I am reading about cup products and am stuck on this exercise in Hatcher (3.2.5). Taking as given that $H^*(\mathbb{R}P^\infty,\mathbb{Z}_2)\simeq\mathbb{Z}_2[\alpha]$, how does one show $H^*(\mathbb{R}P^\infty,\mathbb{Z}_4)\simeq \mathbb{Z}_4[\alpha,\beta]/(2\alpha,2\beta,\alpha^2)$ with $\deg(\alpha)=1,\deg(\beta)=2$?
The cohomology groups are $\mathbb{Z}_4$ in position 0 and $\mathbb{Z}_2$ in every other position. I want to find the cup product structure. To do that, I need to find a cochain map $f:C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_4)\rightarrow C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_2)$ induced by the ring map $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$.
The two cochain complexes look like
$ \cdots {\leftarrow}\mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\stackrel{2}{\leftarrow} \mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\leftarrow 0 $
and
$ \cdots {\leftarrow}\mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\leftarrow 0 $
Here are my questions:
$\bullet$ what is the induced cochain map?
$\bullet$ Can someone tell me why $\alpha\cup \alpha=0$?
$\bullet$ Why does $\alpha\cup \beta$ generate in degree 3?
Is there a quick way to answer these geometrically?
Looking at complexes we see that the induced map of cohomology groups is an isomorphism in even degrees and zero in odd degrees (so the notation is slightly misleading: $\alpha$ maps to $0$ and not to $\alpha$). So $\alpha^2=0$ (since $f(\alpha^2)=f(\alpha)^2=0)$ and powers of $\beta$ generate even-dimensional cohomology.
The remaining question is why $\alpha\beta^n$ is non-zero. From Gysin exact sequence (see Hatcher 4.D) for the sphere bundle $S^1\to S^\infty\to\mathbb RP^\infty$, $$ \ldots\to H(S^\infty)=0\to H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)\to H(S^\infty)=0\to\ldots, $$ we see that multiplication by $\beta$ gives (for $k>0$) an isomorphism $H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)$. QED. (But certainly Hatcher had some other argument in mind...)