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Let $f\colon A\to B$ be a group homomorphism between finite abelian groups.

For abelian group $G$, let $G^\wedge=\operatorname{Hom}_\mathbb{Z}(G,\mathbb{Q}/\mathbb{Z})$ be its Pontryagin dual.

Since $A,B$ are finite abelian groups, we have $A^\wedge=A, B^\wedge=B$.

Now, my question is :

How can we describe $f^\wedge\colon B\to A$ in terms of $f$?

Added: Can we relate $\operatorname{Ker} f^\wedge$, $\operatorname{Coker} f^\wedge$ with $\operatorname{Ker} f $, $\operatorname{Coker} f$?

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It is important to note that while a finite group is isomorphic to its Pontryagin dual, the isomorphism is non-canonical. To get a map $f^{\wedge} : B \to A$ using $f$ you need to fix an isomorphism $\phi$ from $A$ to $A^{\wedge}$, and $\psi$ from $B$ to $B^{\wedge}$. Then we have

$f^{\wedge} = \phi^{-1} \circ f^{*} \circ \psi $

where $f^{*}$ is the canonical map $B^{\wedge} \to A^{\wedge}$.

Added: As for kernels and cokernels, note that once we have fixed the above isomorphisms, we have a homomorphism $\text{Ker }f \to \text{Coker } f^{\wedge}$ sending $k$ to $k + \text{Im } f^{\wedge}$. A back-of-envelope diagram chase suggests that this map is an isomorphism, and the same works for the other two groups.