdorff space $X\times Y$ by a closed equivalence relation (the equivalence relation is closed because it is the union of the diagonal in $(X\times Y)^2$ and the set $((X\setminus \{*\}) \times (X \times \{*\} \cup Y \times \{ * \})^2$, both of which are closed in $(X\times Y)^2$). Write $Z=(X\setminus\{*\}\times Y\setminus\{*\})$; then composing the inclusion $Z\to X\times Y$ with the quotient map $\pi:X\times Y\to X\wedge Y$ gives an injection $Z\to X\wedge Y$ whose image is the complement of the basepoint. It now suffices to show this map $Z\to X\wedge Y$ is an open map, so that it is a homeomorphism onto its image and so $X\wedge Y$ is a compact Hausdorff space obtained by adjoining one point to a space homeomorphic to $Z$. So suppose $U\subseteq Z$ is open. Then $U$ is also open as a subset of $X\times Y$, since $Z$ is open in $X\times Y$. Also, $\pi^{-1}(\pi(U))=U$. By definition of the quotient topology, this implies $\pi(U)$ is open in $X\wedge Y$.

Eric Wofsey
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