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

Suppose given a diagram X -> Z <- Y of G-spaces (G a discrete group). Is (X ×hZ Y)hG weakly equivalent to XhG ×hZhG YhG?

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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

-> pi_* (U) -> pi_*(U_{hG}) -> pi_*(BG) -> ...

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