Is a CW complex, homeomorphic to a regular CW complex?

Question

Is a CW complex, homeomorphic to a regular CW complex ?

Regular means that the attaching maps are homeomorphisms (1-1). In particular, an open (resp. closed) $n$-cell, is homeomorphic to an open (resp. closed) $n$-ball.

Answer 1

Something that you could employ in answering this is to note that every regular CW-complex is in fact triangulable.

It turns out that every closed topological manifold of dimension other than four is homeomorphic to a CW-complex [Kirby-Siebenmann, On the triangulation of manifolds and the Hauptvermutung]. On the other hand Ciprian Manolescu has recently proved an important result that there exist non-triangulable manifolds in any dimension above four (before this work, the only known examples of non-triangulable manifolds were of dimension four).

Ok, the above employs some extremely high-powered (and modern) machinery, but the idea take a non-triangulable manifold which is a CW-complex is pretty simple, at least! And I guess this also shows that the space can still be "nice", in a particular sense.

Answer 2

I don't believe so. Consider a CW complex formed by gluing a $2$-cell to a circle, where the attaching map $\varphi:S^1\to S^1$ has a point with infinite preimage. There are many ways to construct such a map. For example, take the Peano space filling curve and project to the first coordinate. This gives such a map from $[0,1]$ to $[0,1]$, and you can join the ends to form a circle. There are also easier constructions, but I don't have time to describe them well. If this space is a regular CW complex, then this bad point on the boundary needs to be infinitely many different $2$-cells, which is a contradiction since infinite CW complexes are not compact.

I bet your statement is true if you replace "homeomorphic" with "homotopy-equivalent." Then you can iron out local pathologies like this.

EDIT (8/30/2013): In order to make this argument more rigorous, let's pick a nicer function $\varphi\colon S^1\to S^1$. Start with the continuous function $x\mapsto x\sin(1/x)$ which sends $[0,1]$ to $[-\sin(1),\sin(1)]$. Convert this to a self map of $[0,1]$ by mapping $[-\sin(1),\sin(1)]$ homeomorphically onto $[0,1]$, and convert this to a map $\varphi\colon S^1\to S^1$ by joining the endpoints. Note that $x\mapsto x\sin(1/x)$ has the property that for all $n$, there is a subinterval of $(\epsilon_1,\epsilon_2)\subset [0,1]$ where the horizontal line $y=\epsilon$ hits the graph in exactly $2n$ points for all $\epsilon\in(\epsilon_1,\epsilon_2)$. Now, in the CW complex $K=S^1\cup_{\varphi} D^2$, each of these subintervals gives an open subset of the complex (if you consider the part of $D^2$ attaching along these pieces, homeomorphic to $*_{2n}\times (0,1)$, where $*_{2n}$ is the open "star" graph with one central vertex, $2n$ peripheral vertices, and an edge connecting center to each peripheral vertex, with peripheral vertices removed. So we have a core edge with $2n$ "fins" sticking out. The core edge is part of the $1$-skeleton since it is not locally homeomorphic to an open subset of the plane for $n>2$. Fixing a point $p$ in the core edge that is not a vertex of the $1$-skeleton, the parts of the fins near $p$ must each be part of some $2$-cell. But in fact all of these will need to be different $2$-cells, since the attaching map would have a double point at $p$ otherwise. So we have at least $2n$ 2-cells. Since $n$ was arbitrary, we have infinitely many, which is a contradiction of compactness.