Suppose we are given a diagram $X \to Z \gets Y$ of $G$spaces ($G$ a discrete group). Let $( \times^h )$ denote homotopy pullback. Is $(X \times^h_Z Y)_{hG}$ weakly equivalent to $X_{hG} \times^h_{Z_{hG}} Y_{hG}$?
1 Answer
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Yes. A sketch:
Taking products with the free $G$space $EG$ commutes with the pullback diagram (because product is also a limit) and so you can assume they're free, and one of the maps is a fibration.
Having done this, there is a natural long exact sequence of homotopy groups
$\to \pi_* (U) \to \pi_*(U_{hG}) \to \pi_*(BG) \to \dots$
and applying this to the pullback diagram you can deduce (from the 5lemma) that the natural map from the orbit of pullbacks to the pullback of the orbits is a weak equivalence.

1$\begingroup$ Thanks. Seems the key is the fibration U > U_hG > BG which I hadn't considered. You can use it to give a proof that doesn't need any algebra: consider the 3x3 square diagram with rows [, BG, X_hG], [, BG, Z_hG], [*, BG, Y_hG], and all maps pointing "inward"; taking horizontal homotopy pullbacks commutes with taking vertical homotopy pullbacks. $\endgroup$ Commented Oct 17, 2009 at 7:20

$\begingroup$ That's right, and that works much more cleanly. Note G doesn't need to be discrete for these arguments. $\endgroup$ Commented Oct 17, 2009 at 12:13