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


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.

  • $\begingroup$ 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

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