Invertible functions in $ R^m$

Question

The definition of an invertible function that my book (Apostol's Mathematical Analysis) gives is:

A function $f:S \to\mathbf{R}^n$, where $S$ is open in $\mathbf{R}^n$, has a unique inverse if $f$ is $C^1$ on $S$ and $J_f(\mathbf{a})\not = 0$ on $S$ (where $J_f(\mathbf{a})$ is the Jacobian determinant of $f$ at $\mathbf{a}$).

This definition is fine for functions from $\mathbf{R}^n$ to $\mathbf{R}^n$, but what about for $g:\mathbf{R}^n\to\mathbf{R}^m$? In this case the determinant does not exist, so this theorem does not apply! How do I determine whether such a $g$ is invertible?

Also, for a function $f$ between topologies we know that $f$ has an inverse if it is bijective - does that apply here as well? Is it sufficent to show that $g$ is bijective? Is the inverse of $g$ always unique?

Answer 1

In the case of different dimensions, you can't normally hope for an inverse function, if you are thinking of the problem in an analytic/differential set-up. In fact, it is a simple consequence of Sard's theorem that if $g$ is $C^1$, then the image of $g$ will have Lebesgue measure zero if $m > n$, so it will be rather "small". In particular, $g^{-1}$ won't make sense on any open set.

On the other hand, if the derivative of $g$ (the Jacobian matrix, or the tangent map, if you prefer) has full rank, i.e. $n$ (assuming $m > n$), then the image of $g$ will locally be manifold, and $g$ will locally be invertible. By this I mean that for a point $x \in R^n$, you can find a small neighbourhood, such that $g$ restricted to that neigbourhood maps it to bijectively and smoothly to its image.

Answer 2

Consider a nice smooth map from $\mathbb{R}^1$ into $\mathbb{R}^2$ whose image is like a loop de loop. You would have a two-to-one point.

Consider a nice smooth map from $\mathbb{R}^2$ into $\mathbb{R}^1$. Locally you will have smashed a two-dimensional vector space into a one-dimensional (or lower) space, so there are many-to-one points.