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It it the exercise in Massey's book

Massey, William S., Singular homology theory, Graduate Texts in Mathematics, 70. New York Heidelberg Berlin: Springer- Verlag. XII, 265 p. DM 49.50; $ 29.20 (1980). ZBL0442.55001.

chapter IX, section 5, exercise 5.1, he states that if $M$ is a compact oriented $n$-manifold, then the torsion subgroup of $H_q (M)$ is isomorphic to the torsion subgroup of $H_{n-q-1} (M)$.

I use the fact that a compact oriented $n$-manifold $M$ must have finitely generated homology groups, $\mathrm{Hom} (G,\mathbb{Q}/\mathbb{Z} )$ is the torsion part of $G$ if $G$ is finitely generated and $\mathbb{Q} /\mathbb{Z} $ is injective to obtain the isomorphism $T(H_q (M))\cong H^q (M;\mathbb{Q} /\mathbb{Z} )$, and use the Poincare duality to give the isomorphism $T(H_q (M))\cong H_{n-q} (M;\mathbb{Q} /\mathbb{Z} )$. However, by the Universal Coefficient Theorem for Tor, I must state that $H_q (M)\otimes\mathbb{Q} /\mathbb{Z} =0$. Will that be true?

TheWildCat
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  • You cannot apply Poincare duality like this because $\Bbb Q/\Bbb Z$ is not a ring. You need to dualize first, then apply UCT. –  Jan 03 '19 at 14:21
  • But Poincare duality can be used here because of theorem 4.1 in this book. I don't find any error on his proof except a lemma for commutativity, it is quite lengthy and I finally choose to believe it. – TheWildCat Jan 03 '19 at 14:39
  • Sorry, you're quite right! (For context, I was thinking of using the product structure on $R$ to take the cap product with the $R$-fundamental class, but you may as well just take cap productwith $\Bbb Z$-fundamental class.) Instead the error is in your isomorphism $T(H_q(M)) \cong H^q(M;\Bbb Q/\Bbb Z)$. I will write an answer. –  Jan 03 '19 at 14:54

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It is not true that $\text{Tors}(H_q(M)) \cong H^q(M;\Bbb Q/\Bbb Z)$. You are right that $\Bbb Q/\Bbb Z$ is injective, so the Ext-term vanishes, but if $H_q(M;\Bbb Z) = \Bbb Z^{b_q} \oplus T$, then $$H^{q}(M;\Bbb Q/\Bbb Z) \cong \text{Hom}(H_q(M;\Bbb Z), \Bbb Q/\Bbb Z) \cong T \oplus (\Bbb Q/\Bbb Z)^{b_q},$$ noncanonically.

So your approach shows that $T \oplus (\Bbb Q/\Bbb Z)^{b_q} \cong H^q(M;\Bbb Q/\Bbb Z) \cong H_{n-q}(M;\Bbb Q/\Bbb Z)$. What you want to do now is understand the last term. Again we have the universal coefficient sequence,

$$0 \to H_{n-q}(M;\Bbb Z) \otimes \Bbb Q/\Bbb Z \to H_{n-q}(M;\Bbb Q/\Bbb Z) \to \text{Tor}(H_{n-q-1}(M;\Bbb Z), \Bbb Q/\Bbb Z) \to 0.$$

The first term is the divisible subgroup of $H_{n-q}(M;\Bbb Q/\Bbb Z)$, and is isomorphic to $(\Bbb Q/\Bbb Z)^{b_{n-q}} = (\Bbb Q/\Bbb Z)^{b_q}$. The last term is $\text{Tors}(H_{n-q-1}(M;\Bbb Z))$.

Because we have an isomorphism of groups $H_{n-q}(M;\Bbb Q/\Bbb Z) \cong H^q(M;\Bbb Q/\Bbb Z)$, and isomorphisms preserve the maximal divisible subgroup, we see that they preserve the quotient by this subgroup; in both cases it is isomorphic to the finite abelian group $T$.

Note that $H_q(M;\Bbb Z) \otimes \Bbb Q/\Bbb Z$ is indeed zero if and only if the claim that $H^q(M;\Bbb Q/\Bbb Z) \cong \text{Tors}(H_q(M;\Bbb Z))$ is true.