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.