# What is enough to conclude that something is a CW complex?

This question was something I considered when looking into CW-structures on Grassmannians, but I found no general treatment of this in the literature:

Question: Assume that $X$ is an $n-1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Assume also that $X'$ is compact Hausdorff. Is $X'$ homeomorphic to a CW complex given by attaching a single $n$ cell to $X$?

Remark: Maybe I am missing some obvious counter example!

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Is your $x$ supposed to be $X$? You only mention it once. If true, your question is vacuous. A space which has a path-component an open cell can't be compact. – Ryan Budney Mar 17 '10 at 17:02
I have changed the $x$ to $X$. I write explicitly that the disjoint union is as sets NOT as spaces. So indeed any $n$ dimensional finite CW complex with one $n$ cell satisfies the assumptions. – Thomas Kragh Mar 17 '10 at 17:10

Let $X$ be the closed subspace of $R^2$ which is the union of $0\times [-1,1]$ and the point $(1, \sin(1))$. Let $e$ be the graph in the plane of $f(x)=\sin(1/x)$ for $x\in (0,1)$ (the "topologists sine curve"), and let $X'=X\cup e$, viewed as a subspace of $R^2$.

I believe that $X'$ is closed, and so is compact Hausdorff, $e$ is an open subset of $X'$ homeomorphic to the open interval, and $X$ is a CW-complex. But $X'$ is not obtained by attaching a $1$-cell to $X$: you can't produce a map $\Phi\colon[0,1]\to X'$ which restricts to a homeomorphism $(0,1)\to e$.

This is not quite a counterexample to your problem, because you wanted the new cell $e$ to have greater dimension than the cells of $X$. But it seems like you could use this kind of idea to make a counterexample.

Added. Let $S^2$ be the unit sphere in $R^3$, pick a point $p$ on $S^2$, and consider the function $f\colon (S^2-\{p\})\to R$ given by $f(x)=\sin(1/|x-p|)$. Let $X'$ be the closure of the graph of $f$ inside $S^2\times [-1,1]$; this should consist of $X=\{p\}\times [-1,1]$ (a CW complex) and $e=$ graph of $f$ (homeomorphic to an open $2$-disk). This would seem to be the counterexample you want.

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Though that might be harder than I thought at first ... – Charles Rezk Mar 17 '10 at 17:40
Consider the revolution surface in $\mathbb R^3$ obtained by rotating the graph of $t\in(-\sqrt{\pi/2},\sqrt{\pi/2})\mapsto\sin(\tan(t^2))$, add a cylinder around it to get a compact subspace, and next collapse all parallels in that cylinder to a point. – Mariano Suárez-Alvarez Mar 17 '10 at 17:50
Excuse me, but what do you mean by "around it"? – J. Fabian Meier Mar 17 '10 at 17:56
@Fabian, I meant «take the closure»: when you do that, a cylinder appears. The example resulting from my construction is homeomorphic to Charles'. – Mariano Suárez-Alvarez Mar 17 '10 at 18:26