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
$$
\epsilon^\vee(\alpha_1,\alpha_2) c \in \mathbb{R}/\mathbb{Z} \ , \ \ \ \ \ \alpha_1,\alpha_2 \in A^\vee \ , \ \ c\in C
$$
?
Best Answer
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.)