Is every CW-complex is a Cellular space? Is its converse true? If it is true then what is the difference between them?

We include the definition of CW-complex in algebraic topology given by Whitehed in 1949:

Definition. A CW complex is a Hausdorff space $X$ together with a partition of $X$ into open cells (varying dimension) that satisfies two additional properties:

  1. For each $n$-dimensional open cell $C$ in the partition of $X$, there exists a continuous map $f$ from the $n$-dimensional closed ball to $X$ such that
    1. the restriction of $f$ to the interior of the closed ball is a homeomorphism onto the cell $C$, and
    2. the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than $n$.
  2. A subset of $X$ is closed if and only if it meets the closure of each cell in a closed set.

Definition. A cellular space is a topological space $X$, with a sequence of subspaces $$X^0\subset X^1\subset X^2\subset \cdots \subset X,$$ such that $X=\bigcup\limits_{n=0} X^n$, with the following properties:

  • CS(1) $X^0$ is a discrete space.
  • CS(2) for each positive integer $n$, there is an index set $A_n$, and continuous map $\psi_i^n: S^{n-1} \to X^{n-1}$ for each $i\in A_n$ and disjoint copies $D^n_i$ of $D^n$ (one for each $i\in A$) by identifying the points $x$ and $\psi_i^n(x)$ for each $x\in S_i^{n-1}$ and each $i\in A_n$.
  • CS(3) A subset $Y$ of $X$ is closed iff $Y\cap X^n$ is closed in $X^n,$ for each $n\geq 0$.

closed as unclear what you're asking by Eric Wofsey, Paul Frost, user91500, Cesareo, Lord_Farin Jan 2 at 12:48

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    $\begingroup$ Why don't you include the definitions you are using, and explain where you are running into issues answering the question from these? $\endgroup$ – vociferous_rutabaga Apr 21 '15 at 9:19
  • $\begingroup$ @MPO my post is updated with definition of CW-complex and Cellular space(Complex) $\endgroup$ – MTMA Apr 21 '15 at 9:45
  • $\begingroup$ Quote: "and explain where you are running into issues answering the question from these". $\endgroup$ – Did Apr 21 '15 at 10:20
  • $\begingroup$ This is almost identical with math.stackexchange.com/q/2946857. $\endgroup$ – Paul Frost Dec 19 '18 at 8:57
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    $\begingroup$ The statement of CS(2) makes no sense and is surely not the intended statement (but I'm not sure what the intended statement is). In particular, you have not said what these "disjoint copies $D_i^n$" are supposed to have to do with the space $X$. $\endgroup$ – Eric Wofsey Jan 2 at 8:05

the brief answer is no, every cellular complex is not a CW-complex. The inclusion goes the other way around: CW-complex are cellular complex satisfying two additional properties:

  • Closure-finiteness $\bf{(C)}$: The closure of each cell is covered by a finite number of cells
  • Weak-topology $\bf{(W)}$: a subset $F$ is closed in $X$ if and only if the intersection between $F$ and every cell of $X$ is closed.

A cellular complex not satisfying $\bf{(C)}$ is for instance the 2-disk made of a 2-cell attached to an infinity of $0$-cell (one for each point of its boundary).

An example of a cellular complex not satisfying $\bf{(W)}$ is:

$$ \{ 1/n \ \ ; \ \ n \leq 1 \} \cup \{ 0 \} \subset \mathbb{R} $$.

To know more:

They both adress this specific question.

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    $\begingroup$ Neither of your counterexamples is a cellular complex by the definition in the question, since the definition requires the $0$-skeleton to be discrete. $\endgroup$ – Eric Wofsey Jan 2 at 8:02
  • $\begingroup$ Indeed, but the definition I rely on (see sources) doesn't involve countability, and as the question did not provide a source for his definition, I could not make sure it was a correct one... $\endgroup$ – TryingToGetOut Jan 2 at 8:05

Both concepts as defined in the question agree. This is proved in the appendix of Hatcher's book "Algebraic Topology". See Proposition A.2.

See also the references given by Jeanne Lefevre.

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    $\begingroup$ There is a crucial difference between Hatcher's definition and "cellular complexes" as defined here, which is that in Hatcher's definition, each $X^n$ is required to have the quotient topology obtained by attaching the $n$-cells to $X^{n-1}$. $\endgroup$ – Eric Wofsey Jan 2 at 7:58
  • $\begingroup$ Although, to be fair, the definition in the question is incredibly vague to the point of incomprehensibility, on the point of what exactly is assumed about the relationship between $X^n$ and $X^{n-1}$. $\endgroup$ – Eric Wofsey Jan 2 at 8:07
  • $\begingroup$ @EricWofsey You are right, it is not clear what condition CS(2) means. Only a very generous interpretation of "by identifying the points" could be read as "$X^n$ has the quotient topology obtained by attaching the $n$-cells to $X^{n−1}$" . $\endgroup$ – Paul Frost Jan 2 at 11:07

A CW-complex is built inductively, with cells of dimension $n$ only allowed to be attached in the $n$-th step. A cell complex is similar but cells of any dimension may be attached in each step, so there exist cell complexes that are not CW-complexes.

  • $\begingroup$ My comment is fair - care to elaborate on the downvote? I'd like to improve this answer if possible. $\endgroup$ – Autolatry Apr 21 '15 at 11:12
  • $\begingroup$ @ Autolatry Please give an example of a Cell complex which is not CW-complex $\endgroup$ – MTMA Apr 21 '15 at 11:44
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    $\begingroup$ Read the definition given in the question -- the cells complexes are built inductively here. $\endgroup$ – Najib Idrissi Apr 21 '15 at 11:48
  • $\begingroup$ @MTMA It is very easy to give a space the structure of a cell complex that is not also a CW complex structure. As an example, take one $0$-cell attach a $2$-cell to create the $2$-sphere then attach an additional $1$-cell to the interior of that $2$-cell. This gives a cell complex which is very easily checked to not be a CW-complex. $\endgroup$ – Autolatry Apr 21 '15 at 11:49
  • $\begingroup$ @Autolatry You should give us your definition of a cell complex. $\endgroup$ – Paul Frost Jan 3 at 23:22

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