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 $f_{j}=f \circ i_{j}$.

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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?

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    $\begingroup$ The comment simply says that a map from coproduct induces a pair of maps, and vice versa: a pair of maps induces a map from coproduct. And indeed, this translates to appropriate Homs being equinumerous (just like you've written). The "$\times$" symbol is the standard Cartesian product, and it is completely valid since $Hom(X,Y)$ is a set by definition. On the other hand "$x_1\in X_1$" is meaningless since $X_1$ need not be a set (even though it is hard to imagine in maths anything that isn't a set, or set-like). $\endgroup$ – freakish Mar 22 at 10:35
  • $\begingroup$ What book is this from? $\endgroup$ – Shaun Mar 22 at 14:57
  • $\begingroup$ @Shaun if you’re asking about my post, then it’s just from my lecture note. $\endgroup$ – ensbana Mar 22 at 15:41
  • $\begingroup$ I was. Thank you nonetheless :) $\endgroup$ – Shaun Mar 22 at 15:42

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.

  • $\begingroup$ I was actually looking at this answer when thinking about the question above: math.stackexchange.com/a/2864701/120141. I’m asked to show that $\text{Ab}(G \ast H) \cong \text{Ab}(G) \times \text{Ab}(H)$, where $\text{Ab}(X)$ is the abelianization of the group $X$. The answer uses heavily universal properties using the language of Hom-sets. How do equalities of Hom-sets imply those groups are isomorphic? $\endgroup$ – ensbana Mar 22 at 12:38

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