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}). $

  • $\begingroup$ I also want to know if these isomorphisms are natural $\endgroup$ – Eduardo Longa Jun 27 at 1:44
  • $\begingroup$ 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. $\endgroup$ – mathphys Jun 27 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.