# Do homotopy pullbacks commute with homotopy orbits (in spaces)?

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 $X_{hG} \times^h_{Z_{hG}} Y_{hG}$?

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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 5-lemma) that the natural map from the orbit of pullbacks to the pullback of the orbits is a weak equivalence.

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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. – Reid Barton Oct 17 '09 at 7:20
That's right, and that works much more cleanly. Note G doesn't need to be discrete for these arguments. – Tyler Lawson Oct 17 '09 at 12:13