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)?
Best Answer
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 $m$.