Complementary Sets in $\mathbb{C}P^2$ 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. 
 A: 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. 
