# Thread: simple question on mapping from R^n -> R^m

1. ## simple question on mapping from R^n -> R^m

I'm studying surface where I came along the following definition:

a coordinate patch $\displaystyle \mathbf{x}$: $\displaystyle D \mapsto \mathbb{R}^3$ is one-to-one regular mapping of an open set $\displaystyle D$ of $\displaystyle \mathbb{R}^2$ into $\displaystyle \mathbb{R}^3$

Now given this definition how would one go about checking if the given map would constitute a patch?

I'm thinking that in order to check if it's 1-1 we simply take the Jacobian and if one of the solutions is not 0 then we have a 1-1 relationship. So for instance if I have:
$\displaystyle \mathbf{x}x,y) = (x^2,y,y^3-y)$ and I define the first set as (u,v) and the second set as (x,y,z) coordinate then the Jacobians would be:

$\displaystyle \frac{\partial(x,y)}{\partial(u.v)}$ = $\displaystyle \det \begin{pmatrix} x_u & x_v\\ y_u & y_v \end{pmatrix}$ $\displaystyle = 2x \neq 0$

$\displaystyle \frac{\partial(x,z)}{\partial(u.v)}=2x*(3y^2-1)$
$\displaystyle \frac{\partial(y,z)}{\partial(u.v)}=0$

so now that at least one of the Jacobians is not 0 then it's 1-1.

is this correct?
In this particular case the last one is 0 does that make any difference?

2. Originally Posted by lllll
I'm studying surface where I came along the following definition:

a coordinate patch $\displaystyle \mathbf{x}$: $\displaystyle D \mapsto \mathbb{R}^3$ is one-to-one regular mapping of an open set $\displaystyle D$ of $\displaystyle \mathbb{R}^2$ into $\displaystyle \mathbb{R}^3$

Now given this definition how would one go about checking if the given map would constitute a patch?

I'm thinking that in order to check if it's 1-1 we simply take the Jacobian and if one of the solutions is not 0 then we have a 1-1 relationship. So for instance if I have:
$\displaystyle \mathbf{x}x,y) = (x^2,y,y^3-y)$ and I define the first set as (u,v) and the second set as (x,y,z) coordinate then the Jacobians would be:

$\displaystyle \frac{\partial(x,y)}{\partial(u.v)}$ = $\displaystyle \det \begin{pmatrix} x_u & x_v\\ y_u & y_v \end{pmatrix}$ $\displaystyle = 2x \neq 0$

$\displaystyle \frac{\partial(x,z)}{\partial(u.v)}=2x*(3y^2-1)$
$\displaystyle \frac{\partial(y,z)}{\partial(u.v)}=0$

so now that at least one of the Jacobians is not 0 then it's 1-1.

is this correct?
In this particular case the last one is 0 does that make any difference?
Are you attempting to appeal to the inverse function theorem?

3. It's not correct, as the example you give shows. The points $\displaystyle (x,y)$ and $\displaystyle (-x,y)$ have the same image.

I think it is true if the open set $\displaystyle U$ is a small enough neighbourhood of a point where the condition you describe holds. There certainly needs to be a restriction on $\displaystyle U$.

4. Originally Posted by Bruno J.
It's not correct, as the example you give shows. The points $\displaystyle (x,y)$ and $\displaystyle (-x,y)$ have the same image.
I didn't really see that. Is there a way to verify injectivity of a given map as a increase/decrease the number of dimensions. I don't think verifying it on a case by case basis is feasible.

5. Originally Posted by lllll
I didn't really see that. Is there a way to verify injectivity of a given map as a increase/decrease the number of dimensions. I don't think verifying it on a case by case basis is feasible.
I don't know of a general method except for the inverse function theorem stated by Drexel, which only applies to differentiable functions $\displaystyle f: \mathbb{R}^n \to \mathbb{R}^n$. There probably exists a generalization of this theorem which would answer your question; I'll think about it.