In the book "The Topology of CW Complexes" p.78,
A CW complex is regular if each closed cell is homeomorphic to a closed Euclidean n-cell. A CW complex is normal if each closed cell is a subcomplex.
We point out that if (X, S) is regular, then we can choose a cell structure on X in which each characteristic map is a homeomorphism and for which the cells are the same subsets of X as those of the cells of S. However, this need not be strictly equivalent in the sense of Chapter I to a cell structure defining S.
If the characteristic maps in a given cell structure are already homeomorphisms, is it also a normal CW complex?
My attempt:
Proof by contradiction, assume there is a $n$-cell, say $e_n$, which is partially covered by a gluing map of a $(n{+}1)$-cell, say $g : \partial D_{n+1} \to X$, where $n \ge 1$. By weak topology, the covered region, written as $g(\partial D_{n+1}) \cap e_n$, is closed under $e_n$. That means there is a boundary point of the covered region which is in the interior of this cell: $x \in \partial (g(\partial D_{n+1}) \cap e_n)$. Find an $n$-disk neighborhood $V$ on the point $x$ under the cell $e_n$, and let $U = g^{-1}(g(\partial D_{n+1}) \cap V)$, which is open because $g(U)$ is open under the subspace $g(\partial D_{n+1})$. In a general sense, $g(U)$ is like half $n$-disk, which must not be homeomorphic to an open set in Euclidean space, but I don't know how to show this.