$f:\mathbb{R}^3 \to \mathbb{R}^3$ , $$f(x_1,x_2,x_3)=(e^{x_2cosx_1},e^{x_2sinx_1},2x_1-cosx_3)$$ and $E=(x_1,x_2,x_3)$ such that there exist an open subset U around $$(x_1,x_2,x_3)$$ such that f restricted to U is an open map.
Then E is?
My approach:
Function is continuously differentiable so I have used inverse function theorem and find the point in which Jacobian is non zero. The point were $R^3/(x_1,x_2,nπ)$.
So at the point there will exist a neighborhood such that, f will be an open map at that neighborhood.
Is there any other point namely on $(x_1,x_2,nπ)$. Is there any result guarantee such that Jacobian is zero at a point p and it will/will not have a neighborhood for which it's an open map?
Suggest me some book for learning this topic, Inverse function theorem and thanks in advance.
Best Answer
I can't think of a general result that would be helpful.
But look at $g(x_1,x_2)=(e^{x_2\cos(x_1)},e^{x_2\sin(x_1)})$, under what conditions is the Jacobian of $g$ invertible?
$\\$
Now, supposing we have a point $(a_1,a_2)$ such that $g$ is invertible,
if we set $(y_1,y_2)$ near $g(a_1,a_2)$, and ask for $(x_1,x_2)$ near $(a_1,a_2)$ such that $g(x_1,x_2)=(y_1,y_2)$, then the choice of $(x_1,x_2)$ is unique, because of the inverse mapping theorem.
Apply that to $f(a_1,a_2,k\pi)$.
If we try to find $(x_1,x_2,x_3)$ near $(a_1,a_2,k\pi)$, such that $f$ maps $(x_1,x_2,x_3)$ to any specific $(y_1,y_2,y_3)$ near $f(a_1,a_2,k\pi)$, then the choice of $x_1$ and $x_2$ would have to be unique.
The trick is to use $y_1=e^{a_2\cos(a_1)}$ and $y_2=e^{a_2\sin(a_1)}$. Then $x_1$ is forced to be $a_1$ and $x_2$ is forced to be $a_2$.
We can think about values of $y_3$ near $a_1-\cos(k\pi)=a_1-(-1)^k$, and asking if $x_1-\cos(x_3)=a_1-\cos(x_3)$ can be equal to $y_3$.
But $\cos(x_3)\in[-1,1]$, no matter the value of $x_3$. So for $k$ even, $a_1-\cos(x_3)$ cannot be less than $a_1-(-1)^k$, and for $k$ odd, $a_1-\cos(x_3)$ cannot be greater than $a_1-(-1)^k$.
Then for any $k$, there are examples of $y_3$ that are arbitrarily close to $a_1-\cos(k\pi)$ but don't allow $a_1-\cos(x_3)=y_3$.
That means that $f$ can't be an open map on a neighbourhood of $(a_1,a_2,k\pi)$.
$\\$
Back to the original question. The Jacobian of $g$ is invertible at $(a_1,a_2)$ if and only if $a_2\neq0$.
How about when $a_2=0$? In this case, $g$ itself is not an open map. [If $g$ is not an open map around $(a_1,a_2)$, then $f$ is not an open map around $(a_1,a_2,k\pi)$.]
Why is $g$ not an open map at $(a,0)$? Because $g(a,0)=(e^0,e^0)=(1,1)$. For each $a\in\mathbb{R}$, either $\cos(a)$ or $\sin(a)$ is non-zero.
If $\cos(a)$ is non-zero, then once again, just like in the proof above: you set a point $(1,y)$ near $(1,1)$ and you want $e^{x\cos(a)}$ to be $1$, which means $x$ would have to be $0$, and you want $e^{x\sin(a)}=e^0=1$ to be a specific $y$, but $y$ near $1$ doesn't have to be equal to $1$. So $g$ is not an open map at $(a,0)$.
Likewise if $\sin(a)$ is non-zero.
$\\$
I don't want to answer too many such questions, so I'll make a general remark. Maybe one needs to practise thinking about specific defined objects. "So-and-so object or group of objects is defined so-and-so, what properties does it have?" Sometimes, it's not immediately obvious.
This is not about theorems, because there may not be any theorem within your aukaat which would be useful in such a situation. This is about investigating an object to which you have to do more than just apply a ready-made theorem.
One way of getting used to this is to find textbooks that have answer keys. Just find different exercises of this type, read through some answers to them, and see if your technique improves.