Given a regular local ring $R$ and an $R$-algebras $S$, which is torsion free and finitely generated (even free if needed) as an $R$-module.


Assume we have a nontrivial surjective map $f: M \rightarrow T$, where $M$ is a projective $S$-module, finitely generated and torsion free, and $T$ is a torsion module over $S$. If $N$ denotes $ker(f)$, we get an exact sequence: $0\rightarrow N\rightarrow M\rightarrow T\rightarrow 0$.


Given another torsion module $Q$, when is the induced map $f^{\\*}: Hom\_S(M,Q)\rightarrow Hom\_S(N,Q)$ non trivial, when is it trivial?


My first idea was to use the long exact $Ext$-sequence: Since $M$ is projective we have $Ext^1\_S(M,Q)=0$, thus if $f^{\\*}=0$, the sequence gives an isomorphism $Hom\_S(N,Q)\cong Ext^1\_S(T,Q)$.


So what can be said about the groups $Ext^1\_S(T,Q)$? Are they always/sometimes/never trivial? Can we compute them if we assume that one of these moudles is a simple $S$-module?
Are there other approaches to this question?



Extensions of torsion modules
Actually, it would be easier to look at the other end of the exact sequence. Namely, your map $f^\*$ is trivial implies the map $g^\*: Hom\_S(T,Q) \to Hom\_S(M,Q)$ is an isomorphism. 


Now, since $M$ is projective, the support of $Hom\_S(M,Q)$ is equal to the support of $Q$. Thus we have $Supp(Hom\_S(T,Q)) = Supp(Q)$, which implies $$Supp(T) \supseteq Supp(Q) \ \ (1)$$


When $Q$ is simple, (so $Q=S/m$ where $m$ is a maximal ideal) as you alluded to in the last paragraph, then $(1)$ is also sufficient, provided that the surjection $M \to T$ is minimal when localizing at $m$, as you can easily check for yourself. 


Another situation when $(1)$ also suffices is when $T,Q$ are both *cyclic* $S$ module and $M=S$ (you always need the map $M\to T$ to be minimal, may be that what you meant by "non-trival" surjection?)


Other than what described above, I think what you want will fail most of the times, even with $(1)$. 


