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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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up vote 7 down vote accepted

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.

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or define f(z)=$z^2$ on the union of |z|<1 in the right-half plane,and f(z)=$z^2$ on the left-half plane, including the axes. –  gary May 14 '11 at 23:11
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Or define f(z)=$z^2$ on the union of |z|<1 on the first quadrant, union |z|<2 on the second quadrant, |z|<3 on the third, and |z|<4 on the fourth. Stretch the domain a bit so that it is defined on the axes and $z^2$ is defined in a region, which you can do since $z^2$ is entire. Then the antipodes of boundary points will not fall on other boundary points, but antipodes of interior points will. –  gary May 14 '11 at 23:30
    
Thanks. How about simply connected regions? –  wzzx May 15 '11 at 0:37
    
Just poke a small open ball around the origin of the defined region, and the same would apply; antipodes of interior points would remain after the removal, but antipodes of boundary points would still not be included in the region. Is that what you meant? –  gary May 15 '11 at 0:50
    
Indeed. Thanks! –  wzzx May 15 '11 at 6:08

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