In the Corollary 3.37 Hatcher proves that for a closed odd-dimensional manifold $M$, its Euler characteristic is zero.

The first part of the proof deals with orientable manifolds, and uses Poincare Duality to show that $\operatorname{rank} H_i(M,\mathbb Z)=\operatorname{rank} H^{n-i}(M,\mathbb Z),$ and then mentions the Universal Coefficients Theorem to justify that the latter is equal to $\operatorname{rank}H_{n-i}(M,\mathbb Z).$ I guess the reasoning here is as follows. The Universal Coefficient Theorem he uses is Theorem 3.2 which stipulates the existence of split exact sequence: $$ 0\longrightarrow \operatorname{Ext}(H_{n-1}(C,\mathbb Z),G)\longrightarrow H^n(C,G)\longrightarrow \operatorname{Hom}(H_n(C,\mathbb Z),G)\longrightarrow 0. $$ Since $\operatorname{Ext}(H_{n-1}(C,\mathbb Z),G)$ is torsion, the rank of $H^n(C,G)$ and rank of $\operatorname{Hom}(H_n(C,\mathbb Z),G)$ are equal, and the latter is equal to the rank of $H_n(C,\mathbb Z)$ when $G=\mathbb Z$, since only free summands of $H_n(C,\mathbb Z)$ can contribute non-zero values to $\operatorname{Hom}(-,\mathbb Z)$. Then in the formula for $\chi$, the terms $(-1)^i\operatorname{rank} H_i(M,\mathbb Z)$ cancel with $(-1)^{n-i}\operatorname{rank} H_{n-i}(M,\mathbb Z)$ for $n$ odd.

In the second part of the proof, Hatcher sketches the proof for non-orientable $M$. He starts with applying the same reasoning for $\mathbb Z_2$ instead of $\mathbb Z$, and needs to show first that $\dim H_i(M,\mathbb Z_2)=\dim H_{n-i}(M,\mathbb Z_2)$ where dimension is taken over $\mathbb Z_2$. Poincare Duality gives $H_i(M,\mathbb Z_2)\cong H^{n-i}(M,\mathbb Z_2)$, and he mentions in passing that $$ H_i(M,\mathbb Z_2)\cong H^i(M,\mathbb Z_2). $$ Question: why are these groups isomorphic? I don't see how the Universal Coefficient theorem in the above form can be applied to prove this claim. (If possible, I would like to see an answer based on the material from the same book of Hatcher.)

EDIT: I am aware that one can prove this fact for non-orientable manifolds by considering the orientable two-fold cover: $\tilde M\to M$ and using the fact that $0=\chi(\tilde M)=2\chi(M)$. I just want to understand the author's reasoning.

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    $\begingroup$ Can you really take the two-fold cover? What if there is no CW structure for the manifold? $\endgroup$
    – Nick
    Commented May 10, 2019 at 14:55

1 Answer 1


With field coefficients you have that cohomology is the dual of homology. By linear algebra, we get an isomorphism.

Proof for my statement is the fact that $Ext_{R}(H_i(M;R),R)=0$ for a field $R$ (because dualizing is exact for vector spaces), which you apply to the universal coefficient theorem.

About the reasoning in the first part, you were completely right. If I were you I would just convince myself again why $rk (G) = rk (Hom (G,\mathbb Z))$.

Because everybody loves characteristic classes it might be worth to mention, that the statement also follows from the fact that the Euler class of odd-dimensional orientable manifolds vanishes. This actually sheds a little geometric intuition on this fact and also generalizes it to closed orientable manifolds which admit an orientable reversing homeomorphism.

  • $\begingroup$ You are talking about some other version of the Universal Coefficient Theorem, than the one Hatcher is using. That is my concern: to deduce everything from the material of his book, as it is supposed to be self-contained. $\endgroup$
    – mathreader
    Commented Apr 22, 2015 at 23:14
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    $\begingroup$ In fact you are right, he does not introduce the universal coefficient theorem for $R$-modules, only for $\mathbb Z$-modules. However he explains in the chapter: "Thus, with field coefficients, cohomology of spaces" that "cohomology is the exact dual of homology". The reason for this is just that $Hom_F(C_n(X;F),F) = Hom(C_n(X),F)$. Try to figure it out yourself now. The relationship between cohomology modules and the duals of homology is very important. $\endgroup$ Commented Apr 22, 2015 at 23:30

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