Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.