A closed manifold of odd dimension has Euler characteristic zero by the <Euler characteristic of an odd-dimensional closed manifold>. Decomposing $M$ into $P$ and the removed closed ball along $S^{2k}$ gives
$$
0=\chi(M)=\chi(P)+\chi(D^{2k+1})-\chi(S^{2k})
=\chi(P)+1-2,
$$
so $\chi(P)=1$.

If the antipodal boundary map extended to a fixed-point-free involution of $P$, the quotient map $P\to P/\langle\tau\rangle$ would be a double covering. The formula for <Euler characteristic under a finite covering> would imply
$$
1=\chi(P)=2\chi(P/\langle\tau\rangle),
$$
which is impossible. Hence no such extension exists.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 2 / c / Solution

Every closed odd-dimensional manifold has Euler characteristic zero. For an orientable manifold this follows by pairing complementary Betti numbers using <Poincare duality>; the general case follows from the orientable double cover and <Euler characteristic under a finite covering>.

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 Incoming links: Euler characteristic of an odd-dimensional closed manifold