Let $M$ be a compact Manifold with boundary. I want to show that $\chi(\partial M)$ is even. This question has been asked before, but unfortunately I don't understand the answer that is always given:
"Let $M$ be your (compact) manifold. You can glue two copies $M_1$, $M_2$ of $M$ along their boundary, getting a closed manifold $2M$. Using the Mayer-Vietoris long exact sequence for the triad $(2M;M_1,M_2)$. It gives us the relation $\chi(2M)=2\chi(M)-\chi(\partial M)$, because $M_1$ and $M_2$ intersect along $\partial M$.
Now, if $\dim M$ is odd, then $\dim 2M$ is also odd and $\chi(2M)=0$, so $\chi(\partial M)=2\chi(M)$ is even. If $\dim M$ is even, then $\dim \partial M$ is odd and therefore $\chi(\partial M)=0$ is also even."
I am supposed to use Mayer Vietoris and collar theorem to get $\chi(2M)=2\chi(M)-\chi(\partial M)$, but I don't know much about the homology groups of $M$, do I? And even If I have this, why is $\chi(2M)=0$, if dim(M) is odd.
Best Answer
Try to fill in the details in the following outline:
Prove this fact by considering the two short exact sequences $$0\to Z_n \to C_n \stackrel{\partial}{\to} B_{n-1}\to 0$$ and $$0\to B_n\to Z_n \to H_n(C_*)\to 0$$ where $B_n\subset C_n$ are the $n$-boundaries and $Z_n\subset C_n$ the $n$-cycles.
Taking that fact for granted, consider the Mayer-Vietoris sequence corresponding to the triple $(2M; M_1, M_2)$:
$$\cdots \to \underbrace{H_1(2M)}_{C_3}\to \underbrace{H_0(\partial M)}_{C_2}\to \underbrace{H_0(M_1)\oplus H_0(M_2)}_{C_1}\to \underbrace{H_0(2M)}_{C_0}\to 0$$
Since it is a long exact sequence, it is in particular a chain complex with vanishing homology. Define chain modules $C_n$ as indicated above. Fact 1 tells you that $\sum_n (-1)^n \operatorname{rank} C_n=0$.
Write out what that means, rearrange the terms and you will obtain the desired formula $\chi(2M)=2\chi(M)-\chi(\partial M)$.
This fact is a corollary of Poincaré Duality. As was hinted in the comment section, use this to prove that $\chi(2M)=0$.