# Infinite cyclic cover corresponding to non-zero cohomology class $\alpha \in H^1(x,\mathbb Z)$

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?

• $H^1(X;G) = \text{Hom}(\pi_1, G)$. – user98602 Jan 1 at 14:03
• Your hypothetical Stiefel-Whitney class would live in $H^1(X,\mathbb Z/2)$, not in $H^1(X,\mathbb Z)$. – Georges Elencwajg Jan 1 at 14:44
• @GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part. – klirk Jan 1 at 15:41
• @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. – klirk Jan 1 at 15:45
• @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. – Balarka Sen Jan 1 at 16:48