I have just started studying simplicial sets, and I was given the definition of geometric realization of a simplicial set X as $$|X| = (\coprod_{n \in \mathbb{N}} X_n \times \Delta_n) / \sim$$ with $(d_i(x), v) \sim (x, \delta_i(v)$ and $(s_i(x), v) \sim (x, \sigma_i(v))$, denoting by $d_i, s_i$ the face and degeneracy maps on $X$ respectively.

I have shown that, given a simplex $x$ of $X$, there exists a unique nondegenerate simplex $\overline{x}$ and a unique sequence $i_1 \leq \cdots \leq i_k$ such that $x = s_{i_1} \cdots s_{i_k} (\overline{x}).$ I would like to use this result to prove the following:

Let $\overline{X_n}\subset X_n$ be the subset of nondegenerate $n$-simplices. Then, there is a homeomorphism $$|X| \cong (\coprod_{n \in \mathbb{N}} \overline{X_n} \times \Delta^n) / \sim$$ with $(x, \delta_i(u)) \sim (\overline{d_i(x)}, \sigma_{i_k} \cdots \sigma_{i_1} (u))$, where $d_i(x) = s_{i_1} \cdots s_{i_k} (\overline{d_i(x)}).$

But I don't know how to proceed. I found online that there is a bijection if I consider $\overline{X_n} \times \delta \Delta^n$, but it is not a homeomorphism.

Any help will be highly appreciated. Thanks!

  • $\begingroup$ There should be a proof of this in Goerss-Jardine. $\endgroup$ Commented Nov 25, 2023 at 21:38
  • $\begingroup$ @VincentBoelens I had a look, but unfortunately I couldn't find anything that helped me (I only saw that there is a result concerning a pushout with nondegenerate simplices and skeleta). $\endgroup$ Commented Nov 25, 2023 at 23:43
  • $\begingroup$ I am referring to proposition 2.3 in the first chapter. Why do you think it doesn't help you? $\endgroup$ Commented Nov 26, 2023 at 8:24

2 Answers 2


As FShrike notes in the comments, you'll need to use the Eilenberg-Zilber Lemma: For any simplex $x$ in $X_n$, there is a unique $\pi : n \to m$, with $m \leq n$, and $y \in X_m$ with $x = \pi(y)$. One way to show what you want is through an argument with the skeleta; I'll give a more conceptual answer as to how we can calculate the geometric realization using just non-degenerate simplices. As an abuse of notation, for $\pi \in \Delta$ I will write $\pi(x)$ for the action on simplices and $pi(u)$ for the action on a point $u$ of a standard simplex under $|-| : \Delta \to \mathsf{Top}$

Recall that one way of defining the geometric realization functor is as the left Kan extension of $|-| : \Delta \to \mathsf{Top}$ along the Yoneda embedding $y : \Delta \to \mathsf{sSet}$. With this description, the geometric realization of $X \in \mathsf{sSet}$ is the colimit of $\int X \xrightarrow{\Pi} \Delta \xrightarrow{|-|} \mathsf{Top}$, where $\int X$ is the category of elements of $X$: objects are $(n,x)$ for $[n] \in \Delta$ and $x \in X_n$, and a morphism $(n,x) \to (m,y)$ is a map $\pi : [n] \to [m] \in \Delta$ such that $\pi(y)=x$ (see Category Theory in Context, definition 2.4.2). The functor $\Pi : \int X \to \Delta$ is the projection given by $\Pi(n,x) = [n]$ and $\Pi(\pi: (n,x) \to (m,y)) = \pi: [n] \to [m]$.

What we can use to show your proposition is the notion of a final functor (sometimes called cofinal). A functor $F : C \to D$ is final if, for any $d \in D$, the comma category $d \downarrow F$ is non-empty. That is, for any $d \in D$, there is some $c \in C$ and morphism $g : d \to Fc$, and moreover, any two morphisms are connected by a zigzag of commutative triangles. Final functors have the property that, for any diagram $G : D \to E$ which admits a colimit, the induced map $\operatorname{colim} GF \to \operatorname{colim} G$ is an isomorphism (see Categories for the Working Mathematician, Chapter IX.3).

Now, let $\int_{nd}X$ is the full subcategory of $\int X$ spanned by those $(c, x)$ for which $x$ is nondegenerate. Using the EZ-lemma and Eilenberg-Zilber axioms, you can show that the inclusion $\iota : \int_{nd} X \to \int X$ is final. Thus, we can calculate the geometric realization of a simplicial set $X$ as the colimit of

