= Solution
Because $f'(a)$ and $f'(b)$ are nonzero, $r=f(v)$ is a valid one-sided local coordinate at each endpoint. Define
$$
h_a(r)=g\bigl(f^{-1}(r)\bigr)
$$
using the inverse near $a$, and define $h_b$ analogously near $b$. The necessary and sufficient conditions are:
* $g(a)\ne g(b)$, so the two collapsed boundary circles give two distinct poles;
* near each endpoint there is a smooth function $H_*$ such that
$$
h_*(r)=h_*(0)+H_*(r^2).
$$
Indeed, the second condition makes the surface near the corresponding pole the graph
$$
z=h_*(0)+H_*(x^2+y^2),
$$
which is a smooth regular graph over the horizontal tangent plane. Together with part a(i), these two pole charts cover the closure by <regular surface> charts. The surface is compact because it is the continuous image of the compact cylinder $[0,2\pi]\times[a,b]$ after each boundary circle is collapsed to its pole.
Conversely, rotational invariance forces the tangent plane at a pole to be horizontal. The surface is therefore locally the graph of a smooth radial function of $(x,y)$. A smooth rotation-invariant function near the origin is a smooth function of $x^2+y^2$, which gives the displayed condition. The poles must be distinct, since otherwise a punctured neighbourhood of their common image would have two components and could not be a surface chart. This is the <smooth endpoint criterion for a surface of revolution>. In particular it implies the familiar first-order conditions $g'(a)=g'(b)=0$, but those conditions alone do not ensure smoothness of every order.
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