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$\begingroup$

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?2012-06-11
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    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 $\text{StarComp}(m;f)$.2012-06-11
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    @Hurkyl: Missing indexes added.2012-06-11
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    @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.2012-06-11
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    I'm not calling $M$ a set of *morphisms*, I'm calling $M$ a set of *objects*. It looks like you're extending $C$ by adding in a set $M$ of formal products of its objects, and defining a morphisms between any two elements of $M$ to be a list of arrows from $C$, one for each component of the domain.2012-06-11
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    P.S. the associativity rule seems to imply arity(m)$\subseteq$arity(Starcomp(m;f)). Are you supposed to have equality?2012-06-11
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    @Hurkyl: Elements of $M$ are generalized morphisms. Each element of $M$ has an indexed family of objects of $C$. These are not products.2012-06-11
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    @Hurkyl: Yes: arity(m)=arity(Starcomp(m;f)).2012-06-11
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    It really, really looks like $m$ is supposed to be object-like and $(m;f)$ morphism-like. If you're sure that $m$ is supposed to be morphism-like, then I guess I don't see anything useful to say.2012-06-11
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    @Hurkyl: I am sure that $m$ is supposed to be morphism-like.2012-06-11
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    I know what's missing! There isn't a composition rule for two elements of $M$. So, if we're supposed to be thinking of $M$ as morphisms, then in terms of what you've described, only the codomain (given by Obj) of an element plays a role, and that's why it looks object-like to me.2012-06-11
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    Looks like a multicategory to me...2012-06-11

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