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?

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    $\begingroup$ 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. $\endgroup$ Mar 26, 2019 at 22:39
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    $\begingroup$ 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 $K$... $\endgroup$ Apr 8, 2019 at 17:45
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    $\begingroup$ @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 $S^1$. $\endgroup$ Apr 9, 2019 at 18:19
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    $\begingroup$ 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 $\operatorname{Sing}(K)$. $\endgroup$ Apr 9, 2019 at 21:38
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    $\begingroup$ @მამუკაჯიბლაძე Yes, very appropriate! After all $X(K)_n$ is precisely the set of morphisms of simplicial complexes from the n-simplex to $K$. $\endgroup$ Apr 9, 2019 at 22:38

2 Answers 2


$\newcommand\geom[1]{\lvert#1\rvert}\newcommand\Geom[1]{\lVert#1\rVert}$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.

  • $\begingroup$ 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. $\endgroup$ Apr 18, 2019 at 17:07
  • $\begingroup$ 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". $\endgroup$ 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.

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

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