A triangulation of $\Delta_p\times I$

Question

While reading the proof of the homotopy axiom for singular homology from Lee's Introduction to Topological Manifolds and Hatcher's Algebraic Topology, I found out that there is a little technical detail that is not further commented. Adopting Lee's notation, let's recall that:

$$\Delta_p=[e_0,\dots,e_p]$$

Denotes the standard $p-$simplex. Then, in these references, it is taken for granted that the union of the $(p+1)$-simplices

$$[E_0,E_0',\dots,E_p'],[E_0,E_1,E_1',\dots,E_p'],\dots,[E_0,\dots,E_p,E_p']$$

is precisely $\Delta_p\times I$, where $E_i=(e_i,0)$ and $E_i'=(e_i,1)$. Moreover, in Hatcher it is implicitly asserted that the intersection of any two consecutive $(p+1)$-simplices

$$[E_0,\dots,E_i,E_i',\dots,E_p']\text{ and }\;[E_0,\dots,E_{i+1},E_{i+1}',\dots,E_p']$$

is precisely

$$[E_0,\dots,E_i,E_{i+1}',\dots,E_p'].$$

So my questions are the following:

Why do the given simplices form a cover of the whole space $\Delta_p\times I$? Why each of those is a subset of the aforementioned space?

Why is the intersection of any of two consecutive simplices just the face they have in common? In general, why is this decomposition a triangulation of the whole space?

I think it is obvious how to check that these assertions are true when we consider the cases $p=1,2$. However, I would like to prove this in its most general form, for any positive integer $p$. I thought about using induction, but the induction step is difficult to apply. Indeed, I still don't know how it should be applied from $p=1$ to prove $p=2$.

Thanks in advance for your time.

Answer 1

When one wonders about these things it is probably a good idea to start learning about abstract simplicial complexes and simplicial sets.

An abstract simplicial complex is a collection of sets $X_n$ (the n-simplices), indexed by the non-negative integers, and boundary maps $d_i: X_{n+1} \rightarrow X_n$ if $0 \leq i \leq n+1$ that satisfy some relations derived from the case of the usual simplicial complexes.

This is an obvious generalization to make if one cares only about simplicial complexes and simplicial maps between them. One can use the "geometric realization" functor and the "forget the topology functor" to go back and forth between simplicial complexes and abstract simplicial complexes. The "forget the topology functor" is straightforward, simply translate the boundary maps and n-simplices into an abstract simplicial complex. The "geometric realization" functor just takes an n-simplex for every object in $X_n$ and glues the boundaries to the previously created (n-1)-simplices via the information of the boundary maps.

A abstract simplicial complex map is just a collection of maps $X_i \rightarrow Y_i$ that respect the boundary maps. If one knows a little category theory, this seems ripe for describing via commuting diagrams, and one can see this is the same as the collection of functors $\operatorname{Bad}\Delta ^{op} \rightarrow \operatorname{Set}$ where $\operatorname{Bad}\Delta$ has objects the sets $\{0,1,\dots,n\}=[n]$ and arrows the strict order preserving maps between them.

Without talking about too much category theory, the reason $\operatorname{Bad}\Delta$ is bad in this context is that it only encodes information about how we can include simplices into one another. We would also like to be able to talk about simplicial maps that collapse n-simplices into smaller simplices. The way to fix this is by defining $\Delta$ to have the same objects as $\operatorname{Bad}\Delta$, but maps between them being any order preserving maps. Then we call the category of functors $\operatorname{Set}^{\Delta^{op}}$ the category $\operatorname{SSet}$. Any object of this category is called a simplicial set. When constructing a simplicial set, it suffices to provide only maps $d_i :X_m \rightarrow X_{m-1}$ and $s_i :X_m \rightarrow X_{m+1}$ provided they satisfy some relations. The first are called face maps, and the second are called degeneracy maps.

