Is an immersion defined on a simply connected region injective?

Question

Let $\Omega$ be a simply connected region in $\Bbb R^n$ (let's say a closed ball), $f\colon \Omega\to\Bbb R^n$ be smooth and $f'$ everywhere non-singular. Does it imply that $f$ is injective?

(Without the simple connectedness assumption, it is easy to find counter-examples, such as $z^2\colon \{1\leq |z|\leq 2\}\to \Bbb C$, but I don't see how something like this can happen for $\Omega=B^2$...)

Answer 1

Let's work in $\Bbb C\cong\Bbb R^2$. Let $\Omega$ be the right half-plane. Then $z\mapsto z^3$ is an immersion, but not injective. Of course, $\Omega$ is simply connected. If you insist on a ball, just take one inside the right half plane containing both of $\exp(\pm i\pi/3)$.

Answer 2

The other standard example is the exponential map $\Bbb C\to\Bbb C$. Or, in real coordinates, consider $$f\colon\Bbb R^2\to\Bbb R^2, \quad f(x,y) = (e^x\cos y,e^x\sin y).$$ The derivative is everywhere nonsingular, but $f$ takes almost every value infinitely many times.