TITLE: A step in computing the cohomology ring of $\mathbb{C}P^n$
QUESTION [6 upvotes]: On page 250 of Hatcher's Algebraic Topology, he uses a certain corollary to compute the cohomology ring of $\mathbb{C}P^n$. The relevant section is below for convenience:

I understand the proof except for his statement that once can deduce $H^{2i}(\mathbb{C}P^n)$ is generated by $\alpha^i$ for all $i<n$. The base case follows trivially from the induced isomorphism between $H^2(\mathbb{C}P^1)$ and $H^2(\mathbb{C}P^2)$. I tried working everything out after that, but in the end I arrive at $H^{2i} (\mathbb{C}P^n)$ is generated by $\alpha^i$ for all $i<n-1$, not $n$, as Hatcher says. I really don't see how one could use the fact that $H^2(\mathbb{C}P^2)$ is generated by $\alpha$ to show that $H^4(\mathbb{C}P^3)$ is generated by $\alpha^2$. How does one see this?

REPLY [4 votes]: To take the example at the end of your question, the fact that $H^4(\mathbb CP^3)$ is generated by $\alpha^2$ is deduced from the fact (part of the induction hypothesis) that $H^4(\mathbb CP^2)$ is generated by $\alpha^2$.  That's why Hatcher first recalls that the inclusion of $\mathbb CP^2$ in $\mathbb CP^3$ induces an isomorphism of cohomology up to dimension $4$, so that the full result for $\mathbb CP^2$ (including $H^4$) is available when proving the result for $\mathbb CP^3$.  (Analogous comments apply for general $n$ in place of $3$.)