 Solution

Constant Gaussian curvature one gives $f^{\prime \prime }+f=0$, so, after a shift of the arc-length parameter, $f=C\cos u$ with $C>0$ on an interval contained in $(-\pi /2,\pi /2)$. To close the surface by adding only two points, the limiting parallels at both ends must collapse to points on the axis. If an end had limiting radius $f>0$, it would contribute a whole missing circle, which two added points cannot supply. Thus the parameter interval is the full interval between consecutive zeros, and its two ends are $u=\pm \pi /2$.

At a smooth axis point the tangent plane of a surface of revolution is invariant under all rotations about the axis, so it must be horizontal. The unit meridian tangent must consequently have $g^{\prime }\to 0$ and $|f^{\prime }|\to 1$ there. But $|f^{\prime }(\pm \pi /2)|=C$, so smoothness forces $C=1$. The arc-length identity also illustrates the alternatives: $C>1$ prevents reaching either zero with real $g^{\prime }$, while $0<C<1$ gives a nonhorizontal limiting meridian and a conical singularity.

Now $g^{\prime 2}=1-f^{\prime 2}={\cos }^{2}u$. On the connected interior, $g^{\prime }$ has one fixed sign because $\cos u>0$. Therefore $g=g_0\pm \sin u$, andThe revolved surface is the unit sphere centered at $(0,0,g_0)$, with only its two poles $(0,0,g_0\pm 1)$ omitted. Those points are antipodal. This proves the claim using smooth endpoint criterion for a surface of revolution, without integrating a general square root.