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arthurhenry
#3
Mar9-12, 01:14 AM
P: 43
I think I understand it now.
I think you are saying:
suppose p is a point inside the disk, i.e. an interior point. Take nbhd around p that is contained in the unit disk still. Then by Open Mapping Theorem, the image of this disk is open, i.e., the f(p) is also contained in a nbhd that is also inside the unit circle, so f(p) cannot be a boundary point. Since the LFT is injective, all interior points is taken as the images of interior points and the only place for a boundary point to be sent is the boundary.
Hope I am right...in the sense that I am not able conclude this without the Open mapping theorem.