If $$ \Theta\colon\quad 0\to A\to B\to C\to 0 $$ is an exact sequence of abelian groups, and $n$ is an integer, then one obtains an exact sequence $$ 0\to A[n] \to B[n] \to C[n] \stackrel{\delta_n(\Theta)}{\to} A/nA, $$ e.g. by applying the snake lemma to the diagram with exact rows given by $\Theta$ and with vertical maps given by multiplication by $n$; or alternatively by applying the functor ${\rm Hom}(\mathbb{Z}/n\mathbb{Z},\bullet)$ to $\Theta$. One can verify by hand, using Baer sums, that the map $$ \delta_n\colon {\rm Ext}^1(C,A) \to {\rm Hom}(C[n],A/nA),\quad\quad \Theta\mapsto \delta_n(\Theta) $$ is a surjective group homomorphism, but it feels to me like messing with Baer sums is not "the right" kind of proof for such a fact.
Can one define the map $\delta_n$ and prove that it's a surjective group homomorphism purely by using the definition of ${\rm Ext}$ as the derived functor of ${\rm Hom}$, without using the interpretation in terms of exact sequences, and in particular without explicit computations with Baer sums?
The surjectivity is not true for modules over more interesting rings, and I am also hoping that "the right proof" will have something to say about the image of $\delta_n$ in greater generality than just for $\mathbb{Z}$-modules.
