= Solution
Choose a <complex projective line> in each <Complex projective plane> summand. Make the <connected sum> at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The <intersection form> is diagonal with entries one, hence
$$
[S]^2=(h_1+h_2)^2=2.
$$
The oriented <normal bundle> of $S$ is a rank-two real bundle with <Euler number> two, because its Euler number equals the <self-intersection number>. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $TS^2$. The boundary of a closed <tubular neighborhood> $N$ is consequently the <unit tangent bundle> of $S^2$.
A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in $\mathbb R^3$. Thus $UTS^2\cong SO(3)\cong\mathbb{RP}^3$, the latter identification coming from the double covering by <unit quaternions>. Therefore \b[$\partial N$ is the required smoothly embedded $\mathbb{RP}^3$]. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus\partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of <separating projective three-space in a positive connected sum>.
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