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

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?)

(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 $$(X \mapsto \mathrm{Hom}_{\mathcal C}(X, A) \times \mathrm{Hom}_{\mathcal C}(X, B))$$.)

## 3 Answers

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.)

• Actually, one can prove that a coproduct of representable is never representable, as in this question. – Arnaud D. Jun 11 '19 at 6:06
• This seems to be the crux of things. Thanks! I'm still surprised about the asymmetry in $Set$ -- for ex, it's trivial to prove that a (co)representation of a terminal set is a (co)terminal object, and that there are no (co)representations of the coterminal set, and I'm still curious as to whether you (or @qiaochu-yuan or @joppy) have any intuitions for what's up with $Set$ being so biased in favor of limits in this fashion. – Nate Jun 13 '19 at 12:25
• @Nate Here's another bias of Set in the direction of limits, which may or may not be connected to the one you asked about. The usual way of specifying a set $S$ is to say what its members are. In categories, "members" should usually be replaced by "generalized members", i.e., morphisms into $S$. And those are the sort of things that are involved in the definition of limit (whereas the definition of colimit involves morphisms out of $S$). – Andreas Blass Jun 13 '19 at 13:24

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.

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 $$X \to B$$).