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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?

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    $\begingroup$ I think the argument you should use to show that $\alpha \cup \beta$ is in the proof of theorem 3.12, where he shows that for a generator of $H^i(P^n)$ and a generator of $H^{n-i}(P^n)$, the cup product is non-zero. $\endgroup$ – user101036 Nov 21 '14 at 20:28
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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...)

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  • $\begingroup$ Thank you! this makes sense. but I'm very curious about what Hatcher has in mind here. I am thinking if one can get an isomorphism $H^1(\mathbb{R}P^\infty,\mathbb{Z}_4)\times H^2(\mathbb{R}P^\infty,\mathbb{Z}_4)\rightarrow H^3(\mathbb{R}P^\infty,\mathbb{Z}_4)$ out of $\mathbb{Z}_2$ isomorphisms and the module map $\mathbb{Z}_2\stackrel{.2}{\rightarrow}\mathbb{Z}_4$. $\endgroup$ – adrido Dec 30 '13 at 18:24

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