Let $R$ be a commutative ring, and $\left\{ M_{i}\right\} _{i\in{\cal I}}$ ,$N$ are $R$-modules. It is elementary that there is a natural module isomorphism
$${\rm Hom}\left(\bigoplus_{i\in{\cal I}}M_{i},N\right)\cong\prod_{i\in{\cal I}}{\rm Hom}\left(M_{i},N\right).$$
Is the dual statement also true? That is, do we also have the following isomorphism?
$${\rm Hom}\left(\prod_{i\in{\cal I}}M_{i},N\right)\cong\bigoplus_{i\in{\cal I}}{\rm Hom}\left(M_{i},N\right)?$$
It is worth mentioning the most famous example: in this question, a positive answer is given in the special case of $\prod_{n=1}^{\infty} \mathbb{Z}$. But it seems pretty ad-hoc, and I suspect that this is not generalizable.
Best Answer
Here is a counterexample which does not require the use of the axiom of choice. Setting $R = \mathbb{Z}$, let $M = \prod_p C_p$ be the product of the cyclic groups of prime order, and let $N = \prod_p C_p / \bigoplus_p C_p$ be the quotient of the direct product by the direct sum. The direct sum $\bigoplus_p C_p$ is in fact the torsion subgroup of $M$, so $N$ is the torsion-free quotient.
So $\text{Hom}(M, N)$ is nontrivial, since it contains at least the quotient map $M \to N$, but $\bigoplus_p \text{Hom}(C_p, N)$ is trivial, since $N$ is torsion-free.