# Splitting induces isomorphisms

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.

$$\newcommand{\tors}{\mathrm{Tors}}$$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 $$\tors(H^q(M)) \cong \mathrm{Ext}(H_{q-1}(M), \mathbb{Z}).$$
• If you're asking about the isomorphisms $\mathrm{Tors}(H^q(M)) \cong \mathrm{Ext}(H_{q-1}(M), \mathbb{Z})$, I suspect they are; $\alpha$ and $\beta$ are natural as you've mentioned, and I think that the maps in the SES $$0 \rightarrow T^q(M) \rightarrow H^q(M) \rightarrow B^q(M) \rightarrow 0$$ are also natural. – mathphys Jun 27 at 1:50