Intuition behind defining prism operator/chain homotopy Two different kind of $\Delta$-complex structures are given to $\Delta_n \times$I in the following two proofs:

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


*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.
 A: 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:


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*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 $\{ 0, 1, 2 \}$.

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:

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 $(0, 0)$, $(1, 0)$, $(1, 1)$.

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):

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.
