Naturality of certain isomorphisms for $\mathrm{Hom}$-functors

Question

Apparently, it is known that there are isomorphisms $$\mathrm{Hom}_R(A,\prod_{i \in I} B_i)\cong\prod_{i \in I} \mathrm{Hom}_R(A, B_i)$$ and $$\mathrm{Hom}_R(\bigoplus_{i \in I} A_i, B)\cong\prod_{i \in I}\mathrm{Hom}_R(A_i,B)$$ which are said to be natural. However, I'm not sure what it means for them to be natural. Generally, a naturality means an existence of a certain natural transformation between two functors. I wonder what are the functors in question are here.

Answer 1

The functors in question are

$$ A \mapsto \mathrm{Hom}_R(A,\prod_{i \in I} B_i) $$

and

$$ A \mapsto \prod_{i \in I} \mathrm{Hom}_R(A, B_i) $$

and similarly for the other case.

We can do a little exploration to see how these functors act on morphisms. The first functor is just the usual (contravariant) $\mathrm{Hom}$ functor. If we have a map $f: A \to A'$, we get a map $\mathrm{Hom}_R(A',\prod_{i \in I} B_i) \to \mathrm{Hom}_R(A,\prod_{i \in I} B_i)$ via precomposition: $g \mapsto g \circ f$.

For the second functor, it helps to have a more categorical description of the product (in the category of sets). It turns out that the product of a collection of sets $(X_i)_{i \in I}$ is the same as a certain natural transformation. Letting $\mathcal{I}$ be the discrete category (no non-identity morphisms) on the set $I$, we can define two functors $\mathcal{I} \to \mathcal{Set}$. The first is the constant functor $\Delta_1 = (i \mapsto 1)$ (where $1$ is the set with one element). The second is $X_\_ = (i \mapsto X_i)$. You can check that the product of the $X_i$ is the same as a natural transformation from $\Delta_1$ to $X_\_$ (note that naturality is easy since $\mathcal{I}$ doesn't have any non-identity morphisms).

So the second functor is $A \mapsto \mathcal{Nat}(\Delta_1, i \mapsto \mathrm{Hom}(A, B_i))$. $\mathcal{Nat}$ is just a specific kind of $\mathrm{Hom}$ functor, so we just need to compose two different functors. If we have a map $f: A \to A'$, we get a map $\mathcal{Nat}(\Delta_1, i \mapsto \mathrm{Hom}(A', B_i)) \to \mathcal{Nat}(\Delta_1, i \mapsto \mathrm{Hom}(A, B_i))$ via pre- and post-composition: $\alpha \mapsto (g \mapsto g \circ f) \circ \alpha$. That is, we compose the precomposition natural transformation $g \mapsto g \circ f$ with $\alpha$.

Alternatively, you could work with the universal property of products directly to organize all the maps $\mathrm{Hom}(A', B_i) \to \mathrm{Hom}(A, B_i)$ into a single map $\prod_{i \in I} \mathrm{Hom}(A', B_i) \to \prod_{i \in I} \mathrm{Hom}(A, B_i)$

You should check that all the functor laws are satisfied. Both of these functors are contravariant, so it makes sense to ask if they're naturally isomorphic.