Locality of Inverse Function Theorem

Question

For the Inverse function theorem, the theorem proved the existence of a inverse relation on a local scale, that is if $Df(x)$ is invertible, $f$ is $C^1$ function and $f$: open set E $\subset R^2$ -> $R^2$ , then there exists an open ball U $\subset E$, such that $f(U) $ is also open, and it is a 1-1 relation.

I am wondering that if I want to prove that a function is 1-1 in its whole domain, can I use Inverse function theorem to prove.

Answer 1

Global invertibility theorems are generally hard. One example of such a theorem is stated in Global invertibility of a map $\mathbb{R}^n\to \mathbb{R}^n$ from everywhere local invertibility, where pointers to literature are given. Proofs tend to involve substantially more topology than the local consideration: e.g., topological degree of a map, path lifting...

Here's a simple result of purely analytic nature: if $f$ is differentiable in a convex domain $U\subset \mathbb R^n$ and $\|Df - I\|<1$ pointwise, then $f$ is invertible. (Here $I$ is the identity matrix, and the norm is the operator norm.) Indeed, let $g(x)=f(x)-x$ and observe that $\|Dg\|<1$, hence $|g(a)- g(b)|<|a-b|$ whenever $a\ne b$. It follows that $f(a)\ne f(b)$.

Here is more interesting result. Suppose that $U$ and $V$ are Jordan domains in the plane, $f: \overline{U}\to \overline{V}$ is continuous, the restriction of $f$ to $U$ is differentiable with nonvanishing Jacobian determinant, and the restriction of $f$ to $\partial U$ is a homeomorphism onto $\partial V$. Then $f$ is a diffeomorphism of $U$ onto $V$.

The proof goes as following:

1. the degree of $f$ with respect to any point $w\in V$ is equal to $1$ (from consideration of boundary values).
2. the aforementioned degree is the sum of $\operatorname{sign} \det Df(z)$ over the points $z$ with $f(z)=w$.
3. The Jacobian determinant $\det Df(z)$ has constant sign.
4. Conclusion: there is exactly one $z$ such that $f(z)=w$.

The proof works in higher dimensions too, and the assumption on the boundary values can be weakened.

Unfortunately, I don't know of a good source for this material in finite-dimensional setting; it seems that people interested in global invertibility tend to work in nonlinear functional analysis. I can recommend the book Nonlinear Functional Analysis by K. Deimling: its first chapter is on finite dimensional spaces, and global invertibility is considered briefly in Chapter 4.

Answer 2

In $\mathbb{R}^2$, I see no way to get anything like this. The geometry of $\mathbb{R}$ is much simpler though, and you can exploit that to characterize global invertibility (i.e, injectivity) via the behaviour of the derivative.

Let $f \,:\, \mathbb{R} \to \mathbb{R}$ be continuously differentiable and $f'(x) \neq 0$ on the whole real line. Then $f$ is injective.

This works because we required that $f'$ is continuous, so $\forall x \, f'(x) \neq 0$ implies either $f'(\mathbb{R}) \subset (0,\infty)$ or $f'(\mathbb{R}) \subset (-\infty,0)$. The function is thus strictly increasing or strictly decreasing, and hence injective.

If, instead of looking at the whole real line, you look only at a subset, you have to be carefull with disconnected subsets. For example, $f \,:\, \mathbb{R}\setminus\{0\} \to \mathbb{R} \,:\, x \to |x|$ is differentiable on it's whole domain, and the derivative is never zero. Yet this $f$ is obviously not injective.