cell structure of $S^2\times S^2$ with $S^2\times \{p\}$ identified to a point I'm preparing for an exam and found this problem: Let $X$ be obtained from $S^2\times S^2$ by identifying $S^2\times \{p\}$ to a point, then what is $H^*(X, \mathbb{Z})$?
I think first of all I'm not sure what the cell structure of this is. I think it's got one 0-cell, one 2-cell, and a 4-cell right?
 A: As was pointed out above, this quotient has a cell structure as (0-cell, 2-cell, 4-cell) letting you compute the homology groups rather quickly:
$$
H_{*}(S^{2}\times S^{2}/S^{2},\mathbb{Z})=\begin{cases}
\mathbb{Z} & * = 0,4\\
\mathbb{Z} & * = 2\\
0 & else
\end{cases}
$$
As good practice for me, I verified the above with the following long winded approach.
$S^{2}\times S^{2}$ has has a cell decomposition of one 0-cell, two 2-cells, and one 4-cell. Hence it's homology is:
$$
H_{*}(S^{2}\times S^{2},\mathbb{Z})=\begin{cases}
\mathbb{Z} & * = 0,4\\
\mathbb{Z}^{2} & * = 2\\
0 & else
\end{cases}
$$
Now, $S^{2}\times \{p\}$ has an open neighborhood in which it is a deformation retraction (consider $S^{2}\times B_{\epsilon}(p)$). As such, we may apply the following long exact sequence of relative homology:
$$
...\rightarrow \tilde{H}_{k}(S^{2})\rightarrow \tilde{H}_{k}(S^{2}\times S^{2})\rightarrow \tilde{H}_{k}(S^{2}\times S^{2}/S^{2})\rightarrow \tilde{H}_{k-1}(S^{2})\rightarrow...
$$
with the obvious identification of $S^{2}\times\{p\}$ with $S^{2}$.
Now we compute: $S^{2}\times S^{2}$ is path connected. Passing to the quotient wont change that. So $$H_{0}(S^{2}\times S^{2}/S^{2})=\mathbb{Z}$$
Now $\tilde{H}_{0}(S^{2})=0$ and $\tilde{H}_{1}(S^{2}\times S^{2})=H_{1}(S^{2}\times S^{2})=0$. Hence by the long exact sequence
$$H_{1}(S^{2}\times S^{2}/S^{2})=0.$$
Now $H_{1}(S^{2})=0$ and $H_{2}(S^{2})=\mathbb{Z}$. Further, $S^{2}\times\{p\}$ is a generator of $H_{2}(S^{2}\times S^{2})$. By the ES
$$H_{2}(S^{2})\rightarrow H_{2}(S^{2}\times S^{2})\rightarrow H_{2}(S^{2}\times S^{2}/S^{2})\rightarrow 0$$
We have that
$$H_{2}(S^{2}\times S^{2}/S^{2})=\mathbb{Z}$$
Now the kernel of the last map in the above ES is trivial (since $H_{2}(S^{2})$ maps injectively into $H_{2}(S^{2}\times S^{2})$). SO we can split our long exact squence as
$$\rightarrow H_{3}(S^{2})\rightarrow H_{3}(S^{2}\times S^{2}) \rightarrow H_{3}(S^{2}\times S^{2}/S^{2})\rightarrow 0.$$
Checking above, all three of these groups must be 0. So
$$H_{3}(S^{2}\times S^{2}/S^{2})=0$$
Finally we have
$$0\rightarrow H_{4}(S^{2}\times S^{2})\rightarrow H_{4}(S^{2}\times S^{2})\rightarrow 0$$
So
$$H_{4}(S^{2}\times S^{2}/S^{2})=\mathbb{Z}$$
