A triangulation of $\Delta_p\times I$ 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.
 A: 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:


*

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

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

*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 $[n]$.

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

*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.

*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.
