Cohomology Groups of Projective Space and Sphere I am stuck on the following question from Switzer’s Algebraic Topology: Homotopy & Homology:
Describe the cohomology groups of $\mathbb{R}P^2 \times \mathbb{R}P^2$ and $\mathbb{S}^m \times \mathbb{C}P^n$ for any integers $m,n$ greater than or equal to 1.
I have been thinking hard about it since last two days, but I cannot think about cohomology groups of the product space of 2-dimensional real projective spaces nor the product of m-dimensional sphere and complex projective space...Could you help me out?
 A: Here's the roadmap you'll want to take for calculating these groups. If you have any questions about the details (for example, about why the tensor product or Tor groups are what they are) feel free to follow up in the comments.
The Künneth formula for cohomology (linked by William in a comment) gives us that
$$H^k(\mathbb RP^2 \times \mathbb RP^2) \cong \left(\bigoplus_{i+j=k} H^i(\mathbb RP^2) \otimes H^j(\mathbb RP^2)\right) \oplus \left(\bigoplus_{i+j=k+1} \text{Tor}(H^i(\mathbb RP^2), H^j(\mathbb RP^2))\right).$$
When $k \geq 3$ all of the summands will be $0$. When $k = 2$ the left big summand is $\mathbb Z/2\mathbb Z$, and the right is $0$. When $k = 1$ the left big summand is $\mathbb Z/2\mathbb Z \oplus \mathbb Z/2\mathbb Z$, and the right big summand is $\mathbb Z/2\mathbb Z$, so the whole thing is $(\mathbb Z/2\mathbb Z)^{\oplus 3}$. Finally when $k = 0$ you obviously just get $\mathbb Z$.
For the other space I'll start with a slightly more general fact, namely that for any space $X$ we have
$$H^k(S^m \times X) \cong H^{k-m}(X) \oplus H^k(X).$$
You can show this by just writing out the formula again:
$$H^k(S^m \times X) \cong \left(\bigoplus_{i+j=k} H^i(S^m) \otimes H^j(X)\right) \oplus \left(\bigoplus_{i+j=k+1} \text{Tor}(H^i(S^m), H^j(X))\right).$$
Now, $H^i(S^m)$ is $\mathbb Z$ when $i = 0, m$ and $0$ otherwise, so the summands on the left are non-zero exactly $i = 0, m$, and then it simplifies to what I wrote above (note that the whole right summand will become $0$ since the cohomology of $S^m$ is always free).
So $H^k(S^m \times \mathbb CP^n) \cong H^{k-m}(\mathbb CP^n) \oplus H^k(\mathbb CP^n)$. The cohomology of $\mathbb CP^n$ is not too bad (see here), but it will be a little bit obnoxious to write out all the possible cases for $k, m, n$. Let me know if you need some more clarification.
