Cell Structure of $S^2 \times S^2$, and more generally applying theorem to product of CW-complexes

Question

My end goal is calculating the homology of $S^2 \times S^2$, comparing it to that of $\mathbb C P^2 \ \# \ \mathbb C P^2$, and using the cup product to show homology is insufficient as a comparison.

I'm struggling with calculating the first of these. I managed to come up with a CW structure on the connected sum, but for the first space I was trying to use Mayer-Vietoris (unsuccessfully - I didn't find the right homology).

I'm wondering how to apply this question on products of CW complexes - we use points $p$ and $q$, and abusing notation using the same letters for the space of that point. After a first attempt, I'm finding that we get all attachment maps the constant map to the point $(p,q)$ in the product, which would yield a decomposition (omitting the attachment map which is the same) $$ (p,q) \cup (D^2 \times q) \cup (p \times D^2) \cup (D^2 \times D^2) $$ which would be equal to $ S^2 \vee S^2 \vee S^4$ but that contradicts this question which says this isn't even homotopy equivalent to $S^2 \times S^2$, let alone homeomorphic.

What is incorrect about my application of this, with which I'm hoping to apply it to other products?

Answer 1

The key is being precise about what $(p,q)\cup (\{p\}\times D^2\cup D^2\times \{q\}) \cup(D^2\times D^2)$ really is.

The second $\cup$ should refer to gluing a $D^4$ onto $S^2\vee S^2$ via an attaching map $S^3\to S^2\vee S^2$. Different such maps yield different spaces, the notation hides this fact.

The following illustrates how different attaching maps lead to different spaces:

1. One gets $S^2\vee S^2\vee S^4$ when picking the constant attaching map.

2. To arrive at $S^2\times S^2$ we use $$ S^3 =\partial D^4 \cong \partial(D^2\times D^2) \cong S^1\times D^2 \cup_{S^1\times S^1}D^2\times S^1. $$ The two maps $\pi \circ \text{pr}_2\colon S^1\times D^2 \to \{p\}\times S^2$ and $\pi \circ \text{pr}_1\colon D^2\times S^1 \to S^2\times \{q\}$ first projecting away from $S^1$, then collapsing the boundary of $D^2$ then yield our attaching map.

In the end you can use e.g. cellular homology to arrive at $H^*(S^2\times S^2)$.