Hatcher exercise 3.2.14 - on the surjectivity of the map $q: \mathbb{R}P^2 \to \mathbb{C}P^1$ in cohomology $H^2$

Question

I'm working through some exercises from Hatcher and I'm having trouble understanding the first part of exercise 3.2.14, which goes as follows:

Let $q: \mathbb{R}P^{\infty} \to \mathbb{C}P^{\infty}$ be the natural quotient map obtained by regarding both spaces as quotients of $S^{\infty}$, modulo multiplication by real scalars in one case and complex scalars in the other. Show that the induced map $q^*:H^{*}(\mathbb{C}P^{\infty}) \to H^{*}(\mathbb{R}P^{\infty})$ is surjective in even dimensions by showing first by a geometric argument that the restriction $q:\mathbb{R}P^2 \to \mathbb{C}P^1$ induces a surjection on $H^2$ and then appealing to cup product structures. Next, form a quotient space $X$ of $\mathbb{R}P^{\infty} \sqcup \mathbb{C}P^n$ by identifying each point $x \in \mathbb{R}P^{2n}$ with $q(x) \in \mathbb{C}P^n$. Show there are ring isomorphisms $H^{*}(X;\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(2\alpha^{n+1})$ and $H^{*}(X; \mathbb{Z_{2}}) \cong \mathbb{Z_{2}}[\alpha, \beta]/(\beta^2 - \alpha^{n+1})$, where $|\alpha| = 2$ and $|\beta| = 2n+1$. Make a similar construction and analysis for the quotient map $q: \mathbb{C}P^n \to \mathbb{H}P^{\infty}$.

There are many things which are unclear to me in this exercise, but I'll start with the most obvious two:

1. What is the "restriction" $q: \mathbb{R}P^2 \to \mathbb{C}P^1$, and what is the map $q:\mathbb{R}P^{2n} \to \mathbb{C}P^n$, and maybe even more generally, what exactly is the map $q: \mathbb{R}P^{\infty} \to \mathbb{C}P^{\infty}$?

For example, my guess for the most general $q$ would be $$q(x_{0} : x_{1} : \cdots : x_{2n} : 0 : 0 : \cdots) = (x_{0} + ix_{1} : x_{2} + ix_{3} : \cdots : x_{2n-2} + ix_{2n-1} : x_{2n} : 0 : \cdots),$$ when the number of "relevant" elements is odd and similarly when the number is even. However, even though this map really maps $\mathbb{R}P^{2n}$ to $\mathbb{C}P^n$, it doesn't hit the imaginary parts of the last dimension! This seems pretty weird to me, is that the way it's supposed to be? It would be more natural for me to consider $q: \mathbb{R}P^{2n+1} \to \mathbb{C}P^n$ for me since those are both quotients of $S^{2n+1}$. This brings me to my second question.

2. Why am I supposed to consider $q: \mathbb{R}P^2 \to \mathbb{C}P^1$ and not $\mathbb{R}P^3 \to \mathbb{C}P^1?$ I want information about the behavior of $q: \mathbb{R}P^{\infty} \to \mathbb{C}P^{\infty}$ in $H^2$, so why shouldn't I consider the three-skeletons instead of the two-skeletons? If I remember correctly, the behavior of a cellular map in $n$th homotopy, homology etc. depends on the $(n+1)$-skeletons, not just the $n$-skeletons!

Obviously, my other questions would be something along the lines of:

3. How do I solve this problem? (probably asking too much!) Can someone give me the outline of the "geometric argument" and a hint for calculating the ring structures? For the ring structure, my basic idea would be to consider $X$ as a CW-complex where the $2n$-cells of $\mathbb{R}P^{2n}$ and $\mathbb{C}P^n$ have been "melded into one" in some way. Is this intuition correct, and how do I use it to calculate the ring structure?

P.S. I'd like an ideal answer to answer the first two questions fully and the third one at least partially, but I'll upvote anything that's helpful! Also, if there are no responses to this question in the next 48 hours, I will be opening a bounty.

Answer 1

$\mathbb{C}P^\infty$ is the quotient of $S^\infty$ by the $S^1$ action given by the complex vector space structure on $\mathbb{C}^\infty$ (i.e. $\mathbb{C}$ acts on all of $\mathbb{C}^\infty$ and the unit circle $S^1 \subset \mathbb{C}$ acts on $S^\infty$). There is a restricted action on $S^\infty$ of the subgroup $\{1, -1\}\subset S^\infty$, and the quotient by this action is $\mathbb{R}P^\infty$. The function $q\colon\mathbb{R}P^\infty \to \mathbb{C}P^\infty$ sends a $\mathbb{Z}/2$-orbit to its $S^1$-orbit; in terms of $1$-dimensional subspaces, if $L$ is a $1$-dimensional subspace of $\mathbb{R}^\infty \cong_\mathbb{R} \mathbb{C}^\infty$, then $q(L) = \mathbb{C} \cdot L$. The map $\mathbb{R}P^{2n} \to \mathbb{C}P^n$ is a "restriction" in the sense that it is $q$ restricted to the $2n$-skeleton.

It's true in general that if $f\colon X \to Y$ is a map of CW complexes then the induced map on cohomology groups $H^n(f)\colon H^n(Y) \to H^n(X)$ is determined by the restriction of $f$ to the $(n+1)$-skeleton. However in this case the situation is nice enough that the full ring homomorphism $q^*\colon H^*(\mathbb{C}P^\infty) \to H^*(\mathbb{R}P^\infty)$ is determined by the restriction of $q$ to the $2$-skeleton, since the single generator of $H^*(\mathbb{C}P^\infty)$ (as a polynomial ring) lives in degree $2$ and the inclusions $\mathbb{R}P^{2n} \to \mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{n} \to \mathbb{C}P^{n+1}$ induce isomorphisms on $H^{2n}$.

I've forgotten the geometric argument the show it's surjective on $H^2$ lol. I will think about it and try to at least give you a hint, because I'd like to remember how this part works.