= Solution
Constant <Gaussian curvature> one gives $f''+f=0$, 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 $g'\to0$ and $|f'|\to1$ there. But $|f'(\pm\pi/2)|=C$, so smoothness forces \b[$C=1$]. The arc-length identity also illustrates the alternatives: $C>1$ prevents reaching either zero with real $g'$, while $0<C<1$ gives a nonhorizontal limiting meridian and a conical singularity.
Now $g'^2=1-f'^2=\cos^2u$. On the connected interior, $g'$ has one fixed sign because $\cos u>0$. Therefore $g=g_0\pm\sin u$, and
$$
\boxed{f^2+(g-g_0)^2=\cos^2u+\sin^2u=1.}
$$
The revolved surface is the unit sphere centered at $(0,0,g_0)$, with only its two poles $(0,0,g_0\pm1)$ omitted. Those points are antipodal. This proves the claim using <smooth endpoint criterion for a surface of revolution>, without integrating a general square root.
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