# Is a $G$-cell complex always a $G$-CW complex?

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?

• I'm pretty sure that any proof of your first statement would tell you whether the equivariant case or not... – Najib Idrissi Jan 25 '19 at 20:51

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.

• 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? – jdc Jan 25 '19 at 23:01
• 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. – jdc Jan 25 '19 at 23:03
• @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. – Denis Nardin Jan 25 '19 at 23:07
• 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). – jdc Jan 25 '19 at 23:10
• @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) – Denis Nardin Jan 27 '19 at 11:02