If a function has a local inverse everywhere does that mean its invertible?

Question

I just learned the inverse function theorem and I immediately began wondering the following:

Let $F:U\to F(U)$ be an $C^1$ function, where $U\subseteq \Bbb R^n$ is an open set.

If, for all $x \in U$, the function $F$ has a local inverse at $x$, does that mean that $F$ has an inverse?

I tried finding a countre example but I wasn't able to think of any. after that I tried to prove this but I also wasn't able to prove this.

Is this true? If so how can this be proved?

https://math.stackexchange.com/questions/2256093/inverse-function-theorem-and-global-inverses This may be relevant — almosteverywhere, Nov 14 '21 at 12:44

score 0 · Answer 1 · answered Nov 14 '21 at 12:55

0

No, not necessarily. As a counterexample, consider $f:U\to\mathbb R^2$ where $U$ is the open rectangle $(1,2)\times(0,7)$: $$ f(x,y) = \begin{bmatrix} x\cos(y) \\ x\sin(y) \end{bmatrix} $$

answered Nov 14 '21 at 12:55

Troposphere

7,158

If a function has a local inverse everywhere does that mean its invertible?

1 Answers1