Is $\bigoplus_{x \in X} \mathbb{Z}$ reflexive?

Question

Basically the title. I am trying to figure out whether or not $Hom(Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z})) \cong \bigoplus_{x \in X} \mathbb{Z}$ with the isomorphism $i(y)(f)=f(y)$.

Here is my intuition as to why that might be true: $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z}) = \times_{x \in X} \mathbb{Z}$, since any "functional" in $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z})$ is determined by its values on each entry of $\bigoplus_{x \in X} \mathbb{Z}$ and any homomorphism $\mathbb{Z} \to \mathbb{Z}$ is just multiplication by some whole number, so there is a natural isomorphism from $Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z}) \to \times_{x \in X} \mathbb{Z}$ (is this reasoning correct?)

So I am left with showing whether or not $Hom(\times_{x \in X} \mathbb{Z}, \mathbb{Z}) = \bigoplus_{x \in X} \mathbb{Z}$. Again, any "functional" would be determined by its' values on each entry, but it cant be nonzero on infinitely many entries since then we could find a corresponding element on which the functional would not be defined. My question is - is this reasoning fine? Turning it into a formal proof isn't hard, but something still feels "fishy" here.

Answer 1

The first step of your argument is false: the Hom functor converts infinite direct sums into infinite products, so we have

$$\text{Hom}(\bigoplus_X \mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z}^X$$

which is much bigger than the infinite direct sum (e.g. if $X$ is countably infinite, the infinite product is uncountable while the infinite direct sum is countable). What this says explicitly is that every homomorphism $\oplus_X \mathbb{Z} \to \mathbb{Z}$ has the form

$$\bigoplus_{x \in X} n_x \mapsto \sum_{x \in X} c_x n_x$$

for some coefficients $c_x$ and that these coefficients are arbitrary, and in particular can all be nonzero. In the infinite direct sum only finitely many of them can be nonzero.

If $X$ is countably infinite, it is nevertheless still true that $\text{Hom}(\mathbb{Z}^X, \mathbb{Z}) \cong \bigoplus_X \mathbb{Z}$, so we still get reflexivity; this is very surprising, e.g. the corresponding statement with $\mathbb{Z}$ replaced by $\mathbb{Q}$ is false given the axiom of choice. This is due to Specker.

For larger $X$ the answer appears to be independent of ZFC.