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

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:

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:

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:

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

(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

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

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.