# G-equivariant Whitehead's Theorem

Suppose $X$ is a CW complex and $Y$ is a subcomplex. Let $G$ be a compact Lie group that acts on $X$ and $Y$. Suppose further that the CW structures on $X$ and $Y$ are $G$-stable. Moreover assume that $\pi_n(X/G)\cong \pi_n(Y/G)$ for all $n\geq 0$ and are induced by the cellular inclusion $Y/G\hookrightarrow X/G$.

Whitehead's Theorem implies that there is a strong deformation retraction (SDR) from $X/G$ to $Y/G$.

In this setting, does there exist a $G$-equivariant SDR from $X$ to $Y$?

If not, what if one further assumes the existence of a SDR from $X$ to $Y$ (not assumed $G$-equivariant). Would that then imply the existence of a $G$-equivariant SDR from $X$ to $Y$?

EDIT: After Tom Goodwillie answered both questions negatively, I have decided to add another assumption; namely, assume that the fixed point set $X^G$ is contained in $Y$ (or perhaps assume that $X^G$ $G$-equivariantly retracts to a subspace of $Y$).

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I am guessing that by $G$-stable you mean that $G$ acts by cellular maps? Which is more or less what they mean by $G$-CW complex. If so, then no. Think of $G$ of order $2$ acting on a circle $X$ by reflection, and $Y$ one of the two fixed points. – Tom Goodwillie Jul 18 '12 at 0:41
The usual statement is that if $X\to Y$ is an equivariant map of $G$-CW complexes and if for every closed subgroup $H$ the induced map of fixed-point spaces $X^H\to Y^H$ is a homotopy equivalence then in fact the map has an inverse up to equivariant homotopy. A reasonable question is, is the analogous statement true with orbits instead of fixed points? I presume not. – Tom Goodwillie Jul 18 '12 at 0:49
Technical care is needed here: $G$-CW complex has a precise meaning, just like CW complexes but cells of the form $G/H \times D^n$'. So when $G$ is compact Lie, $n$-skeleta do not have geometric dimension $n$. Subcomplex must be taken in this equivariant sense, a union of equivariant cells. Then if $X^H \to Y^H$` is a weak homotopy equivalence, $X\to Y$ is the inclusion of a SDR. All bets are off for other guesses as to what a $G$-CW complex means, and for orbits replacing fixed points, and for merely nonequivariant SDR's. Just not in the cards. – Peter May Jul 18 '12 at 2:18
Yes. Here is an example showing that orbits instead of fixed points isn't enough. Take a CW space $Z$ that is acyclic but not contractible. Let $X$ be the suspension of $Z$, and let $G$ of order $2$ act on it by switching the two cones, with $Z$ as fixed point set. Then both $X$ and the orbit space are contractible, but $X$ is not equivariantly contractible because the fixed point set is not contractible. – Tom Goodwillie Jul 18 '12 at 2:53
@Sean Lawton : It is true that smooth $G$-manifolds and $G$-semialgebraic sets admit $G$-CW structures when $G$ is a compact Lie group. See the following paper and the references therein : MR1770606 (2001j:57032) Illman, Sören(FIN-HELS) Existence and uniqueness of equivariant triangulations of smooth proper G-manifolds with some applications to equivariant Whitehead torsion. J. Reine Angew. Math. 524 (2000), 129–183. – Andy Putman Jul 18 '12 at 4:51

## 1 Answer

The usual statement is that if $X\to Y$ is an equivariant map of $G$-CW complexes and if for every closed subgroup $H$ the induced map of fixed-point spaces $X^H\to Y^H$ is a homotopy equivalence then in fact the map has an inverse up to equivariant homotopy.

This is part of the following picture: The category of $G$-spaces has a model structure in which a $G$-map $X\to Y$ is called a weak equivalence (resp. fibration) if and only if for every subgroup $H$ the resulting map $X^H\to Y^G$ is a weak equivalence (resp. fibration). As generating cofibrations you can use the inclusions $G/H\times S^{n-1}\subset G/H\times D^n$ for all $H$ and $n$. In particular $G$-CW complexes are cofibrant. And strong homotopy equivalence in the model category sense becomes equivariant homotopy equivalence in the obvious sense. Thus the statement above becomes an instance of the general principle that for cofibrant objects every weak equivalence is a strong equivalence.

By the way, a reasonable question is, is the analogous statement true with orbits instead of fixed points? That's not true, though. Take a CW space $Z$ that is acyclic but not contractible. Let $X$ be the suspension of $Z$, and let $G$ of order $2$ act on it by switching the two cones, with $Z$ as fixed point set. Then both $X$ and the orbit space are contractible, but $X$ is not equivariantly contractible because the fixed point set is not contractible.

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