The Groove of Math

or in other words, the Devil is in the Details

Filling the Details #3: Homotopy Fibrations and Long Exact sequence in Cohomology

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Currently I’m reading the paper of Prof. Peter Teichner On the Signature of four-manifolds with universal covering spin, and I was stuck on the following passage:

The homotopy fibration \tilde{M} \to M \to K(\pi, 1) induces an exact sequence in cohomology

0 \to H^2(\pi ; Z/2) \to H^2(M;Z/2) \to H^2( \tilde{M} ; Z/2)

I managed to find a solution to it now, and I thought it could be a good idea to write it here:

So let us start with a definition:

Def 1. X \to Y \to Z is an homotopy fibration sequence if a homotopy is given from the composed map to a constant map $latex z$ and the resulting map from $latex X$ to the homotopy fiber of $latex Y \to Z $ over z is a weak homotopy equivalence.

So we start by proving that the sequence we have is really an homotopy fibration.

Claim 1. The sequence is an homotopy fibration sequence.

Proof: Consider the following square:

Screen Shot 2016-08-04 at 15.41.25

where we used the fact that the universal cover of the Eilenberg-Maclane space K(\pi,1) is a model for E\pi. The nullhomotopy for u \circ p is provided by the contractibility of E\pi. Now consider this pullback diagram:

Screen Shot 2016-08-04 at 15.49.10

where the bended arrow \tilde{M} \to \ast is provided by the null-homotopy chosen before. Notice that that fiber of u is really the homotopy fiber of u since we can assume that u is a fibration. Moreover, since monomorphisms are stable under pullback, the map Fib(u) \to M is really the usual inclusion of fibre. A quick application of the l.e.s. of homotopy groups for a Serre fibration, gives us that

Screen Shot 2016-08-04 at 16.04.59

(use the fact that u induces iso on first homotopy group). By the above diagram should be clear that the induced map is a weak equivalence. Therefore we concluded the proof of the fact that the sequence is an homotopy fibration.

Prop 1. Let p: E \to B be a fibration with B path connected and based. Set F = p^{-1}(*). Assume B is r-connected and p is s-connected. Then there’s a exact sequence

0 \to H^0(B) \to H^0(E) \to H^0(F) \to H^1(B) \to ... \to H^{r+s}(F) \to H^{r+s+1}(B)

we will prove this proposition in several steps.

Recall first the following theorem, called the Dual Blakers-Massey Theorem for squares [Munson & Volić, Cubical Homotopy Theory, Thm 4.2.2. page 188]

Theorem 1. Suppose that the following square

Screen Shot 2016-08-04 at 16.21.30

is an homotopy pullback square, and that the maps X \to Z  and Y \to Z are respectively  k_i-connected for i=1,2. Then the canonical map hocolim(X \leftarrow W \to Y ) \to Z is k_1+k_2+1 connected.

Claim 2. In the above setting, the map E \cup Cone F \to B is  (r+s+1)-connected.

Proof. Consider the following square

Screen Shot 2016-08-04 at 16.25.40

we want to apply Theorem 1 to it, therefore we need to show that the square is homotopy pullback (see def. 3.2.4 page 102 in Cubical Homotopy Theory). By definition the homotopy pullback holim (E \to B \leftarrow Cone(F)) is the subspace of E \times B^I \times Cone(F) such that (x,\alpha, y) is s.t. \alpha(0)=p(x) and \alpha(1)= p\circ \pi_0(y), in particular it’s the subspace consisting of points (e,\alpha) where \alpha(1)= \ast. If we denote with B^I_* the subspace of B^I whose paths end in \ast, we have that the homotopy pullback is just E \times_B B^I_*. Now it’s well-known that B^I_* deformation retract to the constant path \ast, therefore, by definition E \times_B B^I_* \simeq E \times_B \ast = F \times \ast \cong F, proving that the square is indeed homotopy pullback. Notice then that the map Cone(F) \to B is r-connected since the cone is contractible and[B is r-connected by assumption.

Therefore by Theorem 1 we have that the map hocolim(E \leftarrow F \to Cone(F) ) \to B is s+r+1 connected. Following definition 3.6.3 page 138 of Cubical Homotopy Theory, it’s immediate to see that hocolim(E \leftarrow F \to Cone(F) )\simeq E \cup Cone(F) which concludes the proof.

Proof of Prop 1:  Consider the cofibration sequence F \to E \to E \cup Cone(F). Consider the Puppe Sequence for [-,K(A,n)] for n large enough. Now use the fact that a r+s+1 equivalence  gives us isomorphism in cohomology up to r+s+1  [tom Dieck, Algebraic Topology. Theorem  9.5.2 page 236] (so H^k(E\cup Cone(F)) \cong H^k(B)) to conclude.

Back to our initial problem: our setting was F=\tilde{M}, E=M and B= K(\pi,1), r=0, s=2 since \pi_2(K(\pi,1))=0. After applying our machinery we have the result. Notice that H^1(\tilde{M},Z)=0  and therefore using UCT for cohomology it’s 0 even in Z/2 coefficient.


Author: RicPed

Young mathematician :)

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