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Let $U,V\subset \mathbb{C}P^2$ be complementary sets (ie $U\simeq V^c$) with $M=U\cap V$ a $3$-manifold. Assume further $M$ is orientable.

My question is, must it be the case that one of $i_*:H_2(U;\mathbb{Q})\to H_2(\mathbb{C}P^2;\mathbb{Q})$ or $i_*:H_2(V;\mathbb{Q})\to H_2(\mathbb{C}P^2;\mathbb{Q})$ are non-zero.

I would be happy with $\mathbb{Z}_2$ in place of $\mathbb{Q}$ as well.

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    $\begingroup$ If they are complementary sets, they do not intersect. :-) $\endgroup$
    – Bruno Martelli
    Commented Feb 12, 2013 at 14:54
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    $\begingroup$ Why not, $U$ is just homotopic to the complement set. They may still interest. $\endgroup$
    – Prasit
    Commented Feb 12, 2013 at 16:40
  • $\begingroup$ @Prasit: That is correct. Also, if it was not clear, I additionally mean that $U\cup V=\mathbb{C}P^2$ $\endgroup$
    – K Matthews
    Commented Feb 12, 2013 at 16:49
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    $\begingroup$ If you mean '$ U \cup V = \mathbb{CP^{2}}$', then apply Mayer-Veitoris. Since $M$ is oriented $H_{2}(M)$ does not have any $2$ torsion. And the answer pops out. But I am sorry to say this not a mathoverflow problem. $\endgroup$
    – Prasit
    Commented Feb 12, 2013 at 17:20
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    $\begingroup$ @Prasit: I don't see how Mayer-Vietoris alone solves the problem - could you elaborate on your argument? $\endgroup$
    – Mark Grant
    Commented Feb 12, 2013 at 20:05

1 Answer 1

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Here is an argument using cup products, which shows that whenever $\mathbb{C}P^2=U\cup V$ for open sets $U$ and $V$, one of the inclusions $U,V\subseteq \mathbb{C}P^2$ must be non-trivial on second homology.

Working with coefficients in an arbitrary field $k$, cohomology is dual to homology. By the naturality of the Universal Coefficient Theorem, for an arbitrary subset $U\subseteq \mathbb{C}P^2$ we have a commutative diagram $$ \begin{array}{ccc} H^2(\mathbb{C}P^2;k) & \cong & \operatorname{Hom}_k(H_2(\mathbb{C}P^2;k),k) \newline \downarrow & & \downarrow \newline H^2(U;k) & \cong & \operatorname{Hom}_k(H_2(U;k),k) \end{array} $$ which shows that if $H_2(U;k)\to H_2(\mathbb{C}P^2;k)$ is zero, then so must be $H^2(\mathbb{C}P^2;k)\to H^2(U;k)$.

Omitting coefficients from now on, it follows that the generator $x\in H^2(\mathbb{C}P^2)$ is the image of a class $x_U\in H^2(\mathbb{C}P^2,U)$. Likewise, $x$ is the image of a class $x_V\in H^2(\mathbb{C}P^2,V)$ if $H_2(V)\to H_2(\mathbb{C}P^2)$ is zero. But then by naturality of cup products, $x\cup x$ must be the image of $x_U\cup x_V\in H^4(\mathbb{C}P^2,U\cup V)=0$, a contradiction.

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  • $\begingroup$ This is a very nice argument, thank you. $\endgroup$
    – K Matthews
    Commented Feb 13, 2013 at 7:59

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