local inverse functions consider $f(x,y)=(x\sin y,x\cos y),\; (x,y)\in (0,\infty)\times (0,3\pi)=U$. f is locally invertible at every point in U, because $\det(Df(x,y))\not= 0$ for all $(x,y)\in U$. I want to know : What are the locally inverse functions?
Set $u=x\sin y, v=x\cos y$. It is $u^2+v^2=x^2$ and therefore $x=\sqrt{u^2+v^2}$, because $x\in (0,\infty)$. $\frac{u}{v}=\tan y$ therefore $\arctan(\frac{u}{v})=y$. If i consider the local inverse functions, I can't take the whole interval $(0,3\pi)$, but I don't know what are the inverse functions and the correct intervals?
 A: A local inverse is local exactly because it doesn't necessarily extend to the whole image and domain of the original function. In this case, there is not a single inverse that will take $(u,v)$ in the image and give you every $(x, y)$ in the domain. This is because your function $f$ is not one-to-one. $f(x, y) = f(x, y + 2\pi)$ for $y \in (0, \pi)$. So the value of $f$ corresponds to two different values of $(x, y)$. An inverse can only choose one of them.
Even with that limitation, there are a couple of problems with what you have. $x$ is fine, but obviously, your formula for $y$ fails when $v = 0$. The other issue is that the arctangent will give you values between $-\frac \pi 2$ and $\frac \pi 2$. Even when adjusted to lie within $(0,3\pi)$, this is only a range of width $\pi$. Yet you know that it requires $y$ to go through a range of width $2\pi$ to cover all directions. So your inverse operation on $y$ is missing half of its range. It is obvious where the problem lies: $\frac{-u}{-v} = \frac{u}{v}$, but $(-u, -v)$ is $\pi$ radians away from $(u, v)$.
So you have to do something a little more complex than just $y = \arctan \frac u v$. In particular, break the points $(u, v)$ down into sets where you can specify a single function for $y$ on each set. On one of those sets, you will actually have two choices for the function $y$. One of those choices will return $y \in (0, \pi)$. The other will return $y \in (2\pi, 3\pi)$. Which one you choose determines which local inverse you get.
