The simplex category as a “free” category

abstract-algebracategory-theorysimplicial-stuffsoft-question

Consider the simplex category $\Delta$. Its objects are the linearly ordered sets of the form $[n]=\{0<1<\dots<n\}$. Its morphisms are nondecreasing functions. There are two special classes of morphisms in this category:

  • the "face maps": $\delta^i\colon [n-1]\to [n]$ (this is the unique nondecreasing strictly increasing function which omits $i$),
  • the "degeneracy maps": $\sigma^i\colon [n+1]\to [n]$ (which duplicates $i$).

While reading I got the impression:

$\Delta$ is the free category generated by the $\delta^i$ and $\sigma^i$'s satisfying the simplicial identities.

Question: How can one make that precise, what does it mean?

What I've seen written out is a proof of the fact that each morphism in $\Delta$ can be written as a composition of $\delta^i$ and $\sigma^i$'s. And of course, one can check for oneself that thesee $\delta^i$ and $\sigma^i$'s satisfy the simplicial identities.

But I want to see I way to make the analogy with, say, free monoids precise. For each monoid $M$, each function $f\colon X\to M$ can be uniquely extended to a function $FX\to M$ on the free monoid generated by $X$, sending $a_1\dots a_n$ to $f(a_1)\circ\dots\circ f(a_n)$. In the same way, the data of a simplicial set (consisting of a family of sets $S_n$ and face maps $d_i\colon S_n\to S_{n-1}$ and degeneracy maps $s_i\colon S_n\to S_{n+1}$ satisfying the simplicial identities) can be uniquely extended to the datum of a presheaf on $\Delta$: a contravariant functor $\Delta\to \mathbf{Set}$. To implement this, use the fact that each morphism in $\Delta$ can be written uniquely as a composition of the $\delta^i$ and $\sigma^i$'s, and then extend using the face and degeneracy maps specified in the datum of a simplicial set.

I know that every directed multigraph $G$ has a free category $FG$ consisting of paths in $G$. Can this notion be used to answer the question above?

Best Answer

Yes, you're close to the idea. What you're missing is relations. For monoids, the monoid $M=\langle S\mid R\rangle$ generated by a set $S$ subject to a set of relations $R\subseteq F(S)\times F(S)$ can be constructed as the coequalizer in the category of monoids of the two maps $F(R)\rightrightarrows F(S)$ induced by the two maps $R\to F(S)$ determined by the fact that $R$ is a subset of $F(S)^2.$

The story is formally identical for categories. Intuitively, a category with object set $A$ can be generated by a (directed multi)graph $G$ with vertex set $A$ together with a set of relations given by equations between some "words" in the edges of $G.$ Formally, the relations $R$ form a subgraph of $F(G)\times_A F(G),$ the graph of all parallel pairs of edges in the graph $G.$ Then we can define the category we're generating as the coequalizer in the category of categories of the two induced functors $F(R)\to F(G).$

Now, how to concretely understand these coequalizers? If we were talking groups, then we would just take the quotient of $F(S)$ by the normal subgroup $N$ generated by the relators. The nice thing is that this quotient can be constructed in the category of sets: it's the quotient of $F(S)$ by the equivalence relation containing $(w,w')$ if and only if $w^{-1}w'$ is in $N.$ For monoids, the story is not so simple, but we can still take the quotient of $F(S)$--in the category of sets--by the congruence generated by all the relations: this is the smallest equivalence relation on $F(S)$ containing the relations which is also a submonoid of $F(S)\times F(S).$ (It's nice to check that this is correct in the special case of groups.)

Again, a parallel story allows us to define the category with some generators and relations solely in terms of coequalizers by equivalence relations in the category of graphs, which is just as easy as in sets since graphs form a presheaf category. In short, the arrows of the category $\mathcal C=\langle G\mid R\rangle$ are equivalence classes of arrows in $F(G)$ under the smallest equivalence relation on the arrows of $F(G)$ containing $R$ and forming a subcategory.

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