Bordism classes of real projective plane

Question

I have to prove that doesn't exist a compact $3$-manifold such that $\partial X= \mathbb{R}P^2$. My book suggests to use Euler characteristic and define $Y = X \cup_{\mathbb{R}P^2} X$. What is $X \cup_{\mathbb{R}P^2} X$ ? How can I make the proof?

Answer 1

The following is just a hint, not a complete answer:

Suppose you have two manifolds with boundary $M_1, M_2$ and a homeomorphism $h: \partial M_1\to \partial M_2$ of their boundaries. Then one can define a new manifold $M_1\cup_h M_2$ by taking the quotient of the disjoint union $M_1\sqcup M_2$, by the equivalence relation $x\sim h(x), x\in \partial M_1$. A special case of this construction is called the double of a manifold along its boundary, where $M_1=M_2$, $\partial M_1=N$, and $h$ is the identity map of the boundary to itself. The result of doubling is usually denoted $M_1\cup_N M_1$. (Here is is very important to use the disjoint union in the construction.) The result of the general gluing construction is a manifold $M$ without boundary. A nice exercise now is to compute (in the case of compact manifolds) $\chi(M)$ in terms of $\chi(M_1), \chi(M_2)$ and $\chi(\partial M_1)$ by first assuming that manifolds in question are triangulated. (The same answer works in general, just requires a bit more work.)

Now, in the case of $N=RP^2$, one needs to compute $\chi(RP^2)$ and use the fact that $\chi(M)=0$, as $M$ is a closed odd-dimensional manifold.