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Why is the Euler characteristic of a boundary even? How can one prove this and is there an geometric way to think about it?

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up vote 25 down vote accepted

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.

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@Marino, for using Mayer-Vietoris we need an open cover from $2M$, how will the boundary show up in the sequence? – PtF Jul 20 '13 at 19:32
The boundary has a collar. Use that. – Mariano Suárez-Alvarez Jul 20 '13 at 19:36
should I use the collar neighbourhood theorem? – PtF Jul 20 '13 at 19:45
Well, that theorem is the one that proves that the boundary has a collar, so if I am suggesting that yu usse a collar, you are surely going to need it! – Mariano Suárez-Alvarez Jul 20 '13 at 19:49
well thanks, I'll try =D What would be a good book that has the proof of this theorem? I'll need to reference it because we didn't see it in class.. – PtF Jul 20 '13 at 19:50

Another perspective: Triangulate $M$; we want to compute $\chi(\partial M) \bmod 2$. Let $T$ be the set of all faces (of all dimensions) in the triangulation of $M$. We will count, modulo $2$, the number of pairs $(\sigma, \tau)$ of elements in $T$ with $\sigma \subsetneq \tau$.

Fixing $\sigma$ If $\sigma$ meets the interior of $M$, I claim that there are an even number of $\tau$ containing $\sigma$. Proof sketch: Working in a neighborhood of $x$, let $x$ be a point in the interior of $\sigma$, let $V$ be a linear space through $x$ transverse to $S$, and let $S$ be a sphere around $x$ in $V$. Then the intersection of $S$ with the triangulation $T$ gives a triangulation of the sphere $S$, with one face for each $\tau$ with $\sigma \subsetneq \tau$. A triangulation of a sphere has an even number of faces.

Similarly, if $\sigma$ is in the boundary of $M$, then $S$ meets $M$ in a closed hemisphere and the triangulation of $S$ has an odd number of faces.

So the number of pairs $(\sigma, \tau)$ as above is congruent $\bmod 2$ to the number of faces in $\partial M$, and thus congruent $\bmod 2$ to $\chi(\partial M)$.

Fixing $\tau$ Let $\tau$ have dimension $d$. Then there are $2^{d+1}-2$ faces $\sigma$ with $\sigma \subsetneq \tau$. In particular, this number is even, so the count is even.

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