Past exam of the geometry course of the University of Oxford / 2010 / A / 2 / iv / Solution

Choose $f(u)=\sin u$ and $h(u)=\cos u$ for $0\le u\le \pi$. Adding the smooth poles at the endpoints gives the unit sphere. Its meridian is parametrized by arc length, and $-f^{\prime \prime }/f=1$ away from the poles, hence everywhere by continuity. Thus the unit sphere has constant Gaussian curvature one.

There is no compact smooth surface of revolution without boundary of constant Gaussian curvature minus one in this setting. On a compact such surface, the distance $f$ from the rotation axis attains a positive maximum. The maximum lies on a regular revolution patch, where the meridian arc length coordinate gives $f^{\prime \prime }\le 0$. Hence $K=-f^{\prime \prime }/f\ge 0$ there, contradicting $K=-1$. Equivalently, constant negative curvature would require $f^{\prime \prime }=f>0$, excluding any positive radial maximum. Compact patches with boundary would not be excluded by this argument, but they are not closed surfaces.