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$\newcommand{\mc}{\mathcal} \newcommand{\set}[1]{\{{#1}\}} \DeclareMathOperator{\im}{im} \newcommand{\disun}{\amalg} \newcommand{\vp}{\varphi}$

Preliminary Material.

Definition. An $n$-cell is a topological space homeomorphic to the open ball $B^n$. A cell is a space which is an $n$-cell for some $n$.

Definition. Let $X$ be a topological space. A cell decomposition of $X$ is a collection $\mc E=\set{e_\alpha}_{\alpha\in J}$ of disjoint subspaces $e_\alpha$ of $X$ such that each $e_\alpha$ is a cell and $X=\bigcup_{\alpha\in J}e_\alpha$.

Definition. Let $X$ be a topological space and $\mc E$ be a cell decomposition of $X$. The $n$-skeleton of $X$ is defined as the union of all the cells in $\mc E$ of dimension $n$ or lower. The $n$-skeleton of $X$ is written as $X^n$.

Definition. Let $X$ be a topological space and $\mc E$ be a cell decomposition of $X$. Let $e$ be an $n$-cell in $\mc E$. We say that a map $\Phi:D^n\to X$ is a characteristic map corresponding to $e$ if i) $\Phi$ maps $B^n$ homeomorphically onto $e$, and ii) $\Phi(S^{n-1})\subseteq X^{n-1}$.

Definition. Let $X$ be a Hausdorff topological space and $\mc E$ be a cell decomposition of $X$. We say that $\mc E$ is a CW-complex structure on $X$ if the following axioms are satisfied:
(CM) Each $e\in \mc E$ admits a characteristic map.
(CF) For each cell $e\in \mc E$, the closure of $e$ intersects only a finite number of cells in $\mc E$.
(WT) A subset $A$ of $X$ is closed in $X$ if and only if $A\cap \bar e$ is closed in $X$ for each $e\in \mc E$.

Theorem 1. Let $X$ be a Hausdorff space and $\mc E$ be a cell decomposition of $X$ satisfying the (CM) axiom. Then for each $n$-cell $e\in \mc E$, we have $\bar e=\Phi_e(D^n)$, where $\Phi_e$ is any characteristic map corresponding to $e$.
Proof. For any continuous map $f:A\to B$ between topological space $A$ and $B$, we have $f(\bar S)\subseteq \overline{f(S)}$ for all subspaces $S$ of $A$. Using this, we get $\Phi_e(D^n)\subseteq \bar e$. On the other hand, since $D^n$ is compact, we have $\Phi_e(D^n)$ is a compact subspace of $X$. Since $X$ is Hausdorff, $\Phi_e(D^n)$ is closed in $X$. But since $e\subseteq \Phi_e(D^n)$, we must have $\bar e\subseteq \Phi_e(D^n)$. So we conclude that $\bar e=\Phi_e(D^n)$.

Theorem 2. (Topology of CW Complexes.) Let $X$ be a Hausdorff topological space and $\mc E$ be a cell decomposition on $X$ satisfying the (CM) and the (WT) axiom. Let $\mc F$ be the set of characteristic maps, one for each cell in $\mc E$. Then the topology on $X$ is final with respect to $\mc F$.
Proof. Write $\mc F=\set{\Phi_\alpha: D_\alpha\to X}_{\alpha\in J}$, where each $D_\alpha$ is a closed unit ball in some Euclidean space and each $\Phi_\alpha:D_\alpha\to X$ is a characteristic map corresponding to some cell in $\mc E$, while insisting that $\mc F$ has exactly one characteristic map for any given cell in $\mc E$.

Let $F$ be a subspace of $X$. We have

$$ \begin{array}{rcl} F \text{ is closed in } X & \stackrel{(1)}{\iff} & F\cap \im(\Phi_\alpha) \text{ is closed in } X \text{ for all } \alpha\in J\\ & \stackrel{(2)}{\iff} & F\cap \im(\Phi_\alpha) \text{ is closed in } \im(\Phi_\alpha) \text{ for all } \alpha\in J\\ &\stackrel{(3)}{\iff} & \Phi_\alpha^{-1}(F\cap \im(\Phi_\alpha))\text{ is closed in } D_\alpha \text{ for each } \alpha\in J\\ &\stackrel{(4)}{\iff} & \Phi_\alpha^{-1}(F)\cap D_\alpha \text{ is closed in } D_\alpha \text{ for each } \alpha\in J\\ &\stackrel{(5)}{\iff} & \Phi_\alpha^{-1}(F) \text{ is closed in } D_\alpha \text{ for all } \alpha\in J \end{array} $$ Since $\mc E$ satisfies the (CM) axiom, (1) is justified by Theorem 1 and the (WT) axiom.
To see (2), we note that if $F\cap \im(\Phi_\alpha)$ is closed in $X$, then $F\cap \im(\Phi_\alpha)$ is closed in $\im(\Phi_\alpha)$ too, for $\im(\Phi_\alpha)$ is a compact and hence closed subspace of $X$.
The other direction is similar. To get $(3)$, note that if $\Phi_\alpha^{-1}(F\cap \im(\Phi_\alpha))$ is closed in $D_\alpha$, then it is in fact a compact subspace of $D_\alpha$ because closed subspaces of compact spaces are compact. Therefore, the image of $\Phi_\alpha^{-1}(F\cap \im(\Phi_\alpha))$ under $\Phi_\alpha$, which is nothing but $F\cap \im(\Phi_\alpha)$, is a compact and therefore closed subspace of $\im(\Phi_\alpha)$. The steps (4) and (5) are obvious.


