# Inverse Function Theorem in R

## Main Question or Discussion Point

Aye...title should say in R^2, sorry about that.
I'm hitting somewhat of a wall in my understanding of a theorem (or rather a special case of a theorem). The theorem as stated in the book is as follows.

The Inverse FUnction Theorem in the Plane
Let O be an open subset of the plane R^2 and suppose that the mapping F: O$$\rightarrow$$R$$^{2}$$ is continuously differentiable. Let (x, y) be a point in O at which the derivative matrix DF(x,y) is invertible.

Then there is a neighborhood U of the point (x,y) and a neighborhood V of its image F(x,y) such that F: U$$\rightarrow$$V is one to one and onto.

The theorem goes on to talk about the inverse functions, but that's not where I'm getting stuck. My problem is this. Consider the function F(r,$$\theta$$) = (r*cos$$\theta$$, r*sin$$\theta$$).

The determinant of the derivative matrix of this function is just r, so the theorem seems that it should only break down at r=0. However consider the point (r, 2*pi)It does not seem to me that it's 1-1 in a neighborhood around this point, which seems to contradict the theorem.

Where am I misunderstanding things. Is the mapping actually 1-1 here? It seems not to me as it seems to me if you just let theta go to infinity it circles the same ring in the image. I understand that if you restrict it to a ring then it invalidates the theorem as the neighborhood then isn't open due to the ring being thin. But I don't see any way to get around the looping problem.

mathwonk
We can only state that it's 1-1 if V is some, perhaps restricted, neighborhood of the image of U. That much I see. However we also must restrict U to the neighborhood in which F:U->V is invertible? Otherwise we have that F$$^{-1}$$(1,0) can be equal to (1, 2*pi*n) for any n. Do we have to choose a range for the inverse from the outset in order to preserve 1-1? Is that correct?