# Freely adding degeneracies does not change the homotopy type

Background: Let $S$ be a simplicial set. By freely adding degeneracies to $S$, I mean first applying the forgetfull functor from simplicial sets to semi-simplicial sets which forget the already existing degeneracies maps and then apply its left adjoint which freely add the degeneracies. One gets a new simplicial set $S'$ which can be described explicitly by:

$$S'_n = \left\lbrace (a,f) |{ a \in S_k \text{ and } f:\{0,\dots,n\} \twoheadrightarrow \{0,\dots,k\} \atop \text{ is an order preserving surjection}}\right\rbrace$$

The simplicial operations being applied to (a,f) as if it was a formal "$f^* a$".

It is know that the canonical map from $S'$ to $S$ (which send $(a,f$) to $f^* a$) is a weak equivalence. Indeed the geometric realization of $S'$ is exactly the same as the "fat geometric realization of $S$" (see def 1 and def 2 in nlab) which is known to be weakly equivalent to the classical geometric realization of $S$.

My question: is it possible to give a purely combinatorial proof of the homotopy equivalence between $S$ and $S'$. The best would be an explicit finite sequence of simplicial homotopy equivalences or trivial Kan fibrations between them.

My motivation is to determine to what extent a result of this kind is likely to be true in more general situations like for example simplicial objects in nice enough combinatorial model categories.

It would also be helpful if someone could point me to a proof that the fat geometric realization is homotopy equivalent to the usual one. I already know two proofs of that in the literature:

• The first is in Segal's "Categories and cohomology theories". It deals with the more general case of the geometric realization of a simplicial space, but it just sketches the proof; and once stripped from all the topological consideration not really relevant for my question, it just say something like "it is true when $S$ is $\Delta^n$" (which is already not really trivial I think) and then there is a not completely clear argument to extend this to an arbitrary simplicial set by colimit.

• An other one that I have not been able to found back but which was just saying that it is a standard exercise to check that this map induce a bijection on homology and fundamental groups. I haven't check if this exercise is easy or not, but I was hoping for a more explicit proof.

The nLab also mention a "more detailed" proof due to Tammo tom Dieck, but there is no link to it and I couldn't find it on google.

## 1 Answer

Here is a purely combinatorial and straightforward proof.

First, you can easily check that $\Delta[m]'$ is the nerve of the category $[m]'$ that is obtained from the poset $[m]$ by freely adjoining one idempotent endomorphism to each object of $[m]$. These categories $[m]'$ are contractible since we can write down a natural transformation from the constant functor at $0$ to the identity functor.

Correction: these idempotents are not exactly freely adjoined. They are supposed to act as identities on the morphisms of $[m].$

Next, we use a general fact (verified by induction over skeleta) that if we have a functor $F$ from simplicial sets to simplicial sets and a natural transformation between $F$ and identity such that $F$ preserves colimits and cofibrations and sends simplices to contractible simplicial sets, then this natural transformation is a weak equivalence. The first two conditions are clear for $FS = S'$ and I have verified the third one above so the conclusion follows.

• Ok, that's good for me, thanks ! After some thought about it, it is in fact exactly the proof in Segal's paper specialized to the discrete case and spelled out a little more ! – Simon Henry Oct 29 '14 at 13:01
• Here's a question: how does $F:=(S\mapsto S')$ interact with the Joyal model structure (quasicategories)? Clearly, $S$ and $S'$ aren't weakly equivalent in this structure, but it seems that $F$ will preserve Joyal weak equivalences. Does $F$ have an interpretation on the level of quasicategories? – Charles Rezk Oct 29 '14 at 14:05
• I don't know, but it seems like semi-simplicial sets should model $(\infty,1)$-categories without identities (or maybe without chosen identities). Then this functor should perhaps forget identites and then adjoin new ones so that the old ones become new idempotents as in my answer. However, modelling $(\infty,1)$-categories by semi-simplicial sets seems subtle and I'm not sure of any of this. – Karol Szumiło Oct 30 '14 at 12:40
• This also what Jacob Lurie say here : mathoverflow.net/questions/75094/… But he didn't make it formal either. – Simon Henry Oct 30 '14 at 13:54