If $0 \to A \to B \to C \to 0$ is split, we say that $B \cong A \oplus C$. Is there a reason we take the direct sum rather than the product? I'm studying some commutative algebra and a solution to an exercise(concerning commutative $R$-modules) relies on the isomorphisms $$\text{Hom}(C, A \oplus C) \cong \text{Hom}(C, A \times C) \cong \text{Hom}(C, A) \times \text{Hom}(C, C)$$
where the first isomorphism takes the canonical $A \oplus C \cong A \times C$ and the second uses the fact that Hom preserves products in the second variable. I figured I could do this because finite direct sums and finite products are naturally isomorphic, right? This kind of operation with Hom seemed too good to be true. Does this actually work?
The reason I ask is that this solution would be greatly simplified if I took the sequence being split to mean $B \cong A \times C$ instead. That way, I wouldn't have to worry about the first isomorphism.
Thank you!
 A: I guess you know that the categorical concepts of product and sum (or coproduct) are defined by universal properties which are dual to each other.
Consider a three term sequence $A \stackrel{f}{\to} B \stackrel{g}{\to} C$ in a category $\mathcal C$.

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*Given a morphism $r : B \to A$, we get a unique morphism $\phi : B \to A \times C$ such that $\pi_A \circ \phi =r$ and $\pi_C \circ \phi = g$. Here $\pi_A : A \times C \to A$ and $\pi_C : A \times C \to C$ are the "projection" morphisms belonging to the categorical product of $A$ and $C$. If $\mathcal C$ is the category of $R$-modules, the sequence $0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ is exact and $r \circ f = id_A$, then one can show that $\phi$ is an isomorphism.


*Given a morphism $i : C \to B$, we get a unique morphism $\psi :  A \oplus C \to C$ such that $\psi \circ \iota_A  = f$ and $\psi \circ \iota_C  = i$. Here $\iota_A : A \to A \oplus C$ and $\iota_C : C \to A \oplus C$ are the "embedding" morphisms belonging to the categorical sum of $A$ and $C$. If $\mathcal C$ is the category of $R$-modules, the sequence $0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ is exact and $g \circ i = d_C$, then one can show that $\psi$ is an isomorphism.
As Qi Zhu mentions in his comment, in additive categories finite products are canonically isomorphic to finite sums. Thus you may interpret $A \oplus C$ and $A \times C$ as the same object. In fact, both can be taken as the set of all pairs $(a,c)$ with $a \in A, c \in C$ with the usual algebraic operations making it an $R$-module. But be aware that the categorical status is different which does not become transparent if we work on the "naive" level of sets.
