Equivalence of cross products Kunneth formula

$\def\diaguparrow#1{\smash{ \raise.8em\rlap{\scriptstyle #1} \lower.3em{\mathord{\diagup}} \raise.52em{\!\mathord{\nearrow}} }}$

The pair $(\mathbb{R}^n,\mathbb{R}^n-\mathbb{R}^j)$ deformation retracts on its projection on $\mathbb{R}^i,$ the orthogonal complement of $\mathbb{R}^j$. This will make this pair into $(\mathbb{R}^i,\mathbb{R}^i-\{0\}).$ Expanding the "hole" $\{0\}$ and contracting $\mathbb{R}^i-\{0\}$ (as on the right) to a circle then gives the desired pair.

To convince oneself that the products are "the same", one can take a look at the diagram (I omit subspaces for simplicity): $\require{AMScd}$ \begin{CD} H^i(\mathbb{R}^n) \times H^j(\mathbb{R}^n) @>{\times}>> H^n(\mathbb{R}^n \times \mathbb{R}^n) @>{\Delta^*}>> H^n(\mathbb{R}^n)\\ @VVV @VVV \\ H^i(\mathbb{R}^i) \times H^j(\mathbb{R}^j) @>{\times}>> H^n(\mathbb{R}^i \times \mathbb{R}^j). \end{CD}

The upper row is by definition what was given initially. The lower row is what has to be shown to be equivalent. The left vertical isomorphism here is by definition how we identify the corresponding groups. What needs to be done is we need to identify those target copies of $H^n(\mathbb{R}^n)$ from both rows in a consistent way (to make this diagram commute when adding the isomorphism between these copies). But for the evident homeomorphism $s: \mathbb{R}^i \times \mathbb{R}^j \to \mathbb{R}^n$ the triangle commutes:

$\begin{CD} \mathbb{R}^n \times \mathbb{R}^n @<{\Delta}<< \mathbb{R}^n \\ @VVV \diaguparrow{s} \\ \mathbb{R}^i \times \mathbb{R}^j, \end{CD}$

meaning that $s^*$ can be taken as the desired identification.

The passage to $(I^i,\partial I^i)$ can be done similarly, fixing the identifications on the left and making sure that the identification on the right is consistent with them.