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For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $u\to v$ are all finite sequences $(a_1,\dots,a_n)$ of morphisms $a_i\in Mor(C)$ such that $a_1:u\to x$ and $a_n:y\to v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,\dots,a_m)(b_1,\dots,b_n)=(a_1,\dots,a_mb_1,\dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?

Update:

1) Motivation: the construction $C\mapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C \cup (C\times C)$ modulo the relations which hold in $C$.

2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:

a) type $w$: they tell me I should compose certain morphisms of $C$ the result of which is the morphism $a_1$, say, and thereby I run through the underlying graph $\Gamma$ of $C$

b) type $w^{\mathfrak m}$: they say I should jump elsewhere in the graph $\Gamma$

The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,\dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $\dots a_i,a_{i+1}\dots$ and $\dots a_ia_{i+1}\dots$.

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  • $\begingroup$ Does the composition have $n+m-1$ terms? $\endgroup$
    – Dog_69
    Jan 15, 2019 at 19:16
  • $\begingroup$ @Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence. $\endgroup$
    – user 59363
    Jan 15, 2019 at 19:23
  • $\begingroup$ And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it? $\endgroup$
    – Dog_69
    Jan 15, 2019 at 20:17
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    $\begingroup$ Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me... $\endgroup$
    – Arnaud D.
    Jan 15, 2019 at 21:45
  • $\begingroup$ @Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly). $\endgroup$
    – user 59363
    Jan 16, 2019 at 10:42

1 Answer 1

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It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.

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  • $\begingroup$ This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms). $\endgroup$
    – user 59363
    Jan 15, 2019 at 20:03
  • $\begingroup$ I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$. $\endgroup$ Jan 15, 2019 at 20:05
  • $\begingroup$ Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct? $\endgroup$
    – magma
    Jan 26, 2019 at 6:10
  • $\begingroup$ @magma No, the empty sequence is not admitted. $\endgroup$
    – user 59363
    Jan 27, 2019 at 22:37
  • $\begingroup$ So what sequence is the identity on A? $\endgroup$
    – magma
    Feb 2, 2019 at 8:02

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