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 ...