(1) Let be a one-to-one analytic function on a domain and suppose that there is an analytic function on with . Show that is also one-to-one.
Not really important but if anyone could help me out, I would appreciate it, thanks!
I may be wrong, but it doesn't depend on analysticity:
If are such that then, squaring both sides, , i.e. , and hence because is one-to-one. This shows that is one-to-one as well.
A related question would be to show that is locally one-to-one if is locally one-to-one. This is easy if you know that is locally one-to-one if and only if, for , . Derivating , we get that hence and is thus locally one-to-one as well.