cell structure of $S^2times S^2$ with $S^2times {p}$ identified to a point

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 $$0rightarrow H_{4}(S^{2}times S^{2})rightarrow H_{4}(S^{2}times S^{2})rightarrow 0$$ So $$H_{4}(S^{2}times S^{2}/S^{2})=mathbb{Z}$$