Local Diffeomorphism Theoerm

Question

Is this correct for the local diffeomorphism theorem:

A multivariable function $F(x_1, \cdots x_n)$ has a local diffeomorphism at a point $a = (a_1, \cdots a_n)$ if the determinant of the Jacobian matrix of $F$ at $a$ is not $0$. I.e $\det ||\frac{\partial F_i}{\partial x_j} ||_{1 \leq i,j \leq n} (a) \neq 0$.

Would you agree with that being the local diffeomorphism theorem? I have a slightly longer bit in my notes and I wanted to try and make it more concise and smaller.

EDIT: In my notes it says:

Assume $b = F(a)$ and the Jacobian matrix $|| \frac{\partial F_i}{\partial x_j} ||_{1 \leq i,j \leq n} (a)$ is invertible. Then there are open sets $U^+ \subset U$, $a \in U^+$ and $b \in W$, such that $F(U^+) = W$ and $F:U^+ \rightarrow W$ is $1 - 1$ with $F^{-1}$ differentiable. $F$ is a local diffeomorphism at $a$.

Answer 1

I don't know if what I'm goint to say will help since I'm not sure is what you're asking about, however I think you're trying to state what's known as "Inverse Function Theorem" whose precise statement as seen in Spivak's Calculus on Manifolds is:

Suppose that $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable in an open set containing $a$ and that $\det f'(a) \neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f : V \to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y \in W$ satisfies:

$$(f^{-1})'(y)=[f'(f^{-1}(y))]^{-1}$$

Now if you recall that a diffeomorphism is a differentiable bijection with differentiable inverse, this is stating that if $f$ is differentiable at $a$ with nonzero jacobian determinant you can find neighborhoods of $a$ and $f(a)$ such that $f$ mapping these neighborhoods one to another is a diffeomorphism.

If I understood correctly your notes just made this a little longer first defining a local diffeomorphism and then proving that if the jacobian determinant is nonzero then the function admits being replaced locally by a diffeomorphism (which by the inverse function theorem is just the function with it's domain and range properly restricted).

Although I yet feel I didn't get your point, I hope this helps somehow. Good luck!

Answer 2

The only thing I would add is the fact that $F$ needs to be continuously differentiable i.e. $F\in C^1(E;\mathbb{R}^n)$ where $E$ is a open subset of $\mathbb{R}^n$. Otherwise it is okay but in your version the consequence of the theorem is a little vague but I guess it is acceptable if it's only to have a remainder.