How to use Mayer-Vietoris to show $\chi(X)=2\chi(M)-\chi(\partial M)$ where $X$ is the double of $M$?

Question

I'm in trouble with the following problem:

Let $M$ be a manifold with compact boundary $N$ and let $X$ be the double of $M$, that is, the manifold without boundary one gets by glueing $M$ with itself along its boundary. How can I use Mayer-Vietoris sequence to show the relation $$\chi(X)=2\chi(M)-\chi(\partial M),$$ where $\chi$ stands for the Euler characteristic.. Any help will be useful...Thanks

Answer 1

Lets suppose that we're working with $M$ a finite dimensional manifold with boundary, and for convenience we'll suppose $\dim M$ is even (If it's not, we would just need to use the fact that $0=-0$ at the end). Let's split $X$ up in the obvious way as the union of two copies of $M$ with a slight thickening in to each other just so their interiors cover $X$ and we satisfy all of the criteria to use Mayer-Vietoris. We'll call them $M_1$ and $M_2$ which intersect on a subspace which is homeomorphic to $\partial M\times [0,1]$. This intersection is homotopy equivalent to the boundary of $M$ so we'll just call it $\partial M$ without confusion.

By Mayer-Vietoris (and the fact that $\partial M$ has dimension strictly less than $n$ so $H_n(\partial M)=0$), we get a long exact sequence $$\dots \to 0 \to H_n(M_1) \oplus H_n(M_2) \to H_n(X) \to H_{n-1}(\partial M) \to \dots$$ $$\dots \to H_1(\partial M) \to H_0(M_1) \oplus H_0(M_2) \to H_0(X) \to H_{0}(\partial M) \to 0$$ (note, I am considering all coefficients of homology groups to be in the rationals).

Now, we need two facts. The first is that the general rank nullity theorem tells us that for a long exact sequence $0\rightarrow V_1\rightarrow\ldots\rightarrow V_r\rightarrow 0$ of finite dimensional vector spaces, we have $$0=\sum_{i=1}^r (-1)^i \dim V_i.$$ The second is that for a topological space $Y$ with definable Euler characteristic, $$\chi (Y)=\sum_{i=0}^\infty (-1)^i\dim (H_i(Y;\mathbb{Q})).$$

Now, because $n$ is even, we get the following sum from the above facts and a little rearranging of the finite sum $$0=[\dim (H_n(M_1))+\dim (H_n(M_1))]-\dim (H_n(X))+\dim (H_{n-1}(\partial M))+\ldots$$ $$\ldots+\dim(H_1(\partial M))-[\dim (H_1(M_1))+\dim (H_1(M_1))]+\dim (H_1(X))$$ $$-\dim (H_0(\partial M))+[\dim(H_0( M_1)) + \dim(H_0(M_2))]-\dim (H_0(X))$$ $$=-\chi (\partial M)+\chi (M)+\chi (M)-\chi (X)$$ $$\Rightarrow \chi (X)=2\chi(M)-\chi(\partial M).$$

Answer 2

If $X$ is covered by open subsets $U,V$, then the Mayer-Vietoris sequence

$$\dots \to H_n(Y) \to H_n(U) \oplus H_n(V) \to H_n(X) \to H_{n-1}(Y) \to \dots$$

(where $Y = U \cap V$)

gives that $\chi(X) = \chi(U) + \chi(V) - \chi(Y)$ (under enough finiteness assumptions that things are well-defined). Tensoring the sequence with $\mathbb{Q}$ preserves exactness, and then this equation is a direct result of dimension counting; in any exact sequence $0 \to A \to B \to C \to 0$ of vector spaces, it is the case that $\dim A + \dim C = \dim B$.

To apply to this specific case, you want $U,V$ to be slightly thickened up versions of $M$ whose intersection is a slightly thickened up version of $\partial M$.