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Is it true that a retract of a CW-complex is a CW-complex ?

I failed to find a counterexample! I was wondering if such spaces have any topological characterization ?

My second question is the following:

Suppose that $X$ is a CW complex and $Y$ is a space such that $X\times Y$ is a CW-complex. Is it true that $Y$ is also a CW-complex ?

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The answer to your first question is "no". Here is a counterexample.

Let $C \subset [0,1]$ be the standard middle-thirds Cantor set.

Let $\{I_n\}_{n \in \mathbb{Z}}$ be an enumeration of the components of $[0,1] - C$, a pairwise disjoint set of open intervals, and let $I_n = (a_n - \epsilon_n,a_n + \epsilon_n)$.

In $\mathbb{R}^2$, let $D_n$ be the closed 2-dimensional disc with center $(a_n,0)$ and radius $\epsilon_n$.

Let $X = \bigl([0,1] \times \{0\}\bigr) \cup \bigl(\bigcup_{n \in \mathbb{Z}} D_n\bigr)$.

This space $X$ is a deformation retract of $\mathbb{R}^2$, but it is not a CW complex, because the closure of the set of points of $X$ that separates $X$ is $C \times \{0\}$, but the closure of the set of separating points of a CW complex is a subcomplex of the $1$-skeleton, and $C$ is not a 1-dimensional CW complex.

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  • $\begingroup$ Apologies for the late comment, I think the counter example is valid but I am not sure about the statement that the separating points of a CW-complex form a CW complex. I find that you can make any closed subset of [0,1] into the separating point of a CW complex. Indeed, by gluing disks to [0,1] with image [x-a_n,x+a_n] for a_n increasingly converging to some number, you "cover" an open interval. Doing this repeatedly "covers" any open set, leaving any closed set of your choosing uncovered, making them the separating points (this could be the cantor set). $\endgroup$ Commented Sep 2 at 19:04
  • $\begingroup$ A separating point of a connected space $X$ means a single point $x \in X$ such that $X-\{x\}$ is disconnected. @Alexis $\endgroup$ Commented Sep 2 at 22:39
  • $\begingroup$ I did understand, is my counter example wrong ? $\endgroup$ Commented Sep 3 at 6:43
  • $\begingroup$ More simply, what if you take a circle and any of its subset S, then for each s in S, glue the endpoint of a segment to that point. You end up with a CW complex who's separating points are S. $\endgroup$ Commented Sep 3 at 6:56
  • $\begingroup$ This proof is correct with a slight change of wording, you can look at the 1-skeleton of $X$. It must contain $\delta X$ and therefore a point of the Cantor set can not be on a 1-cell since locally it is not isomorphic to a line in $X^{(1)}$. Therefore it is a point in $X^{(0)}$ $\endgroup$ Commented Nov 21 at 13:18

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