The title says it all, but let me repeat.
We all learn that the Pontryagin dual of a finite Abelian group is abstractly isomorphic as groups, but there’s no canonical isomorphism.
I think I understand it, but I don’t know how to formalize this statement, maybe using category theory.
Could someone enlighten me?
If many thinks that this is more suitable to Math.SE, please move this there.
Suppose we have a family of isomorphisms $\alpha_A\colon A\to A^*$ for all finite abelian groups $A^*$. If $\phi\colon A\to B$ is an isomorphism, we have a dual isomorphism $\phi^*\colon B^*\to A^*$ and thus an isomorphism $(\phi^*)^{-1}\colon A^*\to B^*$. This makes $A^*$ a covariant functor of $A$ on the category of finite abelian groups and isomorphisms, and it is only reasonable to call $\alpha$ canonical if it is natural with respect to this structure. In other words, we should have $(\phi^*)^{-1}\circ\alpha_A=\alpha_B\circ\phi$ for all $\phi$, or $\alpha_A=\phi^*\circ\alpha_B\circ\phi$. In particular, this must hold when $B=A$ and $\phi=n.1_A$ for some $n$ that is coprime to the order of $A$. This means that $(n^2-1).\alpha_A=0$, but $\alpha_A$ is assumed to be an isomorphism, so the exponent of $A$ must divide $n^2-1$. This fails when $A=\mathbb{Z}/5$ and $n=2$, for example, so there is no natural map $\alpha$ as described.
The same argument also shows that there is no natural isomorphism $V\to V^*$ for finite-dimensional vector spaces $V$ over $\mathbb{Z}/p$ provided that $p>3$. The same conclusion holds for $p=2$ or $p=3$ but one needs to use some different choices of $\phi$ to prove it.
On the other hand, if we restrict even further to elementary abelian groups of order $4$, then you can check that there is a natural choice of $\alpha_A$. It sends each nonzero element $a\in A$ to the unique map $\theta\colon A\to\mathbb{Z}/2$ such that $\theta\neq 0$ but $\theta(a)=0$.
