Why are coproduct objects corepresentations of cartesian products, rather than representations of disjoint unions?

Question

One way we can define the product of two objects $A$ and $B$ in some category $\mathcal C$ is as a representation of the contravariant functor $(\to A) \times (\to B)$ in $[\mathcal C^{\mathrm{op}}, Set]$.

The analogous definition of coproducts is as a corepresentation of the covariant functor $(A \to) \times (B \to)$ in $[\mathcal C, Set]$.

Notably, AFAICT one can't get products by asking for a corepresentation of $(A \to) + (B \to)$, nor coproducts by asking for a representation of $(\to A) + (\to B)$, where here I'm using $+$ to denote coproducts.

I find this asymmetry curious, and my main question is the title question. That question's a bit broad and vague, though. To be a bit more specific, aspects of this question that I'm particularly curious about include:

1. What are these other objects? ((co)representations of $(A \to) + (B \to)$) Are they well-studied? Alternatively, are there good reasons to consider them boring and ignore them?
2. Whence the asymmetry? It's almost as if $Set$ is insisting that $\times$ is "more primary" than $+$. Of course, $Set^{\mathrm{op}}$ will sing the opposite tune, but I still find the situation surprising. Is there a good intuition for why the "correct" way to ask for coproducts in arbitrary categories is equivalent to asking for a corepresentation of Set-theoretic products, rather than (say) a direct representation of Set-theoretic coproducts?
3. Given a construction in $Set$ (such as a product or a coproduct or an exponential), what determines whether that construction in going to be "representable-ish" versus "corepresentable-ish"? We could of course just try both and see which one works and classify our constructions accordingly, but is there any way to look at products and coproducts in $Set$ and notice ab initio that products are going to be the "representy" ones and coproducts the "corepresenty" ones? (Alternatively, what are the keywords I should be searching to read up on this?)

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(lmn if I should expand on my notation more here. Note that I'm following the nlab's convention when distinguishing between representable functors and corepresentable functors, and that by eg $(\to A) \times (\to B)$ I mean $(X \mapsto \mathrm{Hom}_{\mathcal C}(X, A) \times \mathrm{Hom}_{\mathcal C}(X, B))$.)

Answer 1

Suppose we had, for some objects $A$ and $B$, an object $C$ such that $(C,-)$ is naturally isomorphic to $(A,-)+(B,-)$. Then the category in which this happens cannot have a terminal object, since such an object, call it $T$, would have only one element in $(C,T)$ but two in $(A,T)+(B,T)$.

More generally, any functor of the form $(C,-)$ preserves all the products that exist in your category. So $(A,-)+(B,-)$ would have to preserve products, and this leads to lots of problems since $(A,-)$ and $(B,-)$ individually also preserve products. More generally yet, you'll have a problem with preservation of limits.

I'm not sure how far one can carry this line of reasoning, but it certainly prevents the existence of such representing objects $C$ in the categories that people usually want to work with. (At the moment, I can't think of any category where such $A,B,C$ exist, but that may be just a deficiency of my imagination.)

Answer 2

The universal property of products gives a correspondence $$ \operatorname{Hom}_{\mathcal{C}}(X, A) \times \operatorname{Hom}_{\mathcal{C}}(X, B) \cong \operatorname{Hom}_{\mathcal{C}}(X, A \times_{\mathcal{C}} B)$$ while the universal property of coproducts gives a correspondence $$ \operatorname{Hom}_{\mathcal{C}}(A, Y) \times \operatorname{Hom}_{\mathcal{C}}(B, Y) \cong \operatorname{Hom}_{\mathcal{C}}(A \sqcup_{\mathcal{C}} B, Y).$$ (where I am using $\sqcup$ to denote coproducts, since I would rather reserve $+$ for biproducts). Note that the product $\times$ on the left of those equations is a product of hom-sets, rather than a product in the category $\mathcal{C}$. This product of hom-sets matches up with the product in the functor categories $[\mathcal{C}^{\mathrm{op}}, \mathsf{Set}]$ and $[\mathcal{C}, \mathsf{Set}]$, which is why only the product, rather than the coproduct, is appearing.

To ask for a representing object for $(\to A) \sqcup (\to B)$ would be to ask for a natural construction $(A, B) \to A \bullet_{\mathcal{C}} B$ such that there is a natural correpsondence $$ \operatorname{Hom}_{\mathcal{C}}(X, A) \sqcup \operatorname{Hom}_{\mathcal{C}}(X, B) \cong \operatorname{Hom}_{\mathcal{C}}(X, A \bullet_{\mathcal{C}} B)$$ which is a strange condition in some categories (eg in $\mathcal{C} = \mathsf{Set}$, we are asking for maps $X \to A \bullet B$ to correspond bijectively to either a morphism $X \to A$ or a morphism $X \to B$).

Answer 3

Coproducts don't always look much like disjoint unions; consider the coproduct of groups, namely the free product. More formally, there's a condition called disjointness that coproducts can satisfy, and they don't always satisfy it.

You can think of your definition as an attempt to formalize disjointness; unfortunately it doesn't work. You're asking for an object $D$ such that a morphism $f : C \to D$ is either a morphism $C \to A$ or a morphism $C \to B$. Unfortunately, even disjoint unions don't satisfy this property! You don't expect this kind of thing unless $C$ is connected in some sense (say, it is a connected topological space in $\text{Top}$ or a connected graph in $\text{Graph}$); in general some part of $C$ might map to $A$ and some other part can map to $B$.

Nevertheless there is something to this idea; it's one way to try formalizing a sum type, namely a type whose terms are either terms of some type $A$ or some other type $B$. From this perspective it's not obvious that sum types have anything to do with coproducts; I wish I understood this better than I do.

A definition that seems related to bridging this gap is the notion of an extensive category.