Did not get the contradiction in case of $\mathbb C P^n$ Case?

Question

Here is the question I am trying to understand the last part of its solution:

Using the cup product structure, show there is no map $\mathbb R P^n \to \mathbb R P^m$ inducing a nontrivial map $H^1(\mathbb R P^m; \mathbb Z/ 2 \mathbb Z ) \to H^1(\mathbb R P^n; \mathbb Z/ 2 \mathbb Z )$ if $n > m.$ What is the corresponding result for maps $\mathbb C P^n \to \mathbb C P^m.$

I found a solution for it here No map $\mathbb{C}P^n \rightarrow \mathbb{C}P^m$ inducing a nontrivial map $H^2(\mathbb{C}P^m;\mathbb{Z}) \rightarrow H^2(\mathbb{C}P^n;\mathbb{Z})$ the contradiction of the last part is given in the third comment below the question but I did not get its idea. Could someone explain it to me please (what about $\alpha ^ {(m + 1)}$ should it be zero? If so, why?)

Answer 1

I guess I will answer this question because I wrote the comment in the linked post.

The corresponding result for complex projective spaces is the following.

There is no map $\Bbb CP^n\to\Bbb CP^m$ inducing non-trivial maps $H^2(\Bbb CP^m;\Bbb Z)\to H^2(\Bbb CP^n;\Bbb Z)$ if $n>m$.

The proof goes exactly the same as in real projective spaces.

Let $f:\Bbb CP^n\to\Bbb CP^m$ be a continuous map, and $\alpha$ generate $H^2(\Bbb CP^m;\Bbb Z)$, we will consider $f^\ast(\alpha)\in H^2(\Bbb CP^n;\Bbb Z)$.

Suppose $f^\ast$ is non-trivial, then $f^\ast(\alpha)=k\beta$, where $k\neq0$, and $\beta$ is a generator of $H^2(\Bbb CP^n;\Bbb Z)$.

We know that

• $\alpha^{m+1}=0$ Why? because the cohomology ring of $\Bbb CP^n$ is isomoprhic to $\Bbb Z[\alpha]/(\alpha^{m+1})$;
• $\beta^{m+1}\neq 0$. Why? because $m<n\implies m+1<n+1$, and the cohomology ring of $\Bbb CP^n$ is $\Bbb Z[\beta]/(\beta^{n+1})$.

But now we have $$0=f^\ast(0)=f^\ast(\alpha^{m+1})=f^\ast(\alpha)^{m+1}=(k\beta)^{m+1}=k^{m+1}\beta^{m+1}\neq 0$$ This is a contradiction. Therefore, $f^\ast$ must be trivial.

Side note: There is an analogous result for quaternionic projective spaces $\Bbb HP^n$ and can be proved by the same argument.

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To be more explicit on why $\beta^{m+1}\neq 0$, we note that $\beta^{m+1}\in H^{2(m+1)}(\Bbb CP^n;\Bbb Z)$, which is non-trivial because $m<n\implies m+1\le n\implies 2(m+1)\le 2n$. In fact $\beta^{m+1}$ generates $H^{2(m+1)}(\Bbb CP^n;\Bbb Z)$ by cohomology ring structure. The author of the linked question made a small mistake while using the degree of cohomology, which leads a bit of confusion. My comment there was meant to address this subtle point.