I want to understand the following sentence:

Let X a compact (complex) manifold which has a non-zero cohomology class $\alpha \in H^1(X,\mathbb Z)$. Let $\pi: \bar X\to X$ be the corresponding infinite cyclic covering.

What does this mean? It seems that an infinite cyclic covering is a cover with fiber $\mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?

  • $\begingroup$ $H^1(X;G) = \text{Hom}(\pi_1, G)$. $\endgroup$ – user98602 Jan 1 at 14:03
  • $\begingroup$ Your hypothetical Stiefel-Whitney class would live in $ H^1(X,\mathbb Z/2)$, not in $ H^1(X,\mathbb Z)$. $\endgroup$ – Georges Elencwajg Jan 1 at 14:44
  • $\begingroup$ @GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part. $\endgroup$ – klirk Jan 1 at 15:41
  • $\begingroup$ @MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $\alpha: \pi_1(X) \to \mathbb Z = \pi_1(K(\mathbb Z,1))$ and so $\alpha$ is induced by a map from $X\to K(\mathbb Z,1)=S^1$. But this is not the map I look for. $\endgroup$ – klirk Jan 1 at 15:45
  • 1
    $\begingroup$ @klirk It's standard covering space theory that homomorphisms $\pi_1(X) \to \Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3. $\endgroup$ – Balarka Sen Jan 1 at 16:48

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