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Can anyone explain why $RP^2$ is not a boundary of any 3-manifold?

I observe that getting of nonorientable 3-manifold by handle attaching is possible only when attaching 1-handle with orientation preserving and reversing in boundaries of 1-handle. But I don't how to use this to prove this question. If it is not possible to prove using this, please explain why.

Thanks, advance.

Selvakumar A
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    Because the Euler characteristic of the boundary of a compact manifold is even (see this question) but $\chi(\mathbb{RP}^2) = 1$. – Michael Albanese Jul 13 '17 at 16:34
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    This is false without extra conditions such as the compactness condition in the comment of @MichaelAlbanese: $RP^2$ is the boundary of the 3-manifold $RP^2 \times [0,\infty)$. – Lee Mosher Jul 14 '17 at 17:33

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