What is the product in the 2-category of spans? 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 $\operatorname{Span}(\mathcal S)$?

 A: 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).
