Transversality in the proof of the Blakers-Massey Theorem. Is it necessary? - MathOverflow

Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?

2011-01-30T03:15:33Z

Assume one is given a commutative square of spaces

$A \quad \to \quad C$

$ \downarrow \qquad \qquad \downarrow$

$B\quad \to \quad X$

which is a pushout and in which each map is a cofibration. If $A \to B$ is $r$-connected and $A\to C$ is $s$-connected, then the Blakers-Massey theorem says that the square is $(r+s-1)$-cartesian (this means that the map from $A$ into the homotopy pullback of the remaining terms is $(r+s-1)$-connected).

The only proofs of the statement that I know of (at this level of generality) make use of transversality. However, if all spaces are simply connected, there are proofs which avoid transversality (for example, when $B$ is a contractible, one can deduce it using the Serre exact sequence).

Question: Is transversality intrinsic to a proof of the theorem in the general case?

Answer by Charles Rezk for Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?

2011-02-02T19:08:47Z

Take a look at the proof (attributed to Puppe) given in tom Dieck's new algebraic topology texbook (section 6.9). (I believe it also appears in tom Dieck, Kamps, Puppe (Lecture Notes in Mathematics 157).) This argument contains no obvious appeal to transversality.

Answer by Jeff Strom for Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?

2011-02-03T04:51:17Z

I suggest you look at Mather's "Hurewicz Theorems for Pairs and Squares" in which Blakers-Massey is derived from his Cube Theorems. It works for all spaces, but the maps involved have to be $m$- and $n$-connected with $m,n \geq 2$.

The basic inputs to the cube theorems are: (1) Hurewicz/Dold local-to-global criteria for fibrations (or weak fibrations) and (2) the pullback of a cofibration by a fibration is a cofibration.

(Actually, (2) is not used in this proof.)

ADDING ON: The proof is extremely cute and easy.

Given a homotopy pushout square (spaces $A,B,C$ and $D$) pull back from the path fibration $\mathcal{P}(D) \to D$ to get another square. The Second Cube Theorem tells you it is a homotopy pushout, and comparing connectivities of maps shows that it suffices to prove the B-M theorem for the new square.

But the new square is a square of simply-connected spaces (since the spaces involved are fibers of the maps involved), and the homotopy pushout is contractible. Now the comparison map to the homotopy pullback may be identified (after suspension) with the inclusion of the wedge into the product, and that connectivity is easy to determine.

Answer by Ronnie Brown for Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?

2012-01-07T22:37:46Z

I refer here to my recent answer to

http://mathoverflow.net/questions/56435/what-is-the-intuition-behind-the-freudenthal-suspension-theorem/85158#85158

where the results do not require simple connectivity for descriptions of the critical group, basically because the proofs do not use homological, i.e. abelian, methods.

May 30: The original Blakers-Massey results were related to triad homotopy groups, since the exact sequences involving these and relative homotopy groups showed the triad groups as the obstruction to excision. So there was a question of calculating these groups, and homology groups were used for this in the simply connected case, see the book by J.F. Adams A student's guide to algebraic topology. However such calculations in the non simply connected case do follow from a Generalised Van Kampen Theorem proved with J.-L. Loday. I have revised and updated a paper of mine ``Triadic Van Kampen theorems and Hurewicz theorems'', Algebraic Topology, Proc. Int. Conf. Evanston March 1988, Edited M.Mahowald and S.Priddy, Cont. Math. 96 (1989) 39-57.

and made it available as

http://pages.bangor.ac.uk/~mas010/pdffiles/VKTEVAN2.pdf