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Let $I=[0,1]$. I would like to compute the homology of $I$-bundles over the Klein bottle $K^2$. As far as I know there are three $I$-bundles over $K^2$: the trivial bundle $K^2\times I$ (I have no problem computing the homology of this one) and two twisted bundles: $K^2\tilde{\times} I$ and $K^2\hat{\times} I$. The bundle $K^2\tilde{\times} I$ is orientable while $K^2\hat{\times} I$ is non-orientable.

I have no idea how to start to compute the homology groups of such a bundle. I think I'm confused about them being twisted and this makes it confusing to me to even see what technique or theorem should I try to apply to compute the homology.

Sak
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1 Answers1

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Let $E$ be such a bundle and $\pi :E \to K $ be the projection. Let $D_1, \dots, D_n$ be disks covering $K$, such that $E$ is trivial over any $D_i$. On $D_1$, you can deform retracts $\pi^{-1}(D_1) \cong D_1 \times I$ onto $D_1$. By induction, you can deform retract your bundle on $K$, in particular their homology groups coincide.

This works for any bundle with a contractible fiber, e.g a vector bundle.

Nicolas Hemelsoet
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    Your induction argument does not really work, but the idea is correct: The bundle projection $E\to K$ is a retraction homotopic to the identity map. This is actually a general fact about fiber bundles with contractible fibers. – Moishe Kohan Dec 21 '17 at 23:41
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    Dear @MoisheCohen, thanks for your comment ! Could you tell me why it's not working ? I was thinking "retracting in several step", so extend the retraction such that it is the identity map outside a little neighborhood of $D_1$ and then repeat the process. Would this work ? ( My first approach was to take a section by sending $x$ to the middle of the segment $I_x$ but I did realize that "the middle of $I_x$" is not defined if the bundle is just a topological bundle.) – Nicolas Hemelsoet Dec 21 '17 at 23:48
  • After the 1st step you loose the induction assumption, since you no longer have a fiber bundle. (Just try to write a formal proof on a piece of paper and see how you get stuck.) There are several ways to handle this, one is to introduce a fiberwise metric and to observe the existence of a section of the bundle. The other is to observe the existence of a trivial bundle which is a 2-fold cover of your bundle and then argue that the trivial bundle has $Z_2$-invariant deformation retraction. – Moishe Kohan Dec 22 '17 at 00:08
  • Ok I see ! Many thanks for the explanations. – Nicolas Hemelsoet Dec 22 '17 at 00:14
  • I assume standardly this is just an application of the long exact sequence for fiber bundles – Carl Dec 22 '17 at 02:49
  • Thank you very much for your answers! I see the idea and also the problem. Should I try to use obstruction theory then to prove that there is a section? – Sak Dec 22 '17 at 11:58
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    @Sak Yes, this is how it is usually done. – Moishe Kohan Dec 22 '17 at 13:39