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Hatcher's Algebraic Topology Corollary $3.37$ states that a closed manifold of odd dimension has Euler characteristic zero. But to consider about the Euler characteristic, every closed manifold must have a finite CW structure. However, there are no relevant comments in the book. Is this true? (It should be true, though) Where can I find a proof or statements of this?

Remarks.

  1. There is a theorem that the Euler characteristic does not depend on a particular CW structure(thus, well-defined) for a finite CW complex. More precisely, we have for a finite CW complex $X$, $\chi(X)=\sum_i(-1)^i\mathrm{rank}H_i(X)$, where $\chi(X)$ is the Euler characteristic of $X$.

  2. In Hatcher's, a closed manifold means a compact manifold without boundary. In particular, we only need to show that a closed manifold has a CW structure, since a compact CW complex must be finite.

  3. Hatcher does not require second countability defining manifolds. Thus, a manifold is just a Hausdorff space which is locally Euclidean.

J. W. Tanner
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blancket
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    Euler characteristic is defined for any space with finite dimensional homology. It is the alternating sum of the dimensions of the homology with coefficients in the rationals. For compact CW complexes this coincides with the definition based on cells. Hatcher’s statement is true with either definition – Connor Malin Jan 02 '20 at 01:25
  • @Conifold Thanks. I also just found a comment about this in the Appendix of Hatchers, below Corollary A.12. – blancket Jan 02 '20 at 01:38
  • @ConnorMalin Thanks, I see now. (But there is no explicit mention in the book that the Euler characteristic is defined for any space with finite dimensional homology) – blancket Jan 02 '20 at 01:40
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    On point 3, a closed manifold is automatically second-countable (by compactness it is covered by finitely many charts). – Eric Wofsey Jan 02 '20 at 02:24

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