The Main Confusion.

Let $(X, \mc E)$ be a CW-complex structure on a Hausdorff space $X$. We want to express $X^n$ as a space formed by attaching $n$-discs to $X^{n-1}$.

For each $e\in \mc E_n$, let $\Phi_e:D^n_e\to X^n$ be a characteristic map for $e$. By the co-product property of the disjoint union, we get a map $\Phi:\disun_{e\in \mc E_n}D^n_e\to X^n$. We write $Z=\disun_{e\in \mc E_n}D^n_e$. Again, the maps $\Phi:Z\to X^n$ and the inclusion map $i:X^{n-1}\hookrightarrow X^n$ give a map $f:Z\disun X^{n-1}\to X^n$.

For each $e\in \mc E_n$, write $\vp_e$ to denote $\Phi_e|_{S^{n-1}_e}$, that is, $\vp_e$is the restriction of $\Phi_e$ to the boundary of $D^n_e$. Then we have a map $\vp:\disun_{e\in \mc E_n} \partial D^n_e\to X^{n-1}$ again obtained by the co-product property of disjoint union.

Now the map $f$ factors bijectively through the space $Z\disun_{\vp}X^{n-1}$ giving a bijective map $\alpha:Z\disun_{\vp}X^{n-1}\to X^n$.

We will show that the map $\alpha$ is a homeomorphism. We know that the topology on $X^{n}$ is final with respect to the maps $\Phi:Z\to X^n$ and the inclusion map $i:X^{n-1}\hookrightarrow X^n$. Thus $\alpha^{-1}$ is continuous if and only if both $\alpha^{-1}\circ i$ and $\alpha^{-1}\circ \Phi$ are continuous. But this is evident from the commutative diagram and therefore $\alpha $is a homeomorphism.

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Confusion. The above working shows that no matter what were the choices of the characteristic maps $\Phi_e$ for $n$-cells in $X$, we get that $X^n$ is homeomorphic to $Z\disun_{\vp}X^{n-1}$. This is perplexing because two different characteristic maps for an $n$-cell may attach the boundary of the $n$-disc in different ways to $X^n$. But still we end up getting the same space (up to homeomorphism). I do not "directly see" why this should be so. Can somebody shed some light on this? That said, I am not entirely sure I my working is correct. I have not see this in any book. I just recetnly read the definition of a CW complex and thought a bit about the definition, ending up with this question.

Hope somebody can settle this. Thanks.

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First a minor point: what you have written in the first box is wrong unless you already assume that $X$ is a CW complex (I could not tell whether you were assuming that). In general, if $X$ is only a cell decomposition, characteristic maps need not exist. For a counterexample: $$X = \bigl\{\underbrace{(0,0)}_{\text{a "$0$-cell"}}\bigr\} \cup \biggl( \underbrace{(-1,+1) \times \{1\}}_{\text{a "$1$-cell"}} \biggr) \cup \biggl( \underbrace{(-1,+1) \times (2,4)}_{\text{a "$2$-cell"}} \biggr) $$ As you can see, in the space $X$ there is no characteristic map for the $1$-cell nor for the $2$-cell.

Second point, more to the point: Even for a CW decomposition as simple as $X=[0,1]$ with two $0$ cells at $0$ and $1$ and with one 1-cell $(0,1)$, there are many, many different characteristic maps, and some of them attach the boundaries in different ways. For example, $\Phi(x)=x^n : [0,1] \to X$ is a characteristic map for all $n$, as is $\Phi(x) = 1-x^n$. That's why the axiom (CM) only says that $e$ "admits" a characteristic map, i.e. it asserts the existence of a characeristic map; it does not assert uniqueness.

Perhaps one "direct way" to see what is going on, at least for compact CW complexes with finitely many cells, is that for any compact Hausdorff space $X$, and for every continuous surjective function $f:Z \to X$ such that $Z$ is a compact Hausdorff space, the map $f$ is a quotient map. In other words, the target space $X$ can be constructed in zillions of ways as a quotient.

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  • $\begingroup$ You are right. In the first box I intended to say that $(X, \mc E)$ is a CW complex but missed it somehow. $\endgroup$ – caffeinemachine Sep 25 '15 at 21:26
  • $\begingroup$ So you think that my working is correct? Do you agree with the conclusion that no matter what choices we make for characteristic maps, one for each cell, we get the same space $X^n$. In fact we can choose as many characteristic maps as we want as long as we make sure that all the cells have been accounted for. Thank you for reading the long post! $\endgroup$ – caffeinemachine Sep 25 '15 at 21:34
  • $\begingroup$ Yes, I agree with all of those points. The topology of a CW complex $X$ is given a priori. The definition requires the existence of characteristic maps, one for each cell, satisfying various properties. Your argument expresses $X$ as a quotient, and is valid for any choice of characteristic maps, one for each cell, satisfying the same properties. $\endgroup$ – Lee Mosher Sep 26 '15 at 14:26

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