Using Mayer-Vietoris to show $\chi(M) = \chi(U)+\chi(V)-\chi(U \cap V)$

Question

Let $M$ be a manifold, and $U$, $V$ open sub-manifolds in $M$. How would one use the Mayer-Vietoris theorem to show that $\chi(M) = \chi(U)+\chi(V)-\chi(U \cap V)$, where $\chi$ is the Euler characteristic?

Answer 1

Hint 1. Recall that $$\chi(N)=\sum_{i=0}^\infty (-1)^i \operatorname{dim}{H^i(N)}$$ for any manifold $N$ where this makes sense. In particular, this applies when $N$ is $M$, $U$, $V$, or $U\cap V$.

Hint 2. The Mayer-Vietoris long exact sequence relates the cohomology groups of $U,V,M$, and $U\cap V$. Recall that for a long exact sequence of finite dimensional vector spaces, the alternating sum of the dimensions of the vector spaces is zero. Write down what this means in this case, rearrange, and apply Hint 1.