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Let $\mathcal S$ be a category with finite limits. The 2-category $\operatorname{Span}(\mathcal S)$ has the same objects as are in $\mathcal S$. For objects $X,Y$, the hom category in $\operatorname{Span}(\mathcal S)$ between $X$ and $Y$ is the category of diagrams in $\mathcal S$ of the form $X \leftarrow \bullet \rightarrow Y$, and morphisms are natural transformations of such diagrams that restrict to $\operatorname{id}_X,\operatorname{id}_Y\,\,$ at the endpoints. The 1-composition of 1-morphisms is given by the obvious pull-back diagrams: $$\{X\leftarrow A \rightarrow Y\} \circ \{Y\leftarrow B \rightarrow Z\} = \{X\leftarrow A\underset Y \times B \rightarrow Z\}$$ Thinking of $\mathcal S$ as a 2-category with only identity morphisms, the "spanishization" functor (does this functor have another name?) $\mathcal S \to \operatorname{Span}(S)$ is the identity on objects and takes $\{X \overset f \to Y\}$ to $\{X = X \overset f \to Y\}$. There is also an obvious isomorphism $\mathcal S \cong \operatorname{Hom}_{\operatorname{Span}}(1,1)$, where $1 \in \mathcal S$ is the terminal object.

I believe that the correct weakened notion of "cartesian product" in a 2-category is that $X\times Y$ is determined up to equivalence (not isomorphism) as the representing object for the 2-functor $Z \mapsto \operatorname{Hom}(Z,X) \times \operatorname{Hom}(Z,Y)$, where on the right-hand side is the usual product of categories. (Incidentally, what's a good reference for $n$-Yoneda's Lemma?) Even if $\mathcal S$ has finite limits, or even all small limits, then I'm not sure whether $\operatorname{Span}(\mathcal S)$ has finite products. But for good enough categories $\mathcal S$, I feel like $\operatorname{Span}(\mathcal S)$ should also be good.

However, I believe that the product in $\operatorname{Span}(\mathcal S)$ is not the product in $\mathcal S$, i.e. spanishization does not respect limits. Provided all my beliefs are correct:

Is there an easy description of the product in $\operatorname{Span}(\mathcal S)$ in terms of $\mathcal S$? Do any products at all exist in $\operatorname{Span}(\mathcal S)$?

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"Spanification"? –  Qiaochu Yuan Jul 19 '10 at 20:14
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up vote 5 down vote accepted

If $\mathcal S = Set$, it looks to me like the product in $Span(\mathcal S)$ is given by disjoint union.

Define functors $F_Z: Span(Z,X\sqcup Y) \to Span(Z,X)\times Span(Z,Y)$ $$ F_Z(Z \leftarrow A \rightarrow X \sqcup Y) = ((Z \leftarrow A \times _{X\sqcup Y}X \rightarrow X), (Z \leftarrow A \times _{X\sqcup Y}Y \rightarrow Y)) $$

and $ G_Z: Span(Z,X)\times Span(Z,Y) \to Span(Z,X\sqcup Y)$

$$ G_Z((Z \leftarrow B \rightarrow X), (Z \leftarrow C \rightarrow Y)) = (Z \leftarrow A\sqcup B \rightarrow X\sqcup Y). $$

Basically, if you have a span between $Z$ and $X\sqcup Y$, take the preimage of each component to get a pair of spans. I think it should be clear what happens on morphisms. Now, in Set (or maybe we just require that products distribute over coproducts), I think these form an equivalence. This should also behave well under maps $Z \to Z^\prime$.

This seems pretty weird to me, and it may not be correct (my category theory is pretty rudimentary). I am interested to find out more though, as it seems related to a question I was once asked about limits and colimits in cobordism categories (thinking of cobordisms as certain cospans of manifolds).

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This looks correct. It does require that product distribute over coproduct, which does not hold in an arbitrary category. For example, it does not hold in algebraic categories like RINGS or COMMUTATIVE RINGS or GROUPS, where the coproduct is given by a "free product" or "tensor product" construction. –  Theo Johnson-Freyd Jul 20 '10 at 3:46
    
But my intended applications are mostly SETS or some other species of SPACES, where product does distribute over coproduct. –  Theo Johnson-Freyd Jul 20 '10 at 3:47
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Yes, that's right. I think what you actually need is that the category is extensive: ncatlab.org/nlab/show/extensive+category which is a bit stronger than being distributive. Note that Span is equivalent to its opposite via an identity-on-objects functor, and hence these products are also coproducts. –  Mike Shulman Jul 20 '10 at 6:36
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