# Turning simplicial complexes into simplicial sets without ordering the vertices

Given an abstract simplicial complex $$K$$, one can make a simplicial set $$X(K)$$ with $$n$$-simplices given by sequences $$(x_0, \dotsc, x_n)$$ such that $$\{x_0, x_1, \dotsc, x_n\}$$ is a simplex of $$K$$. The face maps delete entries and the degeneracy maps repeat entries. I'd like a reference for the fact that the geometric realization of $$X(K)$$ is homotopy equivalent to the geometric realization of $$K$$ itself. (Note that $$\lvert X(K)\rvert$$ is typically very big: for $$K$$ a single edge, $$\lvert X(K)\rvert$$ is the infinite-dimensional sphere $$S^\infty$$.)

I've sketched a proof of this fact at Turning simplicial complexes into simplicial sets, but hope there is a reference I can just cite since, as I expected, every algebraic topologist I've asked in person already knew the fact. :)

Also, does this $$X(K)$$ have a standard name or notation? Or if not, can someone think of a catchy name or nice notation?

• As for the notation, I think the simplicial set $K(X)$ is strictly linked to symmetric simplicial sets. If $\Upsilon$ denotes the category of symmetric simplices, there is a canonical funtor $v \colon \Delta \to \Upsilon$ which induces a Quillen equivalence pair $(v_!,v^*)$ between the category of presheaves (see §8.3). It seems to me that $K(X)$ is precisely $v_!(K_{\leq}(X))$ which in turn is precisely your $E \otimes_{\Delta} K_{\leq}(X)$. Moreover, you show that the unit $1 \to v^*v_!$ is always a weak homotopy equivalence. Mar 26, 2019 at 22:39
• If one only reads the question title, one foolish way that comes to mind is this: allow only neighboring repetitions. That is, if $x_i=x_{i+j}$ then $x_i=x_{i+k}$ for all $0<k<j$ too. I wonder if this is also equivalent to $K$... Apr 8, 2019 at 17:45
• @მამუკაჯიბლაძე I don't think the simplicial set you describe is homotopy equivalent to the complex you start with. For example if $K$ is a single simplex of dimension $n$, then I believe your simplicial set is the simplicial set obtained by freely adjoining degenerate simplicies to the semi-simplicial set known as the "complex of injective words on $n+1$ letters", which is not contractible. Specifically, for $K$ a single interval your simplicial set has the homotopy type of $S^1$. Apr 9, 2019 at 18:19
• Concerning terminology and notation - it might be natural to call elements of your $X(K)$ singular simplices of $K$ and, accordingly, denote $X(K)$ by $\operatorname{Sing}(K)$. Apr 9, 2019 at 21:38
• @მამუკაჯიბლაძე Yes, very appropriate! After all $X(K)_n$ is precisely the set of morphisms of simplicial complexes from the n-simplex to $K$. Apr 9, 2019 at 22:38

$$\newcommand\geom[1]{\lvert#1\rvert}\newcommand\Geom[1]{\lVert#1\rVert}$$Let $$K$$ be a simplicial complex with vertex set $$V$$. Let $$S_\bullet (K)$$ be the simplicial set whose $$p$$-simplices are the maps $$f:[p]\to V$$ such that $$f([p])$$ is a simplex of $$K$$, or alternatively the set of maps $$\Delta^p \to K$$ of simplicial complexes. There is an obvious map $$\pi_K:\geom{S_\bullet (K)} \to \geom K$$ which you ask to be a homotopy equivalence. Here is an argument. I find it easier to work with the fat geometric realization $$\Geom{S_\bullet (K)}$$ instead, but the difference is minimal, since the quotient map to the ordinary geometric realization is a homotopy equivalence.

Step 1. Consider first the case $$K=\Delta^n$$ (rather, it is the full simplicial complex with vertex set $$[n]$$). I claim that $$\Geom{S_\bullet \Delta^n}$$ is contractible. For sake of notational clarity, let me write $$\nabla^p$$ for the topological $$p$$-simplex. Consider the map $$H_p: S_p (\Delta^n) \times \nabla^p \times [0,1] \to S_{p+1}(\Delta^n) \times \nabla^{p+1}$$ which is given by the formula $$H(f,v,t):= (f \ast n,((1-t)v,t)).$$ Explanation: $$f \ast n: [p+1] \to [n]$$ is the map whose restriction to $$[p]$$ is $$f$$ and which has $$f(p+1)=n$$. Furthermore $$((1-t)v,t) \in \mathbb{R}^{p+1} \times \mathbb{R}$$ is a point of $$\nabla^{p+1}$$. It is easily checked that the different $$H_p$$ glue together to a map $$H:\Geom{S_\bullet (\Delta^n)} \times [0,1] \to \Geom{S_{\bullet}(\Delta^n)}$$ (use that products and quotients commute in this setting, as the interval is compact, or work in the context of compactly generated spaces). It is clear that $$H(0,\_)$$ is the identity, and $$H(1,\_)$$ is the constant map to the vertex $$n$$. So we are done in this case.

