Isomorphism between torsion subgroups of $H_q (M)$ and $H_{n-q-1} (M)$ where $M$ is a compact oriented $n$-manifold 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?
 A: 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.
