Let $S$ be a connected surface where the curvature of all its geodesic curves is constant, $K=1$. Prove that $S$ is contained in a sphere Let we have $S$, a connected surface. Let's suppose that the curvature of all its geodesic curves is constant, $K=1$. I have to prove that $S \subseteq \mathbb{S}^2$.
$\mathbb{S}^2=\{(x,y,z)\in \mathbb{R}^3 | x^2+y^2+z^2=1\}$ and $\subseteq$ means subset.

I don't even know how to start with the proof... could anybody tell me which steps do I have to follow? or give me some ideas about the proof?
 A: Here is a rough sketch of the proof:
Let $p\in S$ be arbitrary, $v\in T_{p}S$ a unit vector and consider the unique geodesic $\gamma\colon I\subseteq \mathbb{R}\to S$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. If $N\colon U\subseteq S\to \mathbb{R}^{3}$ is a unit normal vector field defined in a neighborhood of $p$, then
$$\gamma''(t)=[\gamma''(t)]^{\perp}=\langle \gamma''(t),N(\gamma(t)) \rangle N(\gamma(t)),$$
so the curvature of $\gamma$ is equal to $|\langle \gamma''(t),N(\gamma(t))\rangle|=1$ by hypothesis. By continuity, this means that $\langle\gamma'',N\rangle$ is constant and equal to either $1$ or $-1$ identically, so in particular $\langle\gamma''(0),N(p)\rangle=\pm1$. In particular, $\mathbb{II}_{p}(v,v)=\pm1$ for all $v\in T_{p}S$ such that $||v||=1$ (why?), so by continuity again we deduce that the shape operator at $p$ is $\mathcal{S}_{p}=\pm \operatorname{Id}$ (why?), hence $S$ is a totally umbilic surface.
The rest follows from the theory of totally umbilic surfaces: these are always contained in a plane or a sphere, and in the second case, the radius of the sphere is determined by the (unique) eigenvalue of the shape operator.
Hope this helps!
