Htykuut

Let $f: mathbb{R}^2 to mathbb{R}^2$ with $f(x,y) := (e^xcos y, e^xsin y)$

I have to do the following things:

1. Prove that $f$ is locally reversible everywhere.




2. Is $f$ globally reversible? Why (not)?




3. Give a set $A subset mathbb{R}^2$, so that $f(A) = mathbb{R}^2 setminus { ( 0,0 )}$ and $f$ becomes globally reversible on
   $A$



4. Determine the Jacobian matrix of the inverse image $f^{-1}: mathbb{R}^2 setminus { ( 0,0 )} to A$ in a point $a:=
   (a_1,a_2) := (f_1(x,y),f_2(x,y))$.

The theorem for inverse functions and differentiation goes like that:

Let $D in mathbb{R}^n$ be open, $f:D to mathbb{R}^n$, $f in C^1(D)$, $a in D$ and $b := f(a)$ with $det f'(a) neq 0.$ Then there exists a $delta > 0$ and an open set $B in mathbb{R^n}$, so that we have:

i) $det f'(x) neq 0$ for all $x in U_delta(a)$

ii) $f: U_delta(a) to B$ is bijective

iii) $f^{-1}:B to U_delta(a)$ is continuously differentiable with $(f^{-1})'(y) = (f'(f^{-1}(y)))^{-1}$ (whereby $f^{-1}$ is the inverse function and $(f')^{-1}$ is the inverse Jacobian-matrix)

I know that $f(x,y) := (e^xcos y, e^xsin y)$ is continuously differentiable on $mathbb{R}^n$ and

$$f'(x,y) = (e^xcos y -e^xsin y, e^xsin y e^xcos y)$$

On $mathbb{R}$ it has the determinant $e^x$ and furthermore,

$$f'(x,y)^{-1} = (e^{-x} cos y e^{-x} sin y, -e^{-x} sin y e^{-x} cos y)$$

So $f$ is locally reversible, but not globally reversible, because $f$ is periodic, i.e.

$$f(x,y) = f(x,y + 2kpi)$$ with $k in mathbb{Z}$.

I don't know if that is formally O.K. to answer 1. and 2. and I also don't know how to solve 3. and 4. Any help is appreciated.

It looks like you're on the right lines, though I'm not entirely sure if you typeset out the Jacobian out correctly with what you had in mind. It should be

$$ d_{(x,y)}f =
begin{bmatrix}
e^{x}cos y & e^{x}sin y\
-e^{x}sin y & e^{x}cos y
end{bmatrix}
$$
which still has $det d_{(x,y)}f = e^{x} neq 0, $ for all $(x,y) in mathbb{R}^{2}$ like you said. I also agree with your reasoning between why $f$ is not globally invertible since it provides a counterexample, and also this looks like a question which acts as a hint for the next part.

$f$ is not globally invertible because of the periodic nature, so we can just restrict the domain from $mathbb{R} times mathbb{R}$ to $A := mathbb{R} times (0, 2pi )$, which remains surjective once one removes the origin from the image.

The last part is made a lot easier by noticing that the questions asks you to calculate the Jacobian at the point $(a,b) = (f_{1}(x,y), f_{2}(x,y)) in A$, i.e. that $f^{-1}(a,b) = (x,y)$ (as we have proven now that $f$ is a bijection). So $(f^{-1} circ f)(x,y) = (x,y)$, meaning that $d_{(x,y)}(f^{-1}circ f) = text{Id}$. But from the IFT we also know that
$$
d_{(a,b)}(f^{-1}) = (d_{(x,y)}f)^{-1},
$$
recalling that $(a,b)=f(x,y)$. I will leave the explicit calculations to you, but the final answer should be
$$ d_{(a,b)}(f^{-1}) = (d_{(x,y)}f)^{-1} =
begin{bmatrix}
e^{-x}cos y & -e^{-x}sin y\
e^{-x}sin y & e^{-x}cos y
end{bmatrix}
$$