I am reading papers about cellular automata and for some reason, the same result claim is made in all of them and is never proven; I assume that it might be easy to prove, but I cannot see how. Here is the claim:
Let $G$ be an finite abelian group. We denote by $\widehat{G}$ the set of characters of this group, that is to say homomorphisms $\chi : G \to S^1$.
Let $\mathbb{M}$ be a monoid (what really interests me is the case where $\mathbb{M} = \mathbb{Z}$ so if the proof is simpler in this special case, I'm fine with it !)
Then, $\widehat{G^\mathbb{M}}$ is (apparently) isomorphic to the set of sequences $\widehat{G}^\mathbb{M}$ with finitely-many non trivial characters (i.e which are not constant equal to $1$).
In the case of finite products I know that we have $\widehat{G \times G} \cong \widehat{G} \times \widehat{G}$ but I don't understand why we have to restrict ourselves to finite-support sequences in the case of a general $\widehat{G^\mathbb{M}}$.
Edit: For reference, here is one of the papers https://arxiv.org/pdf/math/0108082.pdf , on the 3rd page.
It's because $\hat G$ is not just a group but a topological group, and it isn't defined to be all homomorphisms, only continuous ones.
This doesn't matter at first because $G$ is finite so its a discrete group, and all homomorphisms are continuous. But $G^I$ is only a compact group, so its dual will be affected. (I'll use $I$ as an index set instead of $\mathbb M$ or $\mathbb Z$, feel free to substitute.)
Note that finitely supported functions is a direct sum (coproduct) whereas all functions is a direct product: $$C = \bigoplus_{i\in I}G, \hspace{1em} P = \prod_{i\in I} G$$ The goal is then to see $\widehat P \cong C$, but by general nonsense you can see $$\widehat C = Hom(\coprod_{i\in I} G, S^1) = \prod_{i\in I} Hom(G,S^1) \cong P$$ and then apply pontrjagin duality.
For more details see these notes https://www.maths.ed.ac.uk/~cbarwick/papers/suppprobsLCA.pdf in which your problem is Exercise 4.10.