Cohomology ring of $\mathbb{R}P^{\infty}$

Question

Here's an exercise from Hatcher: Show the ring $H^{*}(\mathbb{R}P^{\infty};\mathbb{Z}_{2k})$ is isomorphic to $\mathbb{Z}_{2k}[\alpha,\beta]/(2\alpha,2\beta,\alpha^2 - k \beta)$. Using cellular homology and the universal coefficient theorem, I'm able to show that $H^0(\mathbb{R}P^{\infty},\mathbb{Z}_{2k}) = \mathbb{Z}_{2k}$ and $H^i(\mathbb{R}P^{\infty},\mathbb{Z}_{2k}) = \mathbb{Z}_2$ for $i \neq 0$. However, I don't know how to mimic Hatcher's proof of 3.19 to show the relation $\alpha^2 = k \beta$ in this ring.

I know this question was asked before on this site, but I don't quite understand the answer. The problem is that I have hard time analyzing the maps in the commutative diagram

where the author claims that the map $H^i(P^n,P^n-P^j;\mathbb{Z}_{2k}) \to H^i(\mathbb{R}^n,\mathbb{R}^n - \mathbb{R}^j;\mathbb{Z}_{2k})$ is multiplication by $k$. I'd appreciate a detailed analysis of this diagram. And how could one use this information to study the original cup product $H^i(\mathbb{R}P^{\infty}) \times H^j(\mathbb{R}P^{\infty}) \to H^{i+j}(\mathbb{R}P^{\infty})$ where $i,j$ are both odd or both even.

Answer 1

You're asking two questions: implicitly, you're asking how to solve Hatcher's 3.2.5, and explicitly, you're asking about a particular solution from this site. I'm going to address the first of these, not the second.

Let's start with this: for any topological space $X$ and any homomorphism of commutative rings $R \to S$, there is an induced map $H^*(X;R) \to H^*(X;S)$ which is compatible with cup products. We apply this to the ring map $\mathbb{Z}/2k \to \mathbb{Z}/2$. Let's look at cellular cochains: $$ \begin{array}{cccccccccc} \xleftarrow{2} & \mathbb{Z}/2k & \xleftarrow{0} & \mathbb{Z}/2k & \xleftarrow{2} & \mathbb{Z}/2k & \xleftarrow{0} & \mathbb{Z}/2k & \xleftarrow{} & 0 \\ & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ \xleftarrow{2} & \mathbb{Z}/2 & \xleftarrow{0} & \mathbb{Z}/2 & \xleftarrow{2} & \mathbb{Z}/2 & \xleftarrow{0} & \mathbb{Z}/2 & \xleftarrow{} & 0 \\ \end{array} $$ The induced map on cohomology: $$ \begin{array}{ccccccccccc} \mathbb{Z}/2 & & \mathbb{Z}/2 & & \mathbb{Z}/2 & & \mathbb{Z}/2k & & 0 \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ \mathbb{Z}/2 & & \mathbb{Z}/2 & & \mathbb{Z}/2 & & \mathbb{Z}/2 & & 0 \\ \end{array} $$ We know that $H^*(\mathbb{R}P^\infty; \mathbb{Z}/2) \cong \mathbb{Z}/2[x]$ with $x \in H^1$. Let $\alpha$ and $\beta$ be the nonzero elements in $H^1(\mathbb{R}P^\infty; \mathbb{Z}/2k)$ and $H^2(\mathbb{R}P^\infty; \mathbb{Z}/2k)$, respectively. We immediately see that $2\alpha=0$ and $2\beta=0$.

Each vertical map in the cochain diagram sends 1 to 1, and $\alpha$ is represented by $k \in \mathbb{Z}/2k$, so $\alpha \mapsto kx$. The element $\beta$ is represented by $1 \in \mathbb{Z}/2k$, so $\beta \mapsto x^2$. So if $k$ is odd, we find that $\alpha \mapsto x$ and $\alpha^2 = \beta = k \beta$. If $k$ is even, then $\alpha \mapsto 0$ so $\alpha^2 \mapsto 0$. On the other hand, $\beta \mapsto x^2 \neq 0$, so $\alpha^2 \neq \beta$. This means that $\alpha^2 = 0 = k\beta$.

Similar reasoning shows that $b^n \mapsto x^{2n}$ and $\alpha \beta^n \mapsto kx^{2n+1}$ for each $n$.

Edit: maybe some more work is required to fully analyze $\alpha \beta^n$ when $k$ is even. That is, you know that $\alpha \beta^n \mapsto 0$ in this case, so you have to explain why $\alpha \beta^n$ is actually nonzero.