The book "The Topology of CW Complexes" mentions the following result in passing, but does not provide a proof. Let $(X, \mathcal{C})$ be a CW complex, $B_k \subset \mathbb{R}^k$ be the origin-centered open unit-$k$-ball, and $\overline{B_k}$ be the corresponding closed $k$-ball. Let $f, g : \overline{B_k} \to X$ both be characteristic functions of the same cell $C \in \mathcal{C}$. Then there exists a homeomorphism $h : \overline{B_k} \to \overline{B_k}$ such that $f = g \circ h$. How can you prove this statement?

For a regular cell, the statement is of course trivial, so I am specifically interested in the non-regular case.


In the light of the counter-example given by Eric Wofsey, I may have misunderstood what is said in the book. The direct quotes are as follows ($\dot{E}_n$ is the boundary of a closed cell $E_n$).


Page 1:

A closed Euclidean $n$-cell $E^n$ is a homeomorphic image of the Euclidean n-cube $I^n$, the cartesian product of $n$ copies of the closed unit interval $I = \{t \in \mathbb{R} : 0 \leq t \leq 1\}$.

Page 6:

Let $X$ be a set. A cell structure on $X$ is a pair $(X, \Phi)$, where $\Phi$ is a collection of maps of closed Euclidean cells into $X$ satisfying the following conditions.

  1. If $\phi \in \Phi$ and $\phi$ has domain $E_n$, then $\phi$ is injective on $E_n - \dot{E}_n$.

  2. The images $\{\phi(E_n - \dot{E}_n) : \phi \in \Phi\}$ partition $X$, i.e., they are disjoint and have union $X$.

  3. If $\phi \in \Phi$ has domain $E_n$, then $\phi(\dot{E}_n) \subset \bigcup \{\psi(E_k - \dot{E}_k) : \psi \in \Phi \text{ has domain } E_k \text{ and } k \leq n - 1\}$

Page 7:

We say that two cell structures $(X, \Phi)$ and $(X, \Phi')$ are strictly equivalent if there is a one-to-one correspondence between $\Phi$ and $\Phi'$ such that a characteristic function with domain $E_n$ corresponds to a characteristic function with domain $E_n$, and corresponding functions differ only by a reparametrization of their domain. That is, if $\phi$ and $\phi'$ are corresponding functions of $\Phi$ and $\Phi'$, respectively, then $\phi' = \phi \circ h$, where $h: (E_n, \dot{E}_n) \to (E_n, \dot{E}_n)$ is a homeomorphism of pairs. We leave it to the reader to check that this is an equivalence relation on the collection of cell structures on the set X.

If $(X, \Phi)$ is a cell structure, let $\mathcal{S}_\Phi$ consist of all pairs $(\sigma^n, [\phi])$, where $\sigma^n = \phi(E^n)$ and $[\phi]$ is the strict equivalence class of $\phi \in \Phi$. For convenience, we will denote such a pair by $\sigma^n$ or $\phi(E^n)$ with the class of $\phi$ understood. Note that if $(X, \Phi)$ and $(X, \Phi')$ are strictly equivalent cell structures, then $\mathcal{S}_\Phi = \mathcal{S}_{\Phi'}$.

Page 8:

A cell complex on a set $X$ or a cellular decomposition of a set $X$ is an equivalence class of cell structures $(X, \Phi)$ under the equivalence relation of strict equivalence.

A cell complex on $X$ will be denoted by a pair $(X, \mathcal{S})$, where $\mathcal{S} = S_{\Phi}$ for some representative cell structure $(X, \Phi)$. The set $\mathcal{S}$ is called the set of (closed) cells of $(X, \mathcal{S})$.

Page 41:

A Hausdorff space $X$ is a CW complex with respect to a family of cells $\mathcal{S}$ provided:

  1. the pair $(X,\mathcal{S})$ is a cell complex such that each cell $\sigma \in \mathcal{S}$ has a continuous characteristic function;

  2. the space $X$ has the weak topology with respect to $\mathcal{S}$;

  3. the cell complex $(X,\mathcal{S})$ is closure finite.


I'm not sure where my misunderstanding is, but I suspect it has to do with that the strict equivalence is defined between cell-structures and not between characteristic functions.


Referring to the discussions with Paul Frost below, the confusion arises because the definition in the book is more granular than the usual definition of a CW complex. Therefore, if characteristic functions are not related by a homeomorphism in this definition, then they are not the same CW complex.

In contrast, the usual definition of a CW complex requires only the existence of a characteristic function, which makes it possible to provide non-homeomorphically related characteristic functions for the same CW complex, as provided by the counter-example in the accepted answer.

  • $\begingroup$ Do you mean Lundell and Weingram's book? $\endgroup$ – Paul Frost Mar 31 '19 at 8:52
  • $\begingroup$ @PaulFrost Yes. $\endgroup$ – kaba Mar 31 '19 at 8:54
  • $\begingroup$ @PaulFrost I added more quotes to clarify. $\endgroup$ – kaba Mar 31 '19 at 10:25
  • $\begingroup$ Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion. $\endgroup$ – kaba Mar 31 '19 at 10:41
  • $\begingroup$ So we find the definition of the set $\mathcal{S}$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $\mathcal{S}$. Thus, a cell consists of a subspace $\sigma^n \subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $\Sigma \in \mathcal{S}$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent. $\endgroup$ – Paul Frost Mar 31 '19 at 10:43

This is false. If $k>1$, then there exists a map $f:\overline{B_k}\to\overline{B_k}$ which restricts to a homeomorphism $B_k\to B_k$ and maps $\partial B_k$ to itself but which is not injective on $\partial B_k$. So, taking $X=\overline{B_k}$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_{\overline{B_k}}$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:\overline{B_k}\to\overline{B_k}$ such that $f=g\circ h$.

For an explicit example of such an $f$ for $k=2$, let $S=[0,1]\times[0,1]$ and $T=\{(x,y)\in\mathbb{R}^2:0\leq y\leq x\leq 1\}$ and define $f_0:S\to T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,T\cong \overline{B_2}$ we get $f:\overline{B_2}\to\overline{B_2}$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.


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