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.
Best Answer
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 $X\wedge Y$.