I’m reading up about universal properties. The following is the definition of the coproduct.
Definition. The coproduct $X_{1} \coprod X_{2}$ of $X_{1}$ and $X_{2},$ together with the morphisms $i_{j}: X_{j} \rightarrow X_{1} \coprod X_{2},$ is characterized by the following universal property: Given any object $Y$ with morphisms $f_{j}: X_{j} \rightarrow Y,$ there exists a unique $f: X_{1} \coprod X_{2} \rightarrow Y$ such that $f_{j}=f \circ i_{j}$.
The comment in the title follows this definition.
I’m not sure how to understand the term “equivalent”. I’ve been reading around and seen some relevant statements, so I guess it’s the same as $\text{Hom}(X_1 \coprod X_2,Y) = \text{Hom}(X_1,Y) \times \text{Hom}(X_2,Y)$? But again, my grasp on category theory is quite limited, and I’m learning these things from a group-theoretic perspective, so I’m not sure how to interpret the “$\times$” sign.
Say we have $x_1 \in X_1$ and $x_2 \in X_2$. How do we represent the comment above in this setup?
As to the last sentence: in category one doesn't talk about elements, not just objects and arrows (morphisms). In general $X_i$ need not even be sets for a coproduct.
There is indeed a natural bijection between $\text{Hom}(X_1 \coprod X_2,Y) = \text{Hom}(X_1,Y) \times \text{Hom}(X_2,Y)$: for every pair $(f_1,f_2)$ from the right we assign the unique $f$ from the diagram while to some $g: X_1 \coprod X_2)$ we can assign $(g \circ i_1, g \circ i_2)$ on the right, which is (by unicity) the inverse of the first map.
The $\times$ is just the standard Cartesian product in Set, so we have an isomorphism of Hom-sets in that category. So a coproduct in $C$ (the category we're working in) corresponds bijectively to a product in Set via Hom-sets.