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Define the Euler characteristic of a space $X$ to be $$\chi(X)= \sum_i \dim H_i(X, \mathbb Q)$$ This is obviously not necessarily well-defined for an arbitrary space $X$, so let $X$ be a manifold (manifolds have only finitely many nonzero homology groups, and each homology group is finitely generated). I would prefer to keep this question entirely in the realm of closed manifolds.

There's an obvious restatement of this for $\Bbb Q$ replaced by another field $F$, so let $$\chi(M,F) = \sum_i \dim H_i(M, F)$$

Question: When does $\chi(M)=\chi(M,F)$ for all fields $F$? This is true for every finite CW-complex $M$, but it is my impression that not every closed manifold is a finite CW-complex, though I don't have a counter-example. If this is the case (again, for closed manifolds), what is a reference for this fact? If it's false, the question stands. Does the Euler characteristic depend on the field? I'm hoping for either a reference or a counter example.

Edit: The question was resolved below for smooth manifolds via Morse theory, but as far as I can tell this argument is not generally extendable to the topological case (see: Morse theory in TOP and PL categories?). Hopefully there's a known fully topological answer.

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Morse theory contains the fact that every closed manifold has the homotopy type of a finite CW complex. –  Olivier Bégassat May 19 at 4:33
Independence of the field also follows from using the Mayer-Vietoris spectral sequence for a good cover, so if you can tell that all topological manifolds have such a cover, you are done: see mathoverflow.net/questions/165850/good-covers-of-manifolds –  Mariano Suárez-Alvarez May 19 at 5:24
@MarianoSuárez-Alvarez Your link states that every simplicial complex has a good cover. Since the homology can be computed from the singular complex (which according to that question has a good cover) are we not done? The Euler characteristic of the singular complex is invariant under base change so the Euler characteristic of $M$ is invariant under base change. –  PVAL May 19 at 9:00
The singular complex and simplicial complexes are different things! The first is a complex of abelian groups, the others are obtained by glueing geometric simplices. Not even in the same ballback :P –  Mariano Suárez-Alvarez May 19 at 9:02

2 Answers 2

up vote 3 down vote accepted

Every topological compact manifold can be embedded into $\mathbb{R}^{n}$ and since it is locally contractible it follows that it is an Euclidean Neighbourhood Retract. The proofs are elementary.

It follows that it is a retract of a finite simplicial complex and so its integral homology groups are finitely generated. This is enough to ensure that the Euler characteristic is independant from the field of coefficients (for example by universal coefficient theorem).

For references, see Appendix A of Hatcher's "Algebraic Topology", in particular Corollary A.8 and Corollary A.9.

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Yes, this does it. I should have thought of this. Thank you very much! –  Mike Miller May 19 at 20:48
Just to add: Every compact ANR (e.g. a compact topological manifold) manifold is homotopy-equivalent to a finite CW complex. This is known as "Borsuk conjecture" (proven in 1970s). –  studiosus May 19 at 21:29

Every closed manifold is homotopy equivalent to a finite CW-complex. For a proof see Milnor's book Morse Theory in the section titled "Homotopy Type" (pg. 12 in the ancient edition I have).

(The statement of this is a remark at the end of part 1 section 3)

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This answers the question for the smooth case (or at least $C^2$, but these are, afaik, equivalent). Do you know a reference for the topological case? –  Mike Miller May 19 at 4:55

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