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I need some help for showing the following result:

Let $M$ be a compact ($\partial M= \emptyset$) and oriented manifold of dimension $n$ and $Z$ an oriented manifold with boundary such that $\partial Z=M$. Then $$\chi(M)=0\mod(2),$$ that is, the Euler Characteristic of $M$ is even.

A fact that might be useful (although I didn't see how) is that if $M$ is compact then all of the De Rham cohomology groups of $M$ have finite dimension so that the following equality holds, $$\chi(M)=\sum_{k=0}^n(-1)^{k}\textrm{dim}(H^k(M)).$$

Any help will be great.. Thanks

PtF
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  • i'm not sure what framework you're looking at, but in any case i believe this will be relevant http://math.stackexchange.com/questions/99898/euler-characteristic-of-a-boundary – citedcorpse Jul 20 '13 at 16:34
  • Yes that is very relevant, thanks =D – PtF Jul 20 '13 at 16:44
  • You mean $\partial M =\emptyset$. You can also prove this by using Lefschetz duality and basic properties of long exact sequences. – Ted Shifrin Jul 20 '13 at 19:07
  • Yes @Ted it is $\partial M=\phi$, I already fixed that.. Unfortunately I know nothing about Lefschetz duality...Hey I guess we must also ask $Z$ to be compact and connected for the equality $\chi(M\cup N)=\chi(M)+\chi(N)-\chi(M\cap N)$ to be used, right? – PtF Jul 20 '13 at 19:12
  • I don't think you need $Z$ connected because the proof applies to each component. But you definitely need $Z$ compact. – Ted Shifrin Jul 20 '13 at 19:18
  • @Ted I was Thinking about it I concluded that if $Z$ werecompact then the exercise would be too simple for we'd have $\chi(Z\cup Z)=0$.. I guess compactness of $Z$ was dropped from the hypothesis for forcing the use of Meyer-Vietoris sequence.. – PtF Jul 20 '13 at 19:26
  • If $Z$ is not compact, it may well have infinite dimensional de Rham cohomology groups, so things will be slightly more delicate. – Mariano Suárez-Álvarez Jul 20 '13 at 19:30

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