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I recall vaguely once reading that a cell complex—constructed like a CW-complex but without assuming the cells are appended in order of increasing dimension —"is" actually a CW-complex. I cannot remember if this means there is a homotopy equivalent CW-complex or something stronger.

Is there a similar result for $G$-cell complexes vs. $G$-CW-complexes, for $G$ a compact Lie group?

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    $\begingroup$ I'm pretty sure that any proof of your first statement would tell you whether the equivariant case or not... $\endgroup$
    – Najib Idrissi
    Commented Jan 25, 2019 at 20:51

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As Najib says in the comments to the question, the proof of the classical statement can be easily-ish adapted to the equivariant case. Let's see the details

Lemma Let $X$ be a $G$-CW-complex and let $f:G/H\times S^n\to X$ be a $G$-equivariant continuous map. Then $f$ factors up to equivariant homotopy through the $n$-skeleton of $X$

Proof: This is just a special case of the cellular approximation theorem (theorem I.3.4 in May's Equivariant Homotopy and Cohomology Theory).

A $G$-equivariant continuous map from $G/H\times S^n$ is the same thing as a continuous map from $S^n$ to $X^H$ and the same hold for equivariant homotopies. Hence we need to show that, if $X^{(n)}$ is the $n$-skeleton of $X$, the inclusion $(X^{(n)})^H\to X^H$ is $n$-connected. But the cofiber $X^H/(X^{(n)})^H\cong (X/X^{(n)})^H$ is obtained by adding cells of the form $(G/K\times D^m)^H$ for $m>n$, and so it is $n$-connected. $\square$

Armed with the lemma, we can prove that every $G$-complex $Y$ is homotopy equivalent to a $G$-CW-complex. I will prove only the case where $Y$ has finitely many cells, the general case just needs a transfinite induction that I don't want to write down right now.

We will proceed by induction on the number of cells. If $Y$ has only one cell, then it has of course the structure of a $G$-CW-complex, obtained by using a CW structure on the disc.

Suppose now that the thesis is true for all $G$-complexes with $m$ cells. Then, if $Y$ is a $G$-complex with $(m+1)$ cells, we can write it as the pushout $Y=Y'\amalg_{G/H\times S^n} G/H\times D^{n+1}$ for some $H$ and $n$. But, by the previous lemma, the attaching map $G/H\times S^n\to Y'$ factors up to homotopy through the $n$-skeleton of $Y'$. So we can attach it together with the other $(n+1)$-cells without changing the homotopy type of $Y$. Hence $Y$ has the homotopy type of a $G$-CW-complex.

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  • $\begingroup$ Thanks for this. I was really just hoping to a pointer to somewhere in the literature, so this is more than I needed in some sense, but it is also much easier than I had imagined. I did want (I thought this was clear, sorry), though, a compact Lie group; what else does one need to do in that case? $\endgroup$
    – jdc
    Commented Jan 25, 2019 at 23:01
  • $\begingroup$ Not explicit was another question about how strong this result is. It seems clear from your answer that this only produces a homotopy-equivalent $G$-CW-complex, and I am expecting this can't be hoped to preserve the space up to homeomorphism. I encountered a proposition in a preprint seemingly claiming a certain space has a $G$-CW structure but only proving a $G$-cell structure, and was trying to see if the result as stated could be rescued. $\endgroup$
    – jdc
    Commented Jan 25, 2019 at 23:03
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    $\begingroup$ @jdc I never think about compact Lie groups, but basically you need to pay attention at the connectivity of the inclusion of the skeleton of the $n$-fixed points inside the fixed points of the $n$-skeleton (this is for the proof of the lemma). I'll see if I can manage to generalize the proof tomorrow. Constructing a $G$-CW-structure on a space on the nose and not just up to homotopy is a much more difficult problem, and I'm not sure I can say anything intelligent about it. $\endgroup$
    – Denis Nardin
    Commented Jan 25, 2019 at 23:07
  • $\begingroup$ Thanks! Is the difficulty of the second problem alleviated if one instead demands a $G$-invariant CW/cell structure? My intuition is that this would be closer to the difficulty of the non-equivariant problem (which I still think isn't solvable in general). $\endgroup$
    – jdc
    Commented Jan 25, 2019 at 23:10
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    $\begingroup$ @jdc I added a reference for the compact Lie group case. It is a bit subtle, but it can be deduced from the equivariant version of HELP (homotopy extension and lifting property) $\endgroup$
    – Denis Nardin
    Commented Jan 27, 2019 at 11:02

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