= Solution
A parametrization is $X(\theta,z)=(\cosh z\cos\theta,\cosh z\sin\theta,z)$. It gives a <homeomorphism> $S^1\times\mathbb R\to S$, with inverse obtained from the height and angular position. Its <first fundamental form> is
$$
ds^2=\cosh^2z\,(d\theta^2+dz^2).
$$
For a surface of revolution with radius $r(z)$, the <Gaussian curvature> is $-r''/[r(1+r'^2)^2]$. Thus \b[$\boxed{K=-\operatorname{sech}^4z<0}$]. The waist $z=0$ is a closed <geodesic>: $\Gamma^z_{\theta\theta}=-\tanh z$ vanishes there, and $\theta$ can be parametrized at constant speed.
For the <uniqueness of a closed geodesic on a negatively curved cylinder>, use <Gauss-Bonnet theorem> with piecewise geodesic boundary:
$$
\int_DK\,dA+\sum_j(\pi-\alpha_j)=2\pi\chi(D).
$$
A geodesic disk with smooth boundary would require curvature integral $2\pi$. A disk bounded by a geodesic monogon would require $\pi+\alpha>0$, and a geodesic bigon would require $\alpha+\beta>0$. All are impossible when $K<0$.
Here is the topological reduction needed to apply this observation without assuming simplicity. Lift a closed geodesic to the universal covering plane of the cylinder. A self-intersection of its lift supplies an innermost monogon by the <Jordan curve theorem>, so the lift is embedded. A contractible closed geodesic would similarly supply a geodesic disk or monogon, so its winding number is nonzero. Its lift is consequently a proper line, invariant under a nonzero power of the deck translation. Two distinct translates of that line cannot intersect: periodicity would repeat an intersection, and innermost arcs between intersections would bound a bigon. The disjoint translates are ordered transversely, with the deck translation preserving that order. A finite cyclic permutation of this ordered family is impossible, so the lift is invariant under the primitive deck translation too. Thus the original geodesic has a simple image and is at most a repeated traversal of it.
Likewise two distinct simple essential closed geodesics cannot intersect: their periodic lifts would produce a bigon. If disjoint, the <Jordan curve theorem> on the cylinder says they bound a compact annulus. Its two boundaries have zero <geodesic curvature> and $\chi=0$, so Gauss-Bonnet gives $\int K\,dA=0$, again impossible. Therefore \b[there is at most one closed geodesic image]. Reparametrizations, reversed orientation and multiple traversals are understood to describe the same geometric geodesic.
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