Consider the unit two-sphere $S^2$, embedded in $\mathcal{R}^3$. One way to parametrize this surface is to use spherical coordinates, i.e. with a map $$f(u,v) = (\sin u \cos v, \sin u \sin v, \cos u).$$ Around any point except for the poles $(0,0,\pm 1)$ this map gives a local parametrization. In particular, it is invertible.
However, the formula doesn't work at the poles, where $\sin u = 0$, as $v$ is undetermined. A way to get a local parametrization at the poles is of course to reorient the polar axis, so that it coincides e.g. with the $x$-axis.
My question is if there is a formula that works at every point, i.e. it is invertible at every point, when restricted to a suitable neighbourhood.
To better explain the question, consider the torus, $T^2$, also embedded in $\mathcal{R}^3$. It can be parametrized by the map $$g(u,v) = ((R+r\cos u)\cos v, (R+r\cos u)\sin v, r\sin u).$$ This map can be inverted at every point of the torus, if we restrict the domain to (say) $D= [0,2\pi)\times[0,2\pi)$. Of course the inverse will not be continuous, so we don't get a homeomorphism. That is not what I am asking about. I am sking if we can similarly find a single formula that "works", i.e. can be inverted, at every point of the two-sphere.
Had some enforced free time today, so I gave some more thought to this. I know I've seen a much nicer argument, but it has been too long. This is what I came up with:
Suppose we have such a smooth chart $f: \Bbb R^2 \supset U \to S^2$. Find a simply-connected open set $D \subset U$, maximal with respect to $f(D)$ being 1-1. Any such set $D$ is diffeomorphic to the open unit disk, so it is sufficient to assume that $D$ actually is the unit disk. We can extend $f$ to $\overline D$, and necessarily $f(\overline D) = S^2$. $f$ is injective on the interior, but on the unit circle every point must have at least one other point with the same image under $f$ (otherwise $D$ would not be maximal).
Let $p \in f(S^1)$. If $f^{-1}(p)$ is dense in any arc of $S^1$, then by continuity $f$ must be constant on the arc, which would violate that $f$ is a chart. So $f^{-1}(p)$ is nowhere dense. Choose a maximal open arc $\overset{\frown}{AB}$ of $S^1$ with $p \notin f(\overset{\frown}{AB})$ (so $f(A) = f(B) = p$). The image of the chord $\overline {AB}$ under $f$ is a simple closed loop in $S^2$ passing through $p$. As such, it divides the sphere into two separate regions. The portion of the disk on the arc-side of $\overline {AB}$ is mapped to the interior of that loop, while the other side is mapped to the exterior. By continuity, $\overset{\frown}{AB}$ must map to the interior of the loop.
Let $C \in \overset{\frown}{AB}$, and $C' \ne C$ be any other point of $S^1$ with $f(C') = f(C)$. If $C' \notin \overset{\frown}{AB}$, then a line segment connecting a point between $\overline{AB}$ and $\overset{\frown}{AB}$ with $C$, combined with a line segment connecting $C'$ with a point in the open disk on the other side of $\overline{AB}$, is mapped by $f$ to a continuous curve connecting a point interior to the loop to a point exterior to the loop without every crossing the loop, which cannot be. Therefore for any $C \in \overset{\frown}{AB}$, it must be that $f^{-1}(f(C)) \subset \overset{\frown}{AB}$.
Since every point of $f(S^1)$ has multiple pre-images, we can take $A'$ to be the midpoint of $\overset{\frown}{AB}$ and $B\,'$ any other point of $\overset{\frown}{AB}$ with $f(B\,') = f(A')$, and follow the same argument again. Continuing this way we get a decreasing sequence of arcs that must converge to some point $C$ which has to be a singularity of $f$, since moving from $C$ along the unit circle in either direction will under $f$ move in the exactly the same direction.
Therefore no chart $f: \Bbb R^2 \supset U \to S^2$ can be surjective.
I've skimmed over a lot of details, and it is certainly possible that one of them may prove more of a problem than I realize, but I think this works.