To Understand The Cohomology ring $H^*(\mathbb{R}P^n;\mathbb{Z})$

Question

I am trying to understand the Cohomology ring structure of $\mathbb{R}P^n$ with $\mathbb{Z}$ coefficients using the Coholomogy ring of of $\mathbb{R}P^n$ with $\mathbb{Z}/2\mathbb{Z}$ coefficients. The approach seems to be looking at the Ring map $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$, it induces a group homomorphism $C^k(\mathbb{R}P^n;\mathbb{Z})\rightarrow C^k(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})\forall k\geq 0$, This will induce maps between the Cohomology groups, which in turn will give us that for $k$ even, $H^k(\mathbb{R}P^n;\mathbb{Z})$ and $H^K(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})$ becomes isomorphic (Using Naturality which I don't understand much).

Please don't flag it as a duplicate, I have seen the other answers on this site. I don't know Spectral Sequences. Also I am struggling to understand these proofs rigorously. If someone can explain the mentioned part above and if possible sketch the rest of the calculation of $H^*(\mathbb{R}P^n;\mathbb{Z})$, that would be great. Thanks in advance.

Answer 1

If you build $\mathbb{R}P^n$ with one cell in each dimension up to $n$, then the cellular chain complex looks like this: $$ ... \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z}. $$ If you choose your favorite ring $R$ and apply $\text{Hom}(-, R)$, you will get the cellular cochain complex with coefficients in $R$; the multiplication by 2 map in the old chain complex becomes the multiplication by 2 map $R \to R$ in the new cochain complex, and the same for the zero map. Also, if you have a ring map $f: R \to S$, it will induce a map on cellular cochain complexes. (Why? A cochain with coefficients in $R$ is a map $C_i \to R$, so just compose with $f$ to get a map $C_i \to R \to S$.) In particular, if you apply this to the quotient map $\mathbb{Z} \to \mathbb{F}_2$, you will get a commutative diagram $$ \begin{array}{ccccccccc} \cdots & \xleftarrow{2} &\mathbb{Z} & \xleftarrow{0} & \mathbb{Z} & \xleftarrow{2} & \mathbb{Z} & \xleftarrow{0} & \mathbb{Z} \\ & & \downarrow & & \downarrow && \downarrow && \downarrow \\ \cdots & \xleftarrow{0} & \mathbb{F}_2 & \xleftarrow{0} & \mathbb{F}_2 & \xleftarrow{0} & \mathbb{F}_2 & \xleftarrow{0} & \mathbb{F}_2 \end{array} $$ Each vertical map is the quotient map. Now take cohomology of these chain complexes: $$ \begin{array}{ccccc} \cdots & 0 & \mathbb{Z}/2 & 0 & \mathbb{Z} \\ & \downarrow & \downarrow & \downarrow & \downarrow \\ \cdots & \mathbb{F}_2 & \mathbb{F}_2 & \mathbb{F}_2 & \mathbb{F}_2 \end{array} $$ (I'm thinking of the terms in the top row as abelian groups, the terms in the bottom row as vector spaces over the field with 2 elements; hence the different notation for the same groups.) You can check that the vertical map in each even dimension is an isomorphism: use the fact that in the map of cochain complexes, the vertical maps were reduction mod 2, and that in the integral cohomology in dimension $2r$ (for any $r>0)$, the element $1 \in \mathbb{Z}$ represents the nonzero cohomology class.