3
$\begingroup$

I'm reading the paper "On the signature of four-manifolds with universal covering spin" By Peter Teichner, and at page 746 I got stuck in this passage:

The homotopy fibration $\tilde{M} \to M \to K(\pi, 1)$ induces an exact sequence in cohomology $$ 0 \to H^2(\pi; \mathbb{Z}_2) \to H^2(M;\mathbb{Z}_2) \to H^2(\tilde{M};\mathbb{Z}_2)$$

Where $M$ is a $4$-manifolds whose universal cover is spin, and whose fundamental group is $\pi$. The map $M \to K(\pi, 1)$ is the classifying map of the universal cover.

By homotopy fibration I mean (and I hope that the author used the same definition) the following:

DEF: $X→Y→Z$ is a homotopy fibration sequence if the composed map is a constant and the resulting map from $X$ to the homotopy fiber of $Y→Z$ is a weak homotopy equivalence.

I think something on the line of the Serre Exact sequence should work, but the indices are bothering me since the base space is $0$-connected, the fibre is $1$-connected and therefore the Serre Spectral Sequence should stop at $H^2(\pi; \mathbb{Z}_2)$.

The only possibility would be that the fact we are working in $\mathbb{Z}_2$ coefficient, permits us to do something more, but I do't know how.

Any help is appreciate

$\endgroup$
4
$\begingroup$

Have a look at John Klein's answer to this mathoverflow question. In your case, the base is $0$-connected and the fibration is $2$-connected, so you get a long exact sequence as desired (even integrally) and you may also add a term $H^3(\pi,\mathbb{Z}/2\mathbb{Z})$ at the end.

$\endgroup$
  • $\begingroup$ Oh, hey, I gave up on understanding that sentence in Lawson-Michelsohn a long time ago. Nice answer! $\endgroup$ – user98602 Aug 4 '16 at 4:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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