# Homotopy fibration and exact sequence in cohomology

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

## 1 Answer

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.

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