It is known that the cohomology ring of a Zoll manifold---a riemannian manifold all of whose geodesics are periodic with the same minimal period---must be the same as the cohomology ring of a compact rank one symmetric space (see Besse's book *Manifolds all of whose geodesics are closed* for references).

Is there a simple and elementary proof of the following much weaker property?

*The first Betti number of a Zoll manifold is equal to zero.*

**Addendum.**
The comment by Thomas Richard got me thinking and here is something that **should** lead to a proof that the fundamental group of a Zoll manifold is either trivial or isomorphic to $\mathbb{Z}_2$:

*Any two prime closed geodesics in a Zoll manifold are homotopic.* Indeed, if $v_x$ is a unit vector tangent to a geodesic $\gamma$ and $w_y$ is a unit vectors tangent to a geodesic $\sigma$, then a continuous path on the unit tangent bundle joining these two unit vectors, taken as the initial conditions of prime closed geodesics, will define a homotopy between $\gamma$ and $\sigma$.

Note that there is at least one closed geodesic representing each non-trivial homotopy class of loops, but the geodesic doesn't have to be prime. Still ...

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