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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?

George
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    You can (and should) use mathjax in your title, as well as in the text of your post. – Lee Mosher Jan 11 '24 at 15:47
  • I've added that now - what are the recommendations on when not to use MathJax in the title? I feel like i remember seeing that - is that more for big integrals/things that wouldn't fit in a line? – George Jan 11 '24 at 16:34
  • Whoops forgot to tag @LeeMosher – George Jan 11 '24 at 16:49
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    It's not so much an issue of mathjax as it is of readability. Big integrals/things look terrible on the math.stackexchange front page. Formulas not written in mathjax look terrible on the math.stackexchange front page. – Lee Mosher Jan 11 '24 at 17:13

1 Answers1

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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 $H^*(S^2\times S^2)$.

Julius J.
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  • Does the composition come from something natural? For instance attaching an $(n+m)$-cell $D^n \times D^m$, we have the $S^{n-1} \times D^m$ part mapped via the product of the attachment $\partial D^n = S^{n-1} \to X^{n-1}$ and then the natural "inclusion" where the $m$-cell was attached $D^m \to Y^m$ (for $Y^m$ the $m$-skeleton of $Y$, already constructed in our cellular realisation of some product $X \times Y$ when attaching $* \times D^m$ a smaller dimension cell) – George Jan 11 '24 at 17:15
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    Not sure if I understand, but $S^1\to $ is* the attaching map of the 2-cell in $S^2$, and the collapse $D^2\to S^2$ is the characteristic map of $D^2$ to the 2-skeleton of $S^2$. – Julius J. Jan 11 '24 at 17:22