Let $(X,A)$ be a finite CW-pair $m=p^r$ for some prime $p$. Unspecified coefficient is in $\mathbb{Z}$. From the universal coefficient theorem, We know that

$H^1(A;\mathbb{Z}_m)=\textrm{Hom} (H_1(A),\mathbb{Z}_m)$ ---(1) and

$H^2(X,A;\mathbb{Z}_m)=\textrm{Hom}(H_2(X,A);\mathbb{Z}_m)\bigoplus \textrm{Ext}(H_1(X,A),\mathbb{Z}_m)$. ---(2)

From the long exact sequence of pair, we have a coboundary map $\delta\colon H^1(A;\mathbb{Z}_m)\to H^2(X,A;\mathbb{Z}_m)$.

I know that $\pi_1\circ \delta \colon H^1(A;\mathbb{Z}_m)\to \textrm{Hom}(H_2(X,A),\mathbb{Z}_m)$ ($\pi_1$ is a 1st factor projection from (2)) is same as the composition $H^1(A;\mathbb{Z}_m)\cong \textrm{Hom}(H_1(A);\mathbb{Z}_m)\to \textrm{Hom}(H_2(X,A);\mathbb{Z}_m)$ (the first map is an isomorphism from (1) and the second map is obtained by taking $\textrm{Hom}(-,\mathbb{Z}_m)$ to $\partial\colon H_2(X,A)\to H_1(A)$.)

Let's think $\pi_2\circ \delta \colon H^2(A;\mathbb{Z}_{m})\to Ext(H_1(X,A);\mathbb{Z}_m)$, where $m=p^r$ as before.

Question : Is it true that if $r$ is large, then $\pi_2\circ\delta$ is trivial map or is it meaningless question because of unnaturality of splitting of (2)?

• This is not a meaningful question, because the isomorphism (2) is not canonical, not natural, not well-defined. The map that you call $\pi_1$ is natural, but the careful statement of the universal coeff thm says only that that map is split surjective and that its kernel is naturally isomorphic to the Ext group. It is meaningless, for example, to speak of a cohomology class being in the Hom part of the group. Aug 1, 2010 at 12:50
• OK, I modified a question. But I'm wondering that modified question is also menaingless because of unnaturality of Universal coefficient theorem. But I'm expecting that $\pi_2\circ\delta$ in $\mathbb{Z}_m$-coefficient vanishes if $m=p^r$ and r is large enought because $\textrm{Ext}(H_1(X,A);\mathbb{Q}/\mathbb{Z})$ vanishes. Is it true? Aug 1, 2010 at 13:12

There is a map (i.e. a commutative diagram) from the exact sequence $$0\to Ext(H_{n-1}(A),G)\to H^n(A;G)\to Hom(H_n(A),G)\to 0$$ to the exact sequence $$0\to Ext(H_n(X,A),G)\to H^{n+1}(X,A;G)\to Hom(H_{n+1}(X,A),G)\to 0$$All the groups are functors of both the pair $(X,A)$ and the abelian group $G$. All the maps are natural. The exact sequences split, but the splittings are not natural, so there is not a canonical map $Hom(H_n(A),G)\to Ext(H_n(X,A),G)$ to inquire about. However, by the commutative diagram there is a canonical map from a subgroup of the former to a quotient of the latter, namely from $$ker(Hom(H_n(A),G)\to Hom(H_{n+1}(X,A),G)=Hom(coker(H_{n+1}(X,A)\to H_n(A)),G)=Hom(P,G)$$ to $$coker(Ext(H_{n-1}(A),G)\to Ext(H_n(X,A),G))=Ext(ker(H_n(X,A)\to H_{n-1}(A)),G)=Ext(Q,G)$$ where $P=Im(H_n(A)\to H_n(X))$ and $Q=H_n(X)/P$. This is in fact determined by the exact sequence $0\to P\to H_n(X)\to Q\to 0$, part of the six-term exact sequence $$0\to Hom(Q,G)\to Hom(H_n(X),G)\to Hom(P,G)\to Ext(Q,G)\to Ext(H_n(X),G)\to Ext(P,G)\to 0$$
This map can easily be nonzero when $G=\mathbb Z/m$, for example if $H_{n+1}(X,a)\to H_n(A)\to H_n(X)\to H_n(X,A)\to H_{n-1}(A)$ is the exact sequence $0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\to 0$ and $p$ divides $m$.