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The following is a problem from a topology qualifying exam I am studying for:

Show there does not exist a map $S^2\times S^2\to \mathbb{CP}^2$ with odd degree.

I think I am doing something wrong, because I am getting that the degree of any such map must be zero. I argue as follows: By the universal coefficient theorem, the degree is the same whether we compute it on homology or on cohomology. Hence the degree of our map, lets call it $f$, is the same as computing the homomorphism $f^*:H^4(\mathbb{CP}^2)\to H^4(S^2\times S^2)$. We know that the cohomology ring of $\mathbb{CP}^2$ is isomorphic to $\mathbb{Z}[\alpha]$ where $|\alpha|=2$, hence a generator of the top cohomology is $\alpha^2=\alpha\cup \alpha$, hence we need only compute $f^*(\alpha \cup \alpha)$. But the cup product is functorial, so we have $f^*(\alpha\cup \alpha)=f^*\alpha \cup f^*\alpha$.

However, we note that $f^*\alpha$ is an element of $H^2(S^2\times S^2)$, and we claim that the square of any element in this cohomology is zero. Indeed, the projection map $p:S^2\times S^2\to S^2$ by projecting on either coordinate induces the map $p^*$ which is an isomorphism on $H^2$ by considering cellular cohomology. But clearly the square of any element in $H^2(S^2)$ is zero, since it will be an element of $H^4(S^2)=0$. Hence $f^*\alpha \cup f^*\alpha=0$, so the degree is zero.

Am I making some mistake here?

rondo9
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    You are mistaken only in the last paragraph. The square of an element in $H^2(S^2\times S^2)$ is not zero, but even. Let $\alpha_1, \alpha_2$ be generators of the 2nd cohomology of the $S^2$'s pull-backed to $S^2\times S^2$. Then $\alpha_1\cup\alpha_2$ generates $H^4(S^2\times S^2)$ (by Kunneth). Then for all $a_1, a_2\in\mathbb Z$, $(a_1\alpha_1+a_2\alpha_2)^2=2a_1a_2\alpha_1\cup\alpha_2.$ – Gil Bor May 13 '14 at 18:03
  • A map between two orientable manifolds $f:M\to N$ of dimension $n$ induces a map on their top homologies, and acts on generators for those top homologies as multiplication $d$ for some integer $d$: $f^*[M]=d\cdot [N]$. The degree of $f$ is defined to be $d$. – rondo9 May 13 '14 at 18:04
  • Thanks, Gil. That is what I was missing. – rondo9 May 13 '14 at 18:05
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    Related: Minimal Degree of map $S^2\times S^2\to \mathbb{CP}^2$ – Grigory M May 25 '14 at 14:23

1 Answers1

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This is a CW answer to remove this question from the unanswered list -- it was answered in comments by Gil Bor. The correct formula for the square of an element in $H^2$ is $(\beta_1+\beta_2)^2 = \beta_1^2 + 2 \beta_1 \beta_2 + \beta_2^2$, not $\beta_1^2 + \beta_2^2$.

David E Speyer
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