If I have a bounded, connected, open subset of the complex plane, and a function that is holomorphic on it, continuous on its closure, and injective on its boundary, is my function necessarily injective?
It seems it is not true for arbitrary connected regions. Is it true for simply connected regions?
Best Answer
No. For example, $f(z)=z+1/z$, your domain given by $r<|z|<R$ with $r<1<R$, $rR\neq1$, then the boundary circles are mapped injectively to two confocal ellipses, but $f(i)=f(-i)=0$.
edit: If the region is simply-connected and bounded by a Jordan curve then $f$ must be injective: The image (under $f$) of the boundary is a Jordan curve, hence the winding number of the image curve around any point is either $1$ (if the point is inside) or $0$ (if outside). By the argument principle, the number of preimages of any point is the winding number, hence $f$ is injective.