Splitting induces isomorphisms

Question

Let $M^n$ be a closed topological manifold. For $1 \leq q \leq n$, we have the following short exact sequence:

$$ 0 \longrightarrow \operatorname{Ext}(H_{q-1}(M),\mathbb{Z}) \stackrel{\beta}{\longrightarrow} H^q(M) \stackrel{\alpha}{\longrightarrow} \operatorname{Hom}(H_q(M), \mathbb{Z}) \longrightarrow 0 $$

The homomorphisms $\alpha$ and $\beta$ are natural with respect to continuous maps.

Since $M$ is compact, $H^q(M)$ is a finitely generated abelian group, hence it has the form $$H^q(M) = B^q(M) \oplus T^q(M),$$ where $B^q(M)$ is free abelian and $T^q(M)$ is the torsion subgroup. We also know that, since the above sequence splits (not naturally), $$H^q(M) \cong \operatorname{Hom}(H_q(M),\mathbb{Z}) \oplus \operatorname{Ext}(H_{q-1}(M), \mathbb{Z}).$$

Can we conclude that $B^q(M) \cong \operatorname{Hom}(H_q(M), \mathbb{Z})$ and $T^q(M) \cong \operatorname{Ext}(H_{q-1}(M),\mathbb{Z})$? If so, can we say that $\alpha$ induces the first isomorphism and $\beta$ induces the second?

To give some context: I have $f : M \to M$ a continuous map such that $f_*: H_1(M) \to H_1(M)$ is an isomorphism. If the above is true, then $T^2(M) \cong \operatorname{Ext}(H_1(M),\mathbb{Z})$, and if the isomorphism is natural, then I can conclude that $f^* \vert_{T^2(M)} : T^2(M) \to T^2(M)$ is an isomorphism.

Answer 1

$\newcommand{\tors}{\mathrm{Tors}}$Poincaré duality gives that $H^q(M) \cong H_{n-q}(M)$. Then using the fact that in general $\tors(H_q(M)) \cong \tors(H_{n-q-1}(M))$ (which comes from this question), we have \begin{align} \tors(H^q(M)) &\cong \tors (H_{n-1}(M)) \\ &\cong \tors(H_{n-(n-q) -1}(M)) \\ &= \tors(H_{q-1}(M)). \end{align} But now recall $\mathrm{Ext}(H, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H$ when $H$ is finitely generated, by this question. So $\tors(H^q(M)) \cong \mathrm{Ext}(H_{q-1}(M), \mathbb{Z}). $