This is 5-18b of J. Lee's "Introduction to Topological Manifolds".
Suppose $X$ is a regular CW complex. Then let $\mathcal{E}$ be the set of open cells of $X$, $\mathcal{K}$ the collection of all nonempty finite subsets $\{e_0,…,e_k\}\subseteq\mathcal{E}$ with the property that the dimensions of the elements are all distinct, and $e_{i-1}\subseteq\partial e_i$. I have shown that $\mathcal{K}$ is an abstract simplicial complex in part a) of the question. Part b) proceeds as follows:
Suppose $K$ is a Euclidean simplicial complex whose vertex scheme is isomorphic to $\mathcal{K}$. Show that $X$ is homeomorphic to $|K|$ (the polyhedron of $K$) via a homeomorphism that sends the closure of each cell of $X$ onto the polyhedron of a subcomplex of $K$.
The reader is given the following hint: begin by choosing $v_e$ in each simplex $e\in\mathcal{E}$. Then define a homeomorphism inductively, one skeleton at a time, in such a way that it sends each point $v_e$ to the vertex of $K$ corresponding to $e$.
In previous exercises, I proved that any homeomorphism $f:\partial D\to\partial D'$ can be extended to a homeomorphism $F:D\to D'$, for two closed cells $D$ and $D'$. I understand that this fact should help me construct the homeomorphism in some inductive fashion, but I am not sure how to realise this construction. For the $0$-skeletons, it seems clear that the map should take every $e_0$ to $\{e_0
\}$. Then, assume the homeomorphism has been constructed for the $(n-1)$-skeleton. Denoting by $|K|_i$ the $i$-skeleton of the polyhedron $|K|$, we have a homeomorphism $F_{n-1}:X_{n-1}\to |K|_{n-1}$. From this, how would I create the homeomorphism for the $n$-skeleton?
Best Answer
That earlier exercise is not going to be sufficient, because your description of the induction has a flaw.
As you say, the homeomorphism between $X$ and $|K|$ is being constructed inductively on the skeleta $X_0 \subset X_1 \subset X_2 \subset \cdots$. However, your induction hypothesis is incorrect, because the image $f(X_n)$ is not going to be equal to the entire $n$-skeleton of $|K|$; instead it is only going to be a subcomplex of the $n$-skeleton of $|K|$.
For example, $f(X_0)$ is not the entire $0$-skeleton of $|K|$, it is only the subcomplex of the $0$-skeleton corresponding to singleton sets $\{e_0\}$ such that the dimension of $e_0$ is equal to $0$.
More generally, given a cell $e \subset X$ of dimension $d \ge 1$ corresponding to a $0$-simplex $\{e\}$ of $|K|$, the point of $X$ which is mapped to that $0$-simplex by $f$ is going to be the chosen point $v_e$ in the interior of $e$. Thus the $0$-simplex $\{e\}$ is not contained in $f(X_{d-1})$, but it is contained in $f(X_d)$.
With that said, let me strengthen the hint from Lee's book somewhat. The description of $f$, and the induction hypothesis of the construction of $f$, should both state with more precision what is the image under $f$ of each closed cell in $X$:
If you can answer that question, you should be able to reformulate your induction and then proceed with the construction.