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I am trying the solve the following problem:

Let $g:\mathbb{C}P^\infty\longrightarrow \mathbb{C}P^\infty$ and suppose the induced homomorphism

$$g^*:H^2(\mathbb{C}P^\infty)\longrightarrow H^2(\mathbb{C}P^\infty).$$

is a bijection. Show that the induced map on $H^n(\mathbb{C}P^\infty)$ is bijective for all $n$.

My experience with cohomology is still a bit low and I have been trying to proceed just from definitions to solve the problem, but I have not made any progress beyond the fact that we need only check only even dimensions.

I'm not looking for a solution, just hints in the right direction. Any help is appreciated.

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Do you know the cohomology ring structure of $\mathbb{C}P^n$? –  Chris Gerig Apr 28 '13 at 17:00
    
@chrisgerig yes –  James Vincent Apr 28 '13 at 17:05
    
Then how do you get $H^n$ from $H^2$ here? Heard of naturality? –  Chris Gerig Apr 28 '13 at 18:03
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Dear James, I understand what you mean but you should reformulate your question and explain what $n$ is. –  Georges Elencwajg Apr 28 '13 at 21:01
    
@Chris: I have heard of naturality and I don't see how it applies here. –  Georges Elencwajg Apr 28 '13 at 21:06
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1 Answer 1

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The cohomology of $\mathbb CP^\infty$ is given as a ring by

$$H^*(\mathbb CP^\infty)=\mathbb Z[T]$$ with $T$ of degree $2$. Homogenous polynomials of degree $d$ correspond to elements in $H^{2d}(\mathbb CP^\infty)$. The map $g^*$ is a ring homomorphism and you know what it does in degree $1$.

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