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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?

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No. For example, $f(z)=z+1/z$, your domain given by $r<|z| with $r<1, $rR\neq1$, 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.

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    @1736: I am using disks of different sizes in different quadrants: |z|<1 on 1st, |z|<2 on second, and so on, together with a slight "stretching" (more formally, analytically continuing beyond the union), to contain the axes. Then, on the bdry. of 1st quad. f(z)=z^2; on bdry. of 2nd Quad, f(z)=4$z^4$, etc. But the interior will contain antipodes, e.g., z=1/2 and z=-1/2, with same image. Or, imagine starting with |z|=1, and then stretch it outwards on 2nd , 3rd, 4th quads., respectively. I wish I could attach a drawing as an attachment.2011-05-15