Retracts of CW-complexes

Question

I remember hearing that (all) topological spaces are retracts of CW-complexes. Given a topological space $X$, I am trying to construct a CW-complex $Y$ with continuous morphisms $i:X\to Y$ and $r:Y\to X$, such that $r \circ i=\mathrm{id}_X$, but with no luck so far.

Could you please provide a reference, in case the statement is correct, or a counterexample otherwise.

Lee Mosher · Answer 1 · 2017-02-12T06:13:51.693

3

This statement is false. For a simple construction of counterexamples, in any retract of a CW complex, any path component is a clopen subset --- both closed and open. So any space whose path components are not all clopen subsets is a counterexample, e.g.: the Cantor set; or the set $\{1,2,3,...\} \cup \{0\}$.

edited Feb 12 '17 at 06:13

answered Feb 06 '17 at 14:14

Lee Mosher

120,280

This might be a helpful post. http://math.stackexchange.com/questions/523416/is-there-any-example-of-space-not-having-the-homotopy-type-of-a-cw-complex – Kal S. Feb 06 '17 at 14:19

Eric Wofsey · Accepted Answer · 2017-02-12T05:44:28.627

For another simple way to get counterexamples, note that every CW-complex is Hausdorff. Every subspace of a Hausdorff space is Hausdorff, so if $X$ is any non-Hausdorff space then it cannot embed in a CW-complex (and in particular, it cannot be a retract of a CW-complex).

You can similarly get counterexamples using other nice properties of CW-complexes that are inherited by subspaces (or even just which are inherited by closed subspaces, since a retract of a Hausdorff space is a closed subset). For instance, every CW-complex is perfectly normal (a fairly strong separation axiom), and thus so is any retract of a CW-complex.

Retracts of CW-complexes

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