Joyal's CatLab Spans and cospans

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Spans

Recall that a span between two objects $A$ and $B$ of a category $\mathbf{E}$ is a pair of maps

We shall write $(S,u,v):A\Rightarrow B$ to indicate that $u:S\to A$ and $v:S\to B$. The object $A$ is the domain and the object $B$ the codomain of $(S,u,v)$. The map $u$ is the source map and the map $v$ the target map of the span $(S,u,v)$. The transpose of a span $(S,u,v):A\Rightarrow B$ is the span $(S,v,u):B\Rightarrow A$,

If $(S,u,v)$ and $(S',u',v')$ are two spans $A\Rightarrow B$, then a map $f:(S,u,v)\to (S',u',v')$ is an arrow $f:S\to S'$ in $\mathbf{E}$ such that the following diagram commutes,

The spans $A\Rightarrow B$ form a category $Span(A,B)$. If the cartesian product $A\, \times\, B$ exists, then a span $(S,u,v):A\Rightarrow B$ can be described by a single map $(u,v):S\to A\times B$. Moreover, the category $Span(A,B)$ is isomorphic to the category $\mathbf{E}/(A\times B).$

Let us now suppose that the category $\mathbf{E}$ has finite limits. Recall that the composite of a span $S=(S,s,t):A\Rightarrow B$ with a span $T=(T,u,v):B\Rightarrow C$ is the span $T \circ S=(S\times_B T, s pr_1, v pr_2): A \Rightarrow C$, defined by the following diagram with a cartesian square,

The composition operation $(T,S)\mapsto T\circ S$ defines a functor of two variables

The composition of spans is coherently associative. More precisely, if $S=(S,s,t):A\Rightarrow B$, $T=(T,u,v):B\Rightarrow C$ and $U=(U,x,y):C\Rightarrow D$, then the composite of the natural isomorphisms,

obtained from the diagram

is a natural isomorphism $U\circ (T\circ S) \simeq (U\circ T)\circ S.$

Cospans

References

Stephen Lack, R.F.C. Walters, R.J. Wood: Bicategory of spans as bicartesian categories. Theory and Applications of Categories. Vol 24 (2010). here