Let $A, B, C$ be finite Abelian groups fitting in a short exact sequence
$$
1 \rightarrow A\overset{\iota}{\rightarrow} B\overset{\pi}{\rightarrow} C\rightarrow 1 
$$
This determines a class $[\epsilon]\in H^2(C,A)$ measuring the failure of the sequence to split:
$$
s(c\_1)+s(c\_2)=s(c\_1+c\_2)+\iota(\epsilon (c\_1,c\_2))
$$
where $s:C\rightarrow A$ is any section of $\pi$ (different sections gives different representatives of $[\epsilon]$). By Pontryagin duality we have a dual sequence
$$
1 \rightarrow C^\vee\overset{\pi ^\vee}{\rightarrow} B^\vee\overset{\iota ^\vee}{\rightarrow} A^\vee\rightarrow 1 
$$
and thus a dual class $\epsilon^\vee \in H^2(A^\vee,C^\vee)$.


I think this should be naturally defined explicitly in terms of $\epsilon$, but I don't see how. Part of my problem is that the only "explicit" equation in which $\epsilon$ enters is through the section $s$, which however does not gives something natural under duality. How can I write down an explicit formula for
$$
\epsilon^\vee(\alpha\_1,\alpha\_2) c \in \mathbb{R}/\mathbb{Z} \ , \ \ \ \ \ \alpha\_1,\alpha\_2 \in A^\vee \ , \ \ c\in C
$$
?



Pontryagin dual of a group-cohomology class
Ok, I think I worked this out based on my last comment, writing down the usual double complex for $\operatorname{Ext}(A^\vee,C^\vee)$ using a projective resolution of $A^\vee$ and an injective resolution of $C^\vee$ at the same time, and doing the diagram chase. Here's how the resulting map works:


Given $\varepsilon: C\times C\to A$, interpret this as pairing $\overline{\varepsilon}: C\times C\times A^\vee \to \mathbb{R}/\mathbb{Z}$, $(c\_1,c\_2,\varphi) = \varphi(s(c\_1,c\_2))$.


Using injectivity of $\mathbb{R}/\mathbb{Z}$ and that $\varepsilon$ is a symmetric cocycle, one sees that there exists $\beta: C\times A^\vee\to \mathbb{R}/\mathbb{Z}$ with
$$
\overline{\varepsilon}(c\_1,c\_2,\varphi) = \beta(c\_1,\varphi) + \beta(c\_2,\varphi) - \beta(c\_1+c\_2,\varphi).
$$
Now we form $\overline{\varepsilon^\vee}: C\times A^\vee\times A^\vee\to \mathbb{R}/\mathbb{Z}$ via
$$
\overline{\varepsilon^\vee}(c,\varphi\_1,\varphi\_2) = \beta(c,\varphi\_1) + \beta(c,\varphi\_2) - \beta(c,\varphi\_1+\varphi\_2).
$$
Then we let $\varepsilon^\vee: A^\vee \times A^\vee \to C^\vee$ be the map $(\varphi\_1,\varphi\_2) \mapsto \overline{\varepsilon^\vee}(-, \varphi\_1,\varphi\_2)$, a diagram chase using that $\overline{\varepsilon}$ was additive in the $A^\vee$ argument shows that this is additive in the remaining argument, i.e. really is an element of $C^\vee$.


A further diagram chase shows that the choice of $\beta$ influences $\varepsilon^\vee$, but not its cohomology class.


EDIT: After reverse engineering everything through homological algebra, let me provide a completely self-contained proof.


For an abelian extension $0\to A\to B\to C \to 0$ with section $s$ and cocycle $\varepsilon(c\_1,c\_2) = s(c\_1)+s(c\_2)-s(c\_1+c\_2)$, there exists $\beta: C\times A^\vee \to \mathbb{R}/\mathbb{Z}$ with $\varphi(\varepsilon(c\_1,c\_2)) = \beta(c\_1,\varphi) + \beta(c\_2,\varphi) - \beta(c\_1+c\_2,\varphi)$, using that every abelian extension by $\mathbb{R}/\mathbb{Z}$ splits (i.e. that it is an injective object of the category of abelian groups).


Explicitly, $B$ is identified with $A\times C$ with group structure $(a\_1,c\_1) + (a\_2,c\_2) = (a\_1+a\_2-\varepsilon(c\_1,c\_2), c\_1+c\_2)$.


Now for every $\varphi\in A^\vee$, note that
$$
(a,c) \mapsto \varphi(a) - \beta(c,\varphi)
$$
is a homomorphism $B\to \mathbb{R}/\mathbb{Z}$, which we denote $s^\vee(\varphi)$. This determines a section $s^\vee: A^\vee\to B^\vee$. We have
$$
(s^\vee(\varphi\_1) + s^\vee(\varphi\_2) - s^\vee(\varphi\_1+\varphi\_2))(a,c) = -\beta(c,\varphi\_1) - \beta(c,\varphi\_2) + \beta(c,\varphi\_1+\varphi\_2).
$$
This is an element of $B^\vee$ which is zero in $A^\vee$, so it comes from $C^\vee$, in particular is additive in $c$ (which we could also check directly). So it determines a cocycle of $A^\vee$ with values in $C^\vee$, and this is $-\varepsilon^\vee$ from above (apologies for the sign error above, I'm sure this comes from the double complex story.)


