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 ... 



Easy proof of topological property of Zoll manifolds
As asked by @alvarezpaiva, I repost my remark as an answer.


After his addendum and answer showing that the fundamental group (if non-trivial) has one generator, given by a prime closed geodesic, you just have to observe that this geodesic is homotopic to itself *with reversed orientation*.


Hence the fundamental group has order at most $2$.


