Start with a category $C$. Form a monoid $M$ whose elements are lists of morphisms in the category $C$ subject to commuting diagrams in $C$. Is there a name for this construction or a better way to categorially understand this?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ if you consider lists of morphisms in a category $\mathcal{C}$, how can you compose these (and get a monid)? Then I guess you consider concatenated morphisms, but this isnt a monoid (if there is more than one object) in general these are the morphims of the free category associated to the graph of $\mathcal{C}$, then if you identify two of its arrow if as sequence of morphisms have some composition in $\mathcal{C}$ (i.e. these two make a commutative diagram) in this way you get (but a isomorphism functor) the category $\mathcal{C}$. $\endgroup$– Buschi SergioCommented Nov 6, 2011 at 11:52
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
Do you mean take the monoid with the following presentation? Take the arrows of C as generators and add the relations that f.g = fg if f and g are composable and that each identity of C be equivalent to 1? I would call this the universal monoid U(C) of C. It has the universal property that there is a functor $F:C\to U(C)$ (the obvious one) such that any functor from C to a monoid M, viewed as a one-object category, factors through $U(C)$. So U is the left adjoint to the inclusion of the category of monoids into the category of small categories.
I myself have used the group version of this many times.
answered Nov 6, 2011 at 16:30
-
$\begingroup$ Ah, this seems to fit to the question. $\endgroup$ Commented Nov 6, 2011 at 22:24
-
$\begingroup$ Ah, oh so there is a group version and a monoid version. Indeed universal monoid is a good name for it. Thanks. This is indeed what I was looking for. $\endgroup$– user2529Commented Nov 9, 2011 at 11:55
-
$\begingroup$ Should every identity arrow be made equivalent to 1? Perhaps a pointed version is better, and only the identity arrow of the point be made equivalent to 1. $\endgroup$– user2529Commented Nov 9, 2011 at 12:21
-
$\begingroup$ You could not identify any identity with 1 and get the universal semi functor to a semigroup. $\endgroup$ Commented Nov 9, 2011 at 20:18
$\begingroup$
$\endgroup$
For every $n$ we have the set $N_n(C)$ of $n$-chains $X_0 \to \dotsb \to X_n$ of morphisms in $X$. This set is actually a category, the morphisms are commutative diagrams. A short definition is $N_n(C) = \mathrm{Hom}([n],C)$, where the preorder $[n] = \{0,\dotsc,n\}$ is considered as a category. In fact, $\{N_n(C)\}_n$ is a simplicial set, called the nerve of $C$. It's geometric realization is also called the geometric realization $|C|$ of $C$.
In your question, you obviously vary $n$ and want to get a single monoid. It's not clear to me how you want to achieve this. Perhaps you really mean the nerve, or do you want to consider $\coprod_n N_n(C)$?
answered Nov 6, 2011 at 9:46