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Say I have a map of $G$-spaces $f : X \to Y$ and I know it is a homotopy-equivalence in the plain sense that there exists a map (maybe not equivariant) $g : Y \to X$ such that the two composites are homotopic to the identity $f\circ g \simeq Id_Y$, $g\circ f \simeq Id_X$.

Is there something of an obstruction theory that tells you when you can promote $f$ to a homotopy-equivalence in the category of $G$-spaces?

Ideally the level of generality I care for is $X$ and $Y$ manifolds and $G$ a compact Lie group. But anything in that ballpark interests me.

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3 Answers 3

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When you say $f$ is a map of $G$-spaces I am guessing you mean a morphism in the category of $G$-spaces, i.e. a continuous map satisfying $f(ax)=af(x)$ for $a\in G$. If so, then there is a good answer. (But "promoting" the map to a strong $G$-equivalence is a funny way to say it, because it suggests a piece of extra structure rather than a property.)

The fixed point spaces of closed subgroups are the key to much of equivariant homotopy theory. In the most important model structure for $G$-spaces, a morphism $X\to Y$ is called a weak equivalence (resp. fibration) iff for every subgroup $H$ the induced map $X^H\to Y^H$ of fixed point spaces is a weak homotopy equivalence (resp. Serre fibration). Cofibrations are generated by attaching orbits of cells. For cofibrant objects (and this includes manifolds with smooth action) weak equivalences are then the same as maps that actually have an inverse up to $G$-homotopy.

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Dear Tom, is this identical to the projective model structure on the diagram category $\mathcal{CG}^G$, where we consider $G$ as a one-object category? –  Harry Gindi Jun 29 '10 at 2:14
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No. But it is equivalent to the projective model structure on the category of diagrams indexed by the opposite of the orbit category of $G$, i.e. the category with one object for each subgroup of $G$ with morphisms the $G$-maps $G/H\to G/K$. –  Tom Goodwillie Jun 29 '10 at 2:54
    
Thanks! Good to know. –  Harry Gindi Jun 29 '10 at 2:56

The simplest interesting case would be when G is Z/p, p prime. In this case the main issue is that by Smith Theory (applied to the mapping cylinder rel domain, say) you will only know that the induced map of fixed point sets is a Z/pZ equivalence, and definitely not necessarily a homotopy equivalence. If the G-map that is a homotopy equivalence induces a homotopy equivalence of fixed point sets, then I believe one can prove the desired result by bare hands, assuming, say the spaces are G-CW complexes, or smooth G-manifolds. For more general groups G one would need to assume or arrange that the induced maps on fixed point sets for all subgroups are themselves homotopy equivalences. The idea is the same, but the details seem more daunting.

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A facetious answer would be: sure, but one of the ingredients is going to be the group of homotopy equivalences between $X^G$ and $Y^G$. Since $G$-spaces are diagrams of fixed spaces, the Dwyer-Kan obstruction theory of diagrams (c 1984) probably describes the space of maps, in principle allowing you to answer this question and decide which maps of the given fixed sets are realized. This doesn't sound remotely practical to me.

If you don't understand the fixed points to begin with, Smith theory or bust. The map on fixed sets is going to be an isomorphism on homology with coefficients (possibley $0$) depending on the group and exotic fixed sets satisfying that condition are possible. If you're willing to assume everything is simply connected and the group is the circle, it's probably automatically an equivalence. On the opposite extreme is (binary?) $A_5$ acting on high dimensional manifolds. It has PL actions on the disk with the fixed set arbitrary simplicial complexes. I forget what happens in the smooth category; I think empty fixed set is possible, but I forget how. In between are $p$-groups, where you'll keep control of the mod $p$ homology but not the general homology.

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Yes, I'm in the situation you describe in your 2nd paragraph. My space is a Frechet manifold with an action of $O(n)$ -- the simplest non-trivial case that I care about the group is $O(2)$. But the fixed point sets there's little technology out there that will help to describe their homotopy-type. –  Ryan Budney Jun 29 '10 at 3:00
    
Smith theory is about finite dimensional spaces, so you are in trouble. If $V$ is an infinite dimensional complex Frechet space, it has a natural $U(1)$-action, and $V-\{0\}\to V$ is an equivariant map that is an equivalence on total spaces, though the fixed sets are empty on the left and a point on the right. Maybe you can use some other finiteness assumptions, like control at infinity or finite codimension of the fixed sets, but this is not an off-the-shelf situation. –  Ben Wieland Jun 29 '10 at 3:37

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