Step 2. Now we prove the claim for finite complexes, by induction over both, the dimension and the number of top-dimensional simplices. The induction beginning $$K=\emptyset$$ is trivial. For the induction step, let $$K$$ be $$n$$-dimensional and let $$L$$ be obtained from $$K$$ by deleting one $$n$$-simplex. Then $$\geom K \cong \geom L \cup_{\geom{\partial \Delta^n}} \geom{\Delta^n}$$ and $$\Geom{S_\bullet (K)} \cong \Geom{S_\bullet (L)} \cup_{\Geom{S_\bullet (\partial \Delta^n)}} \Geom{S_\bullet (\Delta^n)}$$. The map $$\pi_K$$ is the pushout of the maps $$\pi_{\Delta^n}$$ and $$\pi_L$$, along $$\pi_{\partial \Delta^n}$$. These maps are homotopy equivalences, by step 1 and by induction hypothesis, respectively. The maps $$\geom{\partial \Delta^n} \to \geom{\Delta^n}$$ and $$\Geom{S_\bullet (\partial \Delta^n)} \to \Geom{S_\bullet (\Delta^n)}$$ are cofibrations, and so the gluing lemma implies that $$\pi_K$$ is a homotopy equivalence.

Step 3. Having shown the claim for finite complexes, it follows by a colimit argument that $$\pi_K$$ is a weak homotopy equivalence for arbitrary $$K$$, and hence a homotopy equivalence, by Whitehead's theorem.

• Thanks Johannes! I think for Step 1 I prefer the argument that goes "$S_\bullet \Delta^n$ is the nerve of the indiscrete category on $n+1$ vertices, and this category is equivalent to the terminal category", but I really like the directness of the rest of the proof. Apr 18, 2019 at 17:07
• I'm not sure how surprised I should be that we now have 3 proofs and 0 references for this fact I always thought was "standard". Apr 18, 2019 at 17:10

$$\DeclareMathOperator\Simp{Simp}\newcommand\geom[1]{\lvert#1\rvert}$$I'm not sure if we need a 4th proof of this fact, but after wondering about this for several years I realized it can be proven in a very formulaic manner: turn everything in sight into (the classifying space of) a category and apply Quillen's Theorem A. Let's see how it goes.

The simplicial complex $$\geom K$$ is more-or-less already a category — if we think of $$K$$ as just its set of simplices, then it's partially ordered by inclusion, and the geometric realization of this poset $$(K, \subseteq)$$ is the usual barycentric subdivision of $$\geom K$$ (and in particular, it's homeomorphic to $$\geom K$$).

Next, every simplicial $$X$$ set can be “turned into a category” by taking the category of simplices $$\Simp(X)$$. I don't know a simple proof that $$\geom{\Simp(X)}$$ is homotopy equivalent to $$\geom X$$, but it is proven in Hirschhorn's book Model Categories and Their Localizations, Theorem 18.9.3.

We want a functor between $$\Simp(X(K))$$ and $$(K, \subseteq)$$. This is straightforward; objects of $$\Simp(X(K))$$ are (in bijection with) simplices of $$X(K)$$, and we can send a list $$(x_0, \dotsc, x_n)$$ to the simplex $$\{x_0, \dotsc, x_n\}$$ in $$K$$. All diagrams in $$K$$ commute (since it's a poset), and this makes it easy to verify that this defines a covariant functor $$s: \Simp(X(K))\rightarrow (K, \subseteq)$$.

Finally, let's try applying Theorem A and see what happens. Fix a simplex $$\sigma\in K$$. The fiber of $$s$$ consisting of all simplices in $$X(K)$$ that map to faces of $$\sigma$$ can be though of as the category of lists in the set $$\sigma$$, with a morphism of lists being a way of embedding one list as a sublist of another. It is more-or-less immediate from the definitions that this fiber category is isomorphic to $$\Simp(N_* (I(\sigma)))$$, where $$I (\sigma)$$ is the indiscrete category on the set $$\sigma$$ (that is, the object set of $$I(\sigma)$$ is $$\sigma$$, and each morphism set has exactly one element). Since $$I(\sigma)$$ is equivalent to the trivial category, $$\geom{N_* (I(\sigma))}$$ is contractible. By the discussion above, so is $$\geom{\Simp(N_* (I(\sigma)))}$$, and Theorem A says that $$\geom s:\, \geom{\Simp X(K)}\stackrel{\simeq}{\rightarrow} \geom K$$ is a homotopy equivalence.

• Nice proof, Dan! (Also: great to hear from you!) Dec 9, 2022 at 2:01