Aye...title should say in R^2, sorry about that.(adsbygoogle = window.adsbygoogle || []).push({});

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

LetObe an open subset of the planeR^2and suppose that the mappingF:O[tex]\rightarrow[/tex]R[tex]^{2}[/tex]is continuously differentiable. Let (x, y) be a point inOat which the derivative matrixDF(x,y) is invertible.

Then there is a neighborhoodUof the point (x,y) and a neighborhoodVof its imageF(x,y) such thatF:U[tex]\rightarrow[/tex]Vis 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 functionF(r,[tex]\theta[/tex]) = (r*cos[tex]\theta[/tex], r*sin[tex]\theta[/tex]).

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.

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# Inverse Function Theorem in R

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