The definition is very similar, but now our models for the n-simplex are maybe not obvious (check that any non-empty simplicial set has simplicies in every dimension!) However, we let categorical intuition guide us a little, and define "the" n-simplex of our category to be the functor $\operatorname{Hom}(-,[n])$, if you are familiar with the Yoneda lemma it might be good to use it here to check that this a good definition.

One last thing about simplicial sets before we can address your problem. We call an n-simplex of a simplicial set degenerate if it is in the image of a map induced by an arrow $\Delta ^{n+1} \rightarrow \Delta^{n}$ (one of the degeneracy maps). It turns out that we can make a similar geometric realization functor that takes a simplicial set and gives us a topological space via gluing, and that the degenerate simplices do not affect the space we get.

Here are some things that are easy enough to prove:

1. The product in $\operatorname{SSet}$ is given by product of the sets of simplices with degeneracy and boundary maps given by their products.

2. Geometric realization commutes with products (at least for simplicial sets with only finitely many nondegenerate simplices).

3. Our model for an n-simplex is equivalent to the simplicial set which has nondegenerate k-simplices the k-simplices of the usual n-simplex with face (boundary) maps $d_i$ given by forgetting the ith coordinate and degeneracy maps $s_i$ given by repeating the ith coordinate. So in all, it is weakly increasing strings of numbers in $[n]$.

4. The realization of our model of an n-simplex is an n-simplex.

5. If every face of a n-simplex is nondegenerate and unique, and each face has the same property, then its corresponding portion of the geometric realization will be topologically an n-simplex. If every nondegenerate simplex has this property and no two have more than one face in common, there is a triangulation of the realization given by forgetting the degenerate simplices.

6. If every nondegenerate k-simplex is in the image of the face map of a nondegenerate (k+1)-simplex for all $k <n$, and there are no nondegenerate simplices higher dimension than $n$, then the geometric realization is a union of the topological n-simplices (which might have some boundaries identified) corresponding to the non-degenerate n-simplices with intersections given by face maps, of possibly different index, that coincide.

Let us show that for $n=p+1$ the hypothesis of statement 6 holds for $\Delta ^p \times \Delta^1$. A k-simplex of this looks like $([v_0,\dots,v_k],[w_0,\dots, w_k])$ where the $v_i$ are in $[p]$ and weakly increasing and the $w_i$ are in $[1]$ and weakly increasing. We would like to classify what all the nondegenerate ones look like. From our earlier considerations, this is the same as asking that for all $0 \leq i < k$ it is not the case that $v_i = v_{i+1}$ and $w_i=w_{i+1}$. Let's consider when this doesn't happen. If the $[v_0,\dots,v_k]$ has two indices $i,j$ such that $v_i = v_{i+1}$ and $v_j = v_{j+1}$, then it must be degenerate. This is because all simplices of $\Delta^1$ look like $[0,0,\dots,0,1,1,\dots,1]$ and if a nondegenerate pair $([v_0,\dots,v_k],[w_0,\dots, w_k])$ has a repetition in the vertices of the first simplex at index $i$, this must come precisely at the index where the second simplex switches from $0$ to $1$ because otherwise $([v_0,\dots, v_i,v_{i+2},\dots v_k],[w_0=0,\dots,w_i=0,w_{i+2}=1,\dots,w_k=0])$ would have it as its ith degeneracy. So the first simplex may never have multiple repetitions; it can have exactly one repetition if it occurs when the second simplex swaps from 0 to 1; and clearly if the first simplex has no repetitions it cannot be in the image of a degeneracy map.

All of the latter are the image of a nondegenerate simplex under a face map because we can directly construct a nondegenerate simplex by repeating an index of the first simplex exactly where the swap from $0$ to $1$ occurs in the second simplex. More can be said, if the former are not the image of a nondegenerate simplex under a face map, these all have $k=p+1$ since otherwise we could insert a new vertex to construct a simplex it is a face of. If $k=p+1$ this cannot happen since we would have to have two repetitions in the first simplex of anything it is a face of.

