For any based topological spaces $X, Y$ does it follow that $X^+ \wedge Y^+ \cong (X \times Y)^+$? In another answer given here (https://math.stackexchange.com/a/17968/266135 ) the following is stated : "There is a natural homeomorphism $(X\times Y)^+ \cong X^+ \wedge Y^+$ for any spaces $X$ and $Y$"
I'm assuming that the writer of that answer notationally means that $(\cdot)^+$ is the one-point compactification of a topological space.  
However the one-point compactification is only defined for locally compact Hausdorff topological spaces and thus in the category of pointed topological spaces it is only defined for locally compact Hausdorff based topological spaces.
In the answer given above it's stated that $(X\times Y)^+ \cong X^+ \wedge Y^+$ for any spaces $X$ and $Y$ which I'm assuming means any based topological spaces. How is this so given that the one-point compactification isn't even defined for general based topological spaces.
Furthemore what exactly is this natural homeomorphism from $(X\times Y)^+ \cong X^+ \wedge Y^+$?
 A: In the answer to Why do we use the smash product in the category of based topological spaces? the notation $X^+$ definitely means the disjoint union of $X$ and a single point space as explained in Tyrone's comment. One-point compactifications exist only for locally compact spaces, so this interpretation  would not make sense for an arbitrary $X$ (as you noticed yourself).
However, for a locally compact $X$ let us denote by $X^\infty$ its one-point compactification $X \cup \{ \infty \}$ which is given $\infty$ as a basepoint. There is canonical continuous map $u : X^\infty \times Y^\infty \to (X \times Y)^\infty$ which maps $X \times Y \subset X^\infty \times Y^\infty$ via the identity onto $X \times Y \subset (X \times Y)^\infty$ and maps $X^\infty \times \{ \infty \} \cup \{ \infty \} \times Y^\infty$ to $\{ \infty \}$. Therefore it induces a continuous bijection  $u' : X^\infty \wedge Y^\infty \to (X \times Y)^\infty$ on the quotient space $X^\infty \wedge Y^\infty$ of $ X^\infty \times Y^\infty$. This space is comapct, therefore $u'$ is a homeomorphism. 
