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I have seen somewhere the following claim (but can't remember where): let $M$ be a connected orientable closed smooth manifold with $b_1(M)=1$, then there exists a connected covering space $p:\tilde{M}\to M$ with covering group $\mathbb{Z}$ such that the induced map in de Rham cohomology $p^*:H^1(M;\mathbb{R})\to H^1(\tilde{M};\mathbb{R})$ is the zero map.

Is this true, and if so what is the proof?

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    $\begingroup$ Welcome to MathOverflow! The assumption $b_1(M) = 1$ means that $H_1(M) = \pi_1(M)^{\text{ab}}$ has rank $1$, so it has a quotient (by a finite subgroup) isomorphic to $\mathbf Z$. The covering corresponding to the quotient $f \colon \pi_1(M) \twoheadrightarrow \mathbf Z$ should do the trick, as the map $\ker(f)^{\text{ab}} \otimes \mathbf R \to \pi_1(M)^{\text{ab}} \otimes \mathbf R$ is zero. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 12 at 16:13
  • $\begingroup$ Is orientability really necessary? $\endgroup$
    – Francesco Polizzi
    Commented Jan 12 at 16:23

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