Two different kind of $\Delta$-complex structures are given to $\Delta_n \times$I in the following two proofs:

  1. Homotopic maps induce chain homotopic maps at chain complex level.

  2. Subdivision of linear chain is chain homotopic to identity(in the proof of excision).

Apparently the main theme is boundary of a prism consists of top and bottom simplices and the vertical wall. What I find difficult to understand is how one can get intuition in partitioning the prism into simplices. Any help will be appreciated.


1 Answer 1


Short answer

This triangulation is a particular case of a general construction for simplicial sets, and triangulations of products of simplices $\Delta^p\times \Delta^q$ are related to combinatorial objects named shuffles. A reference for that is probably Gabriel, Zisman, "Calculus of Fractions and Homotopy Theory", chapter II, or the modern monograph "Simplicial Homotopy Theory" by Goerss and Jardine (though I am not sure they discuss shuffles there).

A longer explanation

Let me try to explain from scratch how these triangulations arise. My explanation is not for the connoisseurs of simplicial stuff, so I apologize for a lengthy post and sweeping under the rug some details.

Let's say that a triangulation is something consisting of the following data:

  • a finite linearly ordered set $V$ of vertices (I will prefer to write "$v\to w$" for $v\le w$)
  • a set of faces $F$, where by definition a face is a nonempty subset $\sigma\subseteq V$.

We also require by definition each face to be triangulated, i.e. if $\sigma\in F$ and $\tau \subset \sigma$ is a nonempty subset of $\sigma$, then $\tau \in F$.

Now for each $n = 0,1,2,\ldots$ let $K_n$ be the set of the $n$-simplices of the triangulation, where by an $n$-simplex we mean an ordered collection of $n+1$ vertices $v_0 \to \cdots \to v_n$: $$K_n = \{ (v_0 \to \cdots \to v_n) \mid \{ v_0,\ldots,v_n \} \in F \}.$$ Here we do not require $v_i$ to be different! They are ordered, but neighboring vertices may repeat. Treating something spanned by $(n+1)$ vertices with repetitions as an $n$-simplex may seem odd, but it will be very useful in a moment.

We have special maps $\sigma_i\colon K_n\to K_{n+1}$, called degeneracy operators. By definition, $\sigma_i$ repeats the $i$-th vertex: $$\sigma_i (v_0 \to \cdots \to v_n) = (v_0 \to \cdots \to v_{i-1} \to v_i \to v_i \to v_{i+1} \to \cdots \to v_n).$$ These operators, together with face operators $\partial_i\colon K_n\to K_{n-1}$ (removing the $i$th vertex) satisfy the so-called simplicial identities, and the sets $K_n$ with these operators form a simplicial set (you can read the details elsewhere), which I will denote by $K$.

When some face lies in the image of a degeneracy operator $\sigma_i$ (i.e. has repeating vertices), we say that it is degenerate.

For example, a triangle with ordered vertices $0 < 1 < 2$ has a triangulation where $F$ consists of all nonempty subsets of $\{ 0, 1, 2 \}$.

enter image description here

Here I enumerated the simplices in dimensions $0,1,2$ and I highlighted the degenerate ones (with repetitions of vertices). Note that starting from dimension $3$, everything will be degenerate (since we have only three vertices).

Now comes the interesting part: if $K$ and $K'$ are two simplicial sets, I can take their product $K\times K'$, where the $n$-simplices in $K\times K'$ will be pairs $$(K\times K')_n = K_n\times K_n' = \{ (x,x') \mid x\in K_n, ~ x'\in K_n' \},$$ and the degeneracy operators will be $$\sigma_i (x,x') = (\sigma_i (x), \sigma_i (x')).$$

Now a simplex $(x,x')$ is degenerate in $K\times K'$ if it is in the image of $(\sigma_i, \sigma_j)$ for some $i=j$, i.e. if both $x$ and $x'$ are degenerate, and via the same degeneracy operator $\sigma_i$. So $K\times K'$ will have more nondegenerate simplices than $K$ and $K'$. Degenerate things in $K$ and $K'$ may give something nondegenerate in $K\times K'$, and intuitively, these are the extra parts we have to add to triangulate $K\times K'$.

Let's work out the easiest example: take the interval $\Delta^1$ which consists of two ordered vertices $0\to 1$. The product $\Delta^1 \times \Delta^1$ will have the following simplices:

enter image description here

Again, I highlight the degenerate simplices. We have four $0$-simplices, five (!) nondegenerate $1$-simplices, and two (!) nondegenerate $2$-simplices. For instance, $(0\to 0\to 0, 0\to 0\to 1)$ is degenerate because it is $\sigma_0 (0\to 0,0\to 1)$. However, $(0\to 0\to 1, 0\to 1\to 1)$ is nondegenerate: though both $0\to 0\to 1$ and $0\to 1\to 1$ are degenerate in $\Delta^1$, they are degenerate via different operators $\sigma_i$ and $\sigma_j$ with $i\ne j$.

From this data I can draw a picture with triangulation, and only nondegenerate simplices will matter. I get a square with vertices $(0,0), (0,1), (1,0), (1,1)$, precisely the $0$-simplices of $\Delta^1\times \Delta^1$. Then the nondegenerate $1$-simplex $(0\to 0, 0\to 1)$ becomes the interval connecting $(0,0)$ with $(0,1)$; the nondegenerate $1$-simplex $(0\to 1, 0\to 1)$ becomes the "diagonal" connecting $(0,0)$ with $(1,1)$, and so on. The nondegenerate $2$-simplex $(0\to 0\to 1, 0\to 1\to 1)$ becomes the triangle spanned by the vertices $(0, 0)$, $(0, 1)$, $(1, 1)$, and $(0\to 1\to 1, 0\to 0\to 1)$ becomes the triangle spanned by the vertices $(0, 0)$, $(1, 0)$, $(1, 1)$.

enter image description here

Similarly one could work out $\Delta^2\times \Delta^1$ (this example looks more exciting, but going through all the possible simplices is kind of tedious):

enter image description here

Though (hopefully) one can understand from the example above how to construct triangulations from a list of nondegenerate simplices in $\Delta^p\times \Delta^q$, I haven't really explained how an abstract simplicial set gives rise to some CW-complex (glued precisely from the nondegenerate simplices). This is called geometric realization, but that's a different story...

And for more details on combinatorics of $\Delta^p\times \Delta^q$, I refer to Gabriel and Zisman. They explain how the nondegenerate simplices in $\Delta^p\times \Delta^q$ precisely come from shuffles.

Triangulations of prisms and chain homotopies

You mentioned that to prove that homotopic maps $X\to Y$ induce chain homotopic maps between complexes $C_\bullet (X) \to C_\bullet (Y)$, you need to triangulate $\Delta^n\times \Delta^1$. In fact, you can construct a chain homotopy inductively, avoiding pulling out of a hat a mysterious combinatorial expression for $h_n$ for all $n$. See Tammo tom Dieck, "Algebraic Topology", §9.3.


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