= Solution
A <geodesic> is a curve with zero covariant acceleration in an affine parameter. For a unit-speed curve on an oriented surface with unit normal $N$, its <geodesic curvature> is $k_g=\langle\gamma'',N\times\gamma'\rangle$: it measures the tangent-plane component of acceleration perpendicular to the tangent. A unit-speed curve is <geodesic> exactly when this component vanishes.
The <exponential map> $\exp_p(v)$ follows for unit time the <geodesic> starting at $p$ with initial velocity $v$, whenever it is defined. Since $\exp_p(tv)=\gamma_v(t)$, its derivative at zero is \b[$\boxed{d(\exp_p)_0=\operatorname{id}_{T_pS}}$]. Choose an oriented <orthonormal basis> of $T_pS$ and let $v(\theta)$ be its unit circle. <Geodesic> polar coordinates are $F(r,\theta)=\exp_p(rv(\theta))$ in a normal neighborhood, with the usual angular singularity at $r=0$.
Put $T=F_r$ and $J=F_\theta$. Radial curves are unit-speed <geodesics>, so $E=\langle T,T\rangle=1$ and $\nabla_rT=0$. The torsion-free connection gives $\nabla_rJ=\nabla_\theta T$, and therefore
$$
\partial_r\langle T,J\rangle=\langle T,\nabla_\theta T\rangle=\tfrac12\partial_\theta\langle T,T\rangle=0.
$$
Since $J(0,\theta)=0$, this proves $F=\langle T,J\rangle=0$, the radial form of the <Gauss lemma>. Moreover $J(r,\theta)=rv'(\theta)+O(r^2)$, with $|v'|=1$, by the derivative of the <exponential map>. Hence $G=|J|^2=r^2+O(r^3)$, giving
$$
\boxed{E=1,\quad F=0,\quad G(0,\theta)=0,\quad(\sqrt G)_r(0,\theta)=1.}
$$
The last derivative is the one-sided radial limit.
The global <Gauss-Bonnet theorem> for a compact oriented surface with piecewise smooth boundary is
$$
\boxed{\int_S K\,dA+\int_{\partial S}k_g\,ds+\sum_j\varepsilon_j=2\pi\chi(S).}
$$
Here $K$ is <Gaussian curvature>, the boundary is positively oriented with the surface on its left, $\varepsilon_j$ are signed exterior turning angles at corners, and $\chi$ is the Euler characteristic. Smooth <geodesic> boundary components contribute neither curvature nor corner terms.
A contractible simple closed <geodesic> on the cylinder would bound a disc, giving $\int K\,dA=2\pi$, impossible when $K<0$. Thus any such <geodesic> is an essential loop. If two distinct essential simple loops on an annulus intersect, the elementary bigon criterion gives two subarcs bounding an innermost disc: lifting the homotopic loops to the covering strip and choosing an innermost region supplies such a bigon. Distinct <geodesics> cannot have tangential intersections by uniqueness of the <geodesic> initial-value problem, so its two interior angles $\alpha,\beta$ are positive. Gauss-Bonnet on this <geodesic> bigon gives $\int K\,dA=\alpha+\beta>0$, another contradiction. Therefore two distinct simple closed <geodesics> would be disjoint. They would then bound a compact annulus with <geodesic> boundary and <Euler characteristic> zero, forcing $\int K\,dA=0$, again impossible. \b[There is at most one simple closed <geodesic>.] Completeness of the noncompact surface is not needed for these compact-region arguments.
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