# Is there a category where products don't exist because uniqueness fails?

I was looking at this question about categories without products, and the main examples are:

1. fields
2. manifolds with boundary
3. posets

But these all seem to fail for either structural reasons (fields/manifolds) or simply no reasonable definition of product in the category is available (posets)

My question is, is there an example of a category that has "pseudo-products", i.e. they match every requirement of the definition of product except they fail the uniqueness condition? Does this question even make sense to ask?

• This may be of interest. – mathphys Jun 24 at 14:44
• @mathphys thanks, that looks interesting, I don't have the background at this point to understand the specific examples but if I understand 'intuition', we can always extend a category without products to a category with products, although we may lose some structural properties. I guess that makes it sound like the issue is always there are too few ways to make products, and not too many (which i guess is another way to ask my question). I will have to come back to that later and see if it makes sense :) – graeme Jun 24 at 15:10
• This is not really worthy of a full answer: but there is the notion of polylimit, there we have a set of cones $\{C_i\}_{i \in I}$ and given any other cone $K$ there is an induced arrow $K \to C_i$ for some $i \in I$, and this induced arrow is only unique up to isomorphism. Although I have to be honest, I just dualized the notion of polycolimit, for which I actually do know a natural example: the algebraic closures of prime fields are a polyinitial object in the category of fields. – Mark Kamsma Jun 24 at 15:17
• Just for the record : what you call "pseudo-product" is known as a weak product (ore more generally, weak limit). – Arnaud D. Jun 24 at 15:42
• Don't posets have products ? with the product order ? (EDIT : oh you didn't mean "the category of posets", you meant "a posetal category" - I'll leave this comment nonetheless to avoid further confusion) – Max Jun 24 at 16:22

Sure. For instance, take the category of sets whose cardinality is not $$4$$. This category obviously has all products except for a product of two $$2$$-element sets (or products of higher arity where two of the factors have two elements and the rest have one, which are essentially the same thing since a singleton is terminal). But a weak product of two $$2$$-element sets (call them $$A$$ and $$B$$) does still exist. For instance, let $$C$$ be any set with more than one element and consider $$P=A\times B\times C$$ with its projections $$p$$ and $$q$$ to $$A$$ and $$B$$. I claim $$(P,p,q)$$ is a weak product of $$A$$ and $$B$$ (that is, it satisfies the definition of a product except for uniqueness of the maps). Indeed, let $$Q$$ be any set and $$f:Q\to A$$, $$g:Q\to B$$. Pick any function $$c:Q\to C$$ (such a function exists since $$C$$ is nonempty), and define $$h:Q\to P$$ by $$h(x)=(f(x),g(x),c(x))$$. Then $$ph=f$$ and $$qh=g$$, as desired.
(Note that no product of $$A$$ and $$B$$ exists in this category, since by considering maps from a singleton set, such a product would need to have $$4$$ elements.)