This is question 3.2.7 in Allen Hatcher, Algebraic Topology.
Here is an answer to it (I found it online):
My questions about this solution:
1- I believe the reduced homology in the first line is incorrect as in the zero dimention, it should be zero, am I correct?
2- for the second line, what is the reduced cohomology of the sphere $S^3$ with coefficients in $\mathbb Z_2$? I also believe that the given formula is incorrect as the reduced cohomology in dimension zero should be zero, am I correct?
3- I also, do not understand why we assumed that $f^*$ is an isomorphism in the given commutative diagram, what is the intuition of building this commutative diaghram like this? could anyone explain this to me please?
4- Why we have $\alpha^3 = 0$ in case of $\mathbb R P^2 \vee S^3$?
6- In the last line in the paragraph before last, should not it be $(0,b)$ instead?
5- Why is the cohomology ring of the sphere looks like this $\mathbf{Z}/2\mathbf{Z}[z]/(z^2)$ and $|z|=3$, according to the following Question about showing $\mathbb{R}P^{3}$ is not homotopy equivalent to $\mathbb{R}P^{2} \vee S^3$. ?
Could someone help me answer this questions please?
Whenever you wedge two spaces, their cohomologies are going to add, with trivial cup product between them. The class $\alpha^2$ is the fundamental cohomology class of $RP^2$ and the next power is zero just by dimension arguments: $\alpha^3=0$. Therefore the same is true in the wedge of the spaces. The argument as stated above seems too wordy. The point is that $RP^3$ has cuplength 3 (with respect to $\mathbb Z_2$ coefficients), whereas the wedge has cuplenth 2 only.