Proving that the simplex category is generated by the face and generacy maps

Question

Note: This is not intended to be a research level question, but concerns graduate level material.

Theorem. The opposite $\Delta^\mathrm{op}$ of the simplex category $\Delta^\mathrm{op}$ (as usually defined via nondecreasing maps) can be presented as follows: it is the category generated by the graph of finite nonempty linearly ordered sets with arrows $\delta_i\colon [n]\to [n-1]$ and $\sigma_i\colon [n]\to [n+1]$ for each $0\leq i\leq n$ subject to the simplicial identities.

Here is a proof sketch using the following well-known fact:

Fact. Any nondecreasing map $f$ in $\Delta^\mathrm{op}$ has a unique representation as $$f=\sigma^{i_1}\dots\sigma^{i_n}\delta^{j_1}\dots\delta^{j_m},$$ where $i_1>\dots > i_n$ and $j_1 <\dots < j_m.$

Here, the $\sigma^i$ and $\delta^j$ are now regarded as nondecreasing maps, while in the above theorem they denote formal arrows. Sorry for the abuse of notation.

The above theorem means that for any category $\mathcal C$ there is a canonical bijection between functors $\Delta^\mathrm{op}\to\mathcal C$ and simplicial objects in $\mathcal C$ (given by face and degeneracy morphisms satisfying the simplicial identities). This is the universal property of being presented by generators and relations. By forgetting information one can a assign a simplicial object to each functor $\Delta^\mathrm{op}\to\mathcal C$. So the challenge is to construct an inverse to that assignment. Let $C_\bullet$ be a simplicial object with face maps $d_i$ and degeneracy maps $s_i$. We want to define a functor $\Delta^\mathrm{op}\to\mathcal C$. On objects this functor is given by $C_\bullet$. Now let $f\colon [n]\to [m]$. We need to find an induced morphism $$C(f)\colon C_n\to C_m.$$ To get one, write $$f=\sigma^{i_1}\dots\sigma^{i_n}\delta^{j_1}\dots\delta^{j_m},$$ and then set $$C(f)=s^{i_1}\dots s^{i_n}d^{j_1}\dots d^{j_m}.$$ One can now check that this defines a functor as desired.

Question: Is there a proof of the above theorem (i.e., a construction of the bijection between functors $\Delta^\mathrm{op}\to\mathcal C$ and simplicial objects of $\mathcal C$) that does not use the above fact? I wonder because the unique representation of a morphism $f$ in $\Delta^\mathrm{op}$ looks a bit random and ad hoc to me. In other settings when wanting to show that some concrete category can be presented by generators and relations, does one always convulsively have to find a way that each morphism can be written in a unique way as a composition of "generators" (satisfying additional properties, like $i_1 > \dots > i_n$ in the above case)?

I feel the theorem should be true nevertheless: just write $f$ as any composition of $\delta^i$ and $\sigma^i$s, and then map it to the same pattern where $\delta$ is replaced by $d$ and $\sigma$ is replaced by $s$. However, in this approach one has to check well-definedness. Can the well-definedness be checked without essentially reproducing the above fact?

The only advantage of the above fact seems to be to have a unique well-defined way to define the mapping. But I strongly feel like the theorem should be true even without particular ad hoc representation.

Answer 1

This is not an answer to the question as stated, but an elaboration on my comment (the OP asked me to clarify, but this would be too long for a comment).

Let me start by pointing out that this theorem becomes really easy to prove once one observes that the composition of $\delta$'s is injective, and that of $\sigma$'s is surjective. Indeed, given our non-decreasing map $f$, if this decomposition is to exist, it must correspond to the decomposition $i\circ s$ of $f$ as a surjection followed by an injection. This reduces both existence and uniqueness to the case of an injection or a surjection.

But both of these are easy again : if $i$ is an injective non-decreasing map between linearly ordered sets (in fact it must then be increasing), then it is entirely specified by its image, and in fact by the complement of its image. Now in the composition $\delta_{j_m}\circ ... \delta_{j_1}$ in $\Delta$, if $j_m>...>j_1$, then the complement of the image is exactly the subset $\{j_1,...,j_m\}$, which proves at once existence and uniqueness.

The argument for surjections is very similar, you just observe that now a surjective nondecreasing map between finite linearly ordered sets is entirely determined by the places where it doesn't "jump".

So first of all, the theorem is not hard and quite conceptual once you think about it. Of course, saying "this theorem is nice, don't ask how to avoid using it" does not answer your question...

The second part of my comment was about why something like this "had" to be done anyways. A "presentation" of an algebraic object is a description of this object in terms of generators and relations. This tells you how to describe morphisms out of it : say where the generators go, and check that their images satisfy these relations. What you want to show amounts (by some abstract nonsense) precisely to finding a presentation for $\Delta$ in terms of these $\delta$'s and $\sigma$'s, with relations the simplicial identities.

Now, when you start with an algebraic structure (an algebra, a module, a category , ...), it always has a tautological presentation : take all elements (/objects + arrows) as generators, and all relations as relations. This is of course useless in practice, so one has to do some specific, "ad hoc" things to get useful presentations - like the one here. Our presentation of $\Delta$ has to... depend on $\Delta$ ! So you have to prove something "ad hoc", or "random" (even though the first part of this answer argued that it was not "random", and was actually very conceptual, and the combinatorics that appear there are quite easy)

In general, it suffices to prove things like " every element can be represented in such a way, and I know how to navigate between these representations". "navigate" means different things in different contexts, but in the context of categories say, it means "I know how to represent the composite of two such representations". If you can explain, via some relations, why representations compose the way they do, then you have your presentation by generators and relations !

In other words (this is where "rewriting" comes in), given two representations of my morphisms in terms of my generators in a nice way that I like, I can represent their composite in a nice way too by following my relations, I can rewrite it using my relations. Rewriting comes up a lot in proving presentations precisely for this reason. If I can prove that every element of my group is of the form, say $a^ib^j$, it'd be nice to know how I can write $a^ib^ja^kb^l$ in these terms. This can get quite complicated, but if I have, say, a simple rule of the form $ba \to a^2 b^5$ then I know I can pass all my $a$'s past the $b^j$ and I will be done. In particular, I will have that the generators $a,b$ and the relation $ba = a^2b^5$ form a presentation of my object.

I hope that clarifies my comment. Let me now maybe explain why I don't think the answer to your question is "there is such a proof", in a reasonable sense.

You need to prove that $\Delta$ has a presentation given by these $\sigma$'s, $\delta$'s, and the simplicial identities. A thing that you have to prove anyways, is that any map can be written as a composite of $\delta$'s and $\sigma$'s. Now, if I'm being honest, I don't know how to do that in a clean way if not for the argument that I gave above.

But let's say you can fiddle around and find a proof of that without that argument. Then you still need to prove that simplicial identities are enough to generate all identities. As Dmitri Pavlov points out in the comments, one way to do so is to show that any two representations of morphisms in terms of the generators can be related by a sequence of "exchanges" allowed by the simplicial identities - he even described such a sequence in general, a "rewriting" scheme if you will. But the way he did it is he rewrote everything to reach the canonical form granted by the theorem !!

I guess the point I'm trying to make is that you can't get away from proving something specific to $\Delta$, and that something will, one way or another, look like the theorem in question. But maybe I'm wrong, and I cannot prove "there is no proof that avoids the theorem", and I'm not well-versed enough in combinatorics to be categorical about this, so don't take this as a definitive answer.