So by our above reasoning and the easy to prove facts, we have that the realization of $\Delta^p \times \Delta ^1$ is equal to $\Delta^p \times I$ as a topological space, and that it is the union of the realizations of the simplices $([0,1,\dots,l,l,l+1,\dots,p],[0,\dots,0,1,\dots,1])$. Each face of a single one of these simplices is clearly unique and, by our observations, nondegenerate with all faces having the same property with respect to their faces. So in the realization, the simplex corresponding to any nondegenerate simplex is homeomorphically a simplex. The only nondegenerate (p+1)-simplices that share faces are then those of the form $([0,1,\dots,l,l,l+1,\dots,p],[0,\dots,0,1,\dots,1])$ and $([0,1,\dots,l+1,l+1,l+2,\dots,p],[0,\dots,0,1,\dots,1])$ because via the sequence of 0's and 1's along with the knowledge of which, if any, indices of the first simplex are repeated, from any nondegenerate p-simplex we can reconstruct the nondegenerate (p+1)-simplex it is a face of to be one of the above. In addition, if the p-simplex is a face of both of these (p+1)-simplices, it is the only such face.

I hope this is satisfactory: we have come up with a decomposition of a space homeomorphic to $\Delta ^p \times I$ into (p+1)-simplices with the property that under the obvious ordering the only intersections are given by adjacent simplices which intersect in a p-simplex, and it has an overall triangulation given by the nondegenerate simplices of the simplicial set.

Answer 2

$\newcommand{\sset}{\mathsf{sSet}}\newcommand{\set}{\mathsf{Set}}\newcommand{\top}{\mathsf{Top}}\newcommand{\op}{^{\mathsf{op}}}$If I understand the proof of homotopy invariance of homology right (which has a direct simplicial analogue) it's not actually necessary that these simplices triangulate the prism or that their literal intersections are equal to the stated simplices. To define the prism operator we need only have that these simplices "exist" within the prism. I would agree with Connor that an approach using simplicial-stuff and category theory is good, not because it's essential - I imagine some elementary convex geometry proof exists somewhere - but because it's far more conceptual and I'm just a fan of category theory generally. I offer a different explanation to Connor's that I find more conceptual, although the detail-checking (which I omit) is arguably more tedious.

I imagine many reading this, motivated by topology alone, might not know about simplicial sets. The existing answer already provides something about this; I'll just add that simplicial sets are a clean way to package the data of intersections, incidence, orientations and operations on simplicial complex. The geometric realisation's precise definition (which I omit) is definitely weird at first glance but intuitively all it is doing is this; it is taking the abstract "combinatorial" data provided by a simplicial set and it is turning this into a space by attaching topological $n$-simplices and gluing them according to the incidence/intersection data. It should be intuitive that it preserves "colimits" - as used and explained further below - by Occam's razor; this is the simplest thing it could do. It itself works by taking the data and "gluing" things according to that data, so it should preserve colimits because all colimits are are formal "gluing" operations.

From now on I have no qualms about using categorical language.

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Write $\sset$ for the category of simplicial sets, the functors $\Delta\op\to\set$ where $\Delta$ is the usual category of finite nonempty ordinals and weakly monotone maps. I write $[n]$ for $\{0,1,2,\cdots,n\}$. $|\cdot|:\sset\to\top$ is the usual geometric realisation. We want to triangulate the prism $|\Delta^n|\times|\Delta^1|\cong|\Delta^n\times\Delta^1|$ (if the simplicial sets are finite-dimensional then $|\cdot|$ preserves products in a natural way). It is, to my mind, much more rigorous and conceptual to "combinatorically" triangulate the simplicial set $\Delta^n\times\Delta^1$ since one doesn't need to worry about points or continuity or geometry or anything so mundane as that.

