Is ${\rm Hom}\left(\prod_{i\in{\cal I}}M_{i},N\right)\cong\bigoplus_{i\in{\cal I}}{\rm Hom}\left(M_{i},N\right)$ for modules?

Question

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.

Answer 1

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.

Answer 2

This should be false. For example, dimension-counting shows that $\hom_{\mathbb{R}}(\prod_{i\in\mathbb{N}}\mathbb{R}, \mathbb{R})\not\cong\bigoplus_{i\in\mathbb{N}}\hom_{\mathbb{R}}(\mathbb{R}, \mathbb{R})$.

The right-hand side has dimension $|\mathbb{N}|$, writing a basis vector for each $i$.

The left-hand side is too big. Even $\prod_{i\in\mathbb{N}}\mathbb{R}$ by itself already has dimension $|\mathbb{R}^\mathbb{N}|$ (See here for example), and taking a dual $\hom(-,\mathbb{R})$ will only make it bigger.