# Morphisms with an arbitrary number of objects

Is this structure familiar for you?

It consists of

1. a category $C$
2. a set $M$
3. a function $\operatorname{arity}$'' defined on $M$
4. a function $\operatorname{Obj}_m$ defined for every $m \in M$, such that $\operatorname{dom} \operatorname{Obj}_m = \operatorname{arity} m$
5. a function (star composition) $\left( m ; f \right) \mapsto \operatorname{StarComp} \left( m ; f \right)$ defined for $m \in M$ and $f$ being an $(\operatorname{arity} m)$-indexed family of morphisms of $C$ such that $\forall i \in \operatorname{arity} m : \operatorname{Src} f_i = \operatorname{Obj}_m i$ ($\operatorname{Src} f_i$ is the source object of the morphism $f_i$) and $\operatorname{arity}\operatorname{StarComp} \left( m ; f \right) =\operatorname{arity}m$.

such that it holds:

1. $\operatorname{StarComp} \left( m ; f \right) \in M$
2. (associativiy law) $\operatorname{StarComp} \left( \operatorname{StarComp} \left( m ; f \right) ; g \right) = \operatorname{StarComp} \left( m ; \lambda i \in \operatorname{arity} m : g_i \circ f_i \right)$

(Here by definition $\lambda x \in D : F \left( x \right) = \left\{ \left( x ; F \left( x \right) \right) \hspace{0.5em} | \hspace{0.5em} x \in D \right\}$.)

The meaning of the set $M$ is an extension of $C$ having as morphisms things with arbitrary (possibly infinite) indexed set $\operatorname{Obj}_m$ of objects, not just two objects as morphims of $C$ have only source and destination).

We may also add the requirement that $\operatorname{StarComp} \left( m ; \lambda i \in \operatorname{arity}m : \operatorname{id}_{\operatorname{Obj}_m i} \right) = m$ (the law of composition with identity).

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Are you missing some indices in your associativity law? Specifically, $f$ and $g$ are indexed families of morphisms, so what does $g \circ f$ mean? – Hurkyl Jun 11 '12 at 19:44
Eyeballing it, it looks like $M$ is intended to be a collection of formal "cartesian products" of objects, and when $(m;f)$ is defined, $f$ represents a formal morphism from $m$ to $\text{StarComp}(m;f)$. – Hurkyl Jun 11 '12 at 19:52
@Hurkyl: Missing indexes added. – porton Jun 11 '12 at 19:58
@Hurkyl: I'm not sure whether every "morphism" in $M$ can be obtained as a product. I rather deem not every morphism in $M$ is a product. – porton Jun 11 '12 at 19:59
Looks like a multicategory to me... – Zhen Lin Jun 11 '12 at 21:23