Fix $n\ge1$. For $k\in[n]$ define $p_k:\Delta^{n+1}\to\Delta^n\times\Delta^1$ via the usual degeneracy maps $\sigma_k:\Delta^{n+1}\to\Delta^n$ and the map $q_k:\Delta^{n+1}\to\Delta^1$ which looks like: $$\begin{pmatrix}x:&0&1&\cdots&k&k+1&k+2&\cdots&n+1\\q_k(x):&0&0&\cdots&0&1&1&\cdots&1\end{pmatrix}$$In Hatcher's notation $p_k$ rigorously describes the simplex $[v_0,v_1,\cdots,v_k,w_k,\cdots,w_n]$ once you take the realisation $|p_k|:|\Delta^{n+1}|\to|\Delta^n|\times|\Delta^1|$.

Connor's answer, in my notation, is essentially this:

• The simplicial prism $\Delta^n\times\Delta^1$ is $\le(n+1)$-dimensional (all simplices of higher dimension are degenerate)
• The $p_\bullet$ constitute all non-degenerate $(n+1)$-simplices of the prism
• All nondegenerate $\le n$-dimensional simplices of the prism are subfaces of at least one $p_\bullet$
• By some well known properties of $|\cdot|$, the realisation of the prism is covered by the maps $|p_\bullet|:|\Delta^{n+1}|\to|\Delta^n\times\Delta^1|$ and so we have a triangulation, as desired (the maps $|p_k|$ and $|p_m|$ are distinct for different $k,m$)

Instead I claim it's possible to view $\Delta^n\times\Delta^1$ as a colimit involving the $p_k$, essentially saying the $p_k$ form a combinatorial triangulation of the simplicial prism. $\Delta^n\times\Delta^1$ is the colimit of the zigzag diagram in $\sset$: $$\Delta^{n+1}\overset{\delta_1}{\longleftarrow}\Delta^n\overset{\delta_1}{\longrightarrow}\Delta^{n+1}\overset{\delta_2}{\longleftarrow}\cdots\overset{\delta_{n-1}}{\longrightarrow}\Delta^{n+1}\overset{\delta_n}{\longleftarrow}\Delta^n\overset{\delta_n}{\longrightarrow}\Delta^{n+1}$$With $(n+1)$ appearances of $\Delta^{n+1}$ and $n$ appearances of an intersection $\Delta^n$. The cocone is provided exactly by the maps $p_k:\Delta^{n+1}\to\Delta^n\times\Delta^1$ for $k=0,1,\cdots,n$.

This is stating that the (simplicial) prism is exactly what you get when you take the $p_\bullet$ and stick them together along their common faces. Notice that $p_k \delta_{k+1}=p_{k+1}\delta_{k+1}$ for all $0\le k\le n-1$ and this is exactly stating, in Hatcher's loose notation, that the topological simplices $[v_0,v_1,\cdots,v_k,w_k,\cdots,w_n]$ and $[v_0,v_1,\cdots,v_{k+1},w_{k+1},\cdots,w_n]$ share a common face $[v_0,v_1,\cdots,v_k,w_{k+1},\cdots,w_n]$. To say that they only intersect here, and nowhere else, follows from the statement of the prism as that colimit; by gluing $p_k$ to $p_{k+1}$ along this common face and this face only, that suffices to describe the prism entirely.

To check this is a true colimit is not too hard if you're experienced with simplicial stuff. Then we can relax and just take the realisation $|\cdot|$; this preserves colimits and gives us, "for free", that the topological prism is (up to homeomorphism) exactly what you get when you glue the $(n+1)$-topological-simplices $|p_0|,|p_1|,\cdots$ along their common faces, giving the desired triangulation. Doing this first and taking geometric realisations later has the advantage that one does not need to worry about the geometry or even the topology of it; this holds for any equivalent definition of realisation up to homeomorphism and is a deeper statement at the ‘combinatorial’ level.