$$\int_{nd} X \to \int X \xrightarrow{\Pi} \Delta \xrightarrow{|-|} \mathsf{Top}$$ Call this composite $S_X$, so $S_X(n,x) = \Delta^n$; I will identify $S_X(n,x) \cong \{x\} \times \Delta^n$ to keep track of which simplex things are paired with. Explicitly, using some standard formulas for colimits, this can be calculated as the coequalizer of: $$\coprod_{\sigma : (n,x) \to (m,y) \in \operatorname{Mor}\int_{nd} X} S_X(n,x) \rightrightarrows \coprod_{(k,z) \in \int_{nd} X}S_X(k,z)$$ where one of the parallel maps is induced by the cone $$\left\{S_X(n,x) \xrightarrow{S_X(\pi)} S_X(m,y) \hookrightarrow \coprod_{(k,z) \in \int_{nd} X} S_X(k,z)\right\}_{\pi: (n,x) \to (m,y)}$$ and the other is induced by the cone $$\left\{S_X(n,x) \xrightarrow{id} S_X(n,x) \hookrightarrow \coprod_{(k,z) \in \int_{nd} X} S_X(k,z)\right\}_{\pi: (n,x) \to (m,y)}$$ We may identify $$\coprod_{(k,z) \int_{nd} X} S_X(k,z) \cong \coprod_{k \in \mathbb{N}} \overline{X_k} \times \Delta^k$$ with $\overline{X_n}$ denoting non-degenerate $n$-simplices, given the discrete topology. In this case, on the component corresponding to $\pi: (n,x) \to (m,y)$, our maps take $(x,v) \in S_X(n,x)$ to $(x, v)$ and $(y, \pi (v))$. Recalling our definition of the category of elements, here $x = \pi(y)$

But we know how to calculate coequalizers in $\mathsf{Top}$: here we'll define $\sim$ to be the smallest equivalence relation for which $(\pi(y), v) = (x,v) \sim (y, \pi(v))$ for each $\pi : (n,x) \to (m,y)$, and then take $|X| = \operatorname{colim} S_X = \coprod_{(k,z) \in \int_{nd} X} \Delta^k/\sim$. Thus, we have formed the geometric realization using just non-degenerate simplices; the end product looks similar to what you had originally suggested but with some modifications, so I hope it's still of use.

One detail worth checking is that the only maps $\pi$ in $\int_{nd} X$ are injective (hence can be written as composites of face maps), as otherwise this contradicts that all simplices involved are nondegenerate. Try thinking about what this would correspond to geometrically, and the differences compared to the "naive" construction taking all simplices of $X$.


Some comments. The levelwise subsets of nondegenerate simplices will not in general assemble to a simplicial (sub)set and so their geometric realisations don't quite make sense. Note that the face of a nondegenerate simplex can be degenerate!

One possible way to make this make sense is exactly the result you mentioned in comments. If you can accept the pushout involving the skeletal filtration (which requires some combinatorical fiddling to rigorously establish) then you immediately obtain the canonical CW structure on the realisation of a simplicial set. This is expressed in terms of the nondegenerate simplices: $|X|$ is the colimit of $|X|_0\subseteq |X|_1\subseteq\cdots$ where each $|X|_n$ is pushed out from the first by: $$\require{AMScd}\begin{CD}\bigsqcup_{\sigma\in X_n^{\mathsf{nd}}}S^{n-1}@>>>|X|_{n-1}\\@VVV@VVV\\\bigsqcup_{\sigma\in X_n^{\mathsf{nd}}}D^n@>>>|X|_n\end{CD}$$What more could you want? :)

Here I have used the fact the geometric realisation is cocontinuous and preserves pushouts. Also, $|\Delta^n|=|\Delta^n|\cong D^n$ by definition and it is not too tricky to argue, using a combinatorial presentation of $\partial\Delta^n$, that $|\partial\Delta^n|\cong\partial|\Delta^n|\cong S^{n-1}$ in such a way that $\partial\Delta^n\hookrightarrow\Delta^n$ induces the canonical inclusion $S^{n-1}\hookrightarrow D^n$.

Well, another thing you might want for finite $n$-dimensional sets is that $|X|\cong|Y|$ where $|Y|$ is the $n$-truncation (or $n$-skeleton) of $X$. This is also true.

  • $\begingroup$ Thank you for your answer. My main difficulty while reading the reference above lies in the fact that I was not introduced to the concept of skeleta - let's say that I only the main definition of simplicial sets and not much more. Hence, my approach to the problem was defining an explicit bijection between $|X|$ and the realization via nondegenerate simplices (call it $Y$). From $Y$ to $|X|$, it seems that the inclusion of $(x,v)$ with $x$ nondegenerate in $|X|$ works. I wanted then to define $|X| \to Y, (x,v) \mapsto (\overline{x}, \sigma_{i_k} \cdots \sigma_{i_1}(v))$ ... $\endgroup$ Commented Nov 27, 2023 at 20:57
  • $\begingroup$ ...where $x=s_{i_1} \cdots s_{i_k} (\overline{x})$, but I'm not able to show that this is actually well-defined, using the given relations on $|X|$ and $Y$! At the same time, I don't see how it could be something different. $\endgroup$ Commented Nov 27, 2023 at 20:59
  • $\begingroup$ @AliceinWonderland I recommend learning a little more abstract theory first. Manual combinatorial computations are sometimes necessary but are always painful; a great success of the categorical manipulation of simplicial stuff is that it minimises this pain. However, it seems the Eilenberg-Zilber lemma may be useful for a manual computation, if you wish. Any simplex is uniquely expressible as $s(a)$ for a nondegenerate simplex $a$ and degeneracy map $s$ $\endgroup$
    – FShrike
    Commented Nov 27, 2023 at 21:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .