Subject: for the list Date: Wed, 28 May 2003 11:02:20 -0400 (Eastern Daylight Time) From: "Nicholas J. Kuhn" To: dmd1@lehigh.edu CC: njk4x@virginia.edu I think the proof of the wedge decomposition question can be finished off rather formally, using the following lemma, which allows the S^1's to be separated from the higher spheres. Lemma Suppose Y = W wedge Y', where W is a wedge of circles and Y' is simply connected. If X is a retract of Y then X = V wedge X', where V is a wedge of circles and X' is simply connected. sketch proof: (i) pi_1(X) is a free group, so there exists j:V --> X giving an iso on pi_1. (ii) Let X' be the cofiber of j. By Van Kampen, it is simply connected. (iii) The composite X -> Y -> Y' -> Y -> X is zero on pi_1, so can be written as X --> X' -f-> X, for some map f. Since the above composite is id on pi_i with i>1, f is epic on pi_i, for i>1. (iv) By construction X' -f-> X --> X' is the identity on homology, so is a homotopy equivalence. Thus f is monic on pi_*. (v) It follows that j wedge f: V wedge X' --> X is an iso on pi_*, and thus a homotopy equivalence. Nick Kuhn Subject: wedges of spheres Date: Tue, 27 May 2003 17:26:26 -0400 From: Tom Goodwillie To: Don Davis >> If X and Y are topological spaces, not homotopy equivalent to >> singletons, >> such that X wedge Y >> has the homotopy type of a wedge of spheres, does it follow that X >> and Y homotopy equivalent to wedges of spheres ?? > I guess it's not hard to see in the simply connected case: > Prop: If X is a retract, in the homotopy category, of a wedge of > spheres of dimension [more than] 1, then X is homotopy equivalent to a wedge of > spheres.