Euler characteristic as alternating sum of cohomology

Question

For the statement that $$\chi=\sum_{i=-1}^d(-1)^i \dim\tilde{H}^i,$$ may I know how to prove it?

(We are considering field coefficients, so the cohomology group is a vector space.)

I know the analogous proof for homology (e.g. Hatcher pg. 146). Is the cohomology case proved by Poincare duality? Any possibility to prove it without Poincare duality? Thanks.

Q2) Why is there a need to consider -1 dimension? Is it just redundant since the -1 dimension is zero?

Thanks a lot.

Answer 1

Let $\chi'$ the new Euler characteristic (i.e yours, computed with cohomology) and $\chi$ the old Euler characteristic.

Using the Mayer-Vietoris long exact sequence you should be able to show the formula $\chi'(M \cup N) = \chi'(M) + \chi'(N) - \chi'(M \cap N)$. So if you know that $\chi = \chi'$ for $M,N,M \cap N$ then you can show that this also holds for $M \cup N$.

Then, you can use induction by finding a covering of $M$ (say $M$ is a manifold or a simplicial complex) which spaces $M_i$, so that you know that $\chi' = \chi$ holds for the various intersections $M_{i_1} \cap M_{i_2} \cap \dots \cap M_{i_n}$.