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.

  • 1
    $\begingroup$ I would say that the $-1$ in the formula is simply a typo, especially in the sense that if there was a need to consider that $-1$ dimension for some reason for the person who tell you the formula, he or she would have given the reason for that. $\endgroup$
    – cjackal
    Mar 1, 2018 at 10:40
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    $\begingroup$ Why not use Mayer-Vietoris to show that $\chi(M \cup N) = \chi(M) + \chi(N) - \chi(M \cap N)$ (with usual hypothesis on $M,N$) ? This shows inductively that the Euler characteristic agree with the sum you wrote. $\endgroup$ Mar 1, 2018 at 10:54
  • $\begingroup$ @NicolasHemelsoet Thanks. Can you provide more details or refer me to a reference? $\endgroup$
    – yoyostein
    Mar 1, 2018 at 10:58

1 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 $M_{i_1} \cap M_{i_2} \cap \dots \cap M_{i_n}$.


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