TITLE: Euler characteristic of a boundary?
QUESTION [26 upvotes]: Why is the Euler characteristic of a boundary even? How can one prove this and is there an geometric way to think about it?

REPLY [7 votes]: 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.