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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?

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  • $\begingroup$ This may be of interest. $\endgroup$ – mathphys Jun 24 at 14:44
  • $\begingroup$ @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 :) $\endgroup$ – graeme Jun 24 at 15:10
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    $\begingroup$ 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. $\endgroup$ – Mark Kamsma Jun 24 at 15:17
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    $\begingroup$ Just for the record : what you call "pseudo-product" is known as a weak product (ore more generally, weak limit). $\endgroup$ – Arnaud D. Jun 24 at 15:42
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    $\begingroup$ 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) $\endgroup$ – Max Jun 24 at 16:22
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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.)

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