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)?



About universal coefficient theorem
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 $m$.


