Define $X \wedge Y := X \times Y/\left(\lbrace x_0\rbrace \times Y \cup X \times \lbrace y_0\rbrace\right)$. I'd like to prove that $I^n/\partial I^n \simeq \mathbb{S}^n$, the unit sphere.
To do so I'd like to using induction with the help of :
Lemma : if $X,Y$, are Hausdorff compact spaces then the Alexandroff compactification of $\overset{\wedge}{\left(X \setminus \lbrace x_0\rbrace \times Y \setminus \lbrace y_0\rbrace \right)}$ (where the wedge simbol indicates the compactification) is homeomorphic to $X \wedge Y$.
This is a neat proof of $I^n/\partial I^n \simeq \mathbb{S}^n$ but I don't get the $n=2$ step, which apparently involves using :
$$I^2/\partial I^2 \simeq I/\partial I \wedge I/\partial I$$
I thought I could do a map from $I/\partial I \times I/\partial I \longrightarrow I^2/\partial I^2$ which is costant on $I/\partial I \times \lbrace x_0\rbrace \cup \lbrace y_0\rbrace \times I/\partial I$ but seems rather complicated.
Any help with the proof or alternative solution to $I^2/\partial I^2 \simeq I/\partial I \wedge I/\partial I$ would be appreciated.
Best Answer
You can apply the lemma also for the $n = 2$ step. Just take $X = Y = I/\partial I$ and $x_0 = y_0 = *$ = equivalence class of $\partial I$. Then $X \setminus \{x_0\} = Y \setminus \{y_0\} = (0,1)$, hence $(X \setminus \{x_0\}) \times (Y \setminus \{y_0\}) = (0,1)^2$. This is an open square and its one-point compactification is clearly $I^2/\partial I^2$.