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It is obvious geometrically, but how one proves with a few words, analytically, the statement above?Additionally, if one has a smaller open disc $D_\epsilon$ of radius $\epsilon$ centered at a point of $D$, how to conclude that $D-D_\epsilon$ is still a manifol with boundary $\partial D \cup\partial D_\epsilon$?

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    Three words: [Inverse Cayley transform.](http://en.wikipedia.org/wiki/Cayley_transform#Conformal_map)2012-05-20
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    Generally the subspaces of Euclidean space which are "obviously" manifolds with boundary locally have the form $f(x_1, \ldots, x_n) \leq C$ (or perhaps they are finite intersections of such objects). For these spaces there are variations on the implicit function theorem which do all the work for you, similarly to the way the ordinary implicit function theorem tells you when $f(x_1, \ldots, x_n) = C$ is a closed submanifold. If I can find a reference I'll post this as an answer.2012-05-20
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    @t.b If I am right, inverse cayley transform is a boundary chart when we see the disc as a complex manifold, but if we want to prove just the real case?2012-05-21
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    @PaulSiegel What do you mean by "a closed manifold"?2012-05-21
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    If you just write out the formulas for the real and imaginary components of the inverse Cayley transform then you get diffeomorphisms between small open subsets of the disk and small open subsets of the closed upper half plane. The complex structure isn't relevant here.2012-05-26
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    "Closed manifold" = compact manifold without boundary. Of course compact isn't really an issue here - I'm just pointing out that you can prove that certain sets described by inequalities are manifolds with boundary using essentially the same tools that are used to prove that certain sets described by equations are manifolds without boundary. Namely, the implicit function theorem and its generalizations.2012-05-26
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    I don't get it! But if $x$ is an interior point of $D$, i.e. $x \in D\setminus\partial D$ we know that $T_xD=\mathbb{R}^2$, and if $x \in \partial D$, then $T_xD$ is isomorphic to $\mathbb{R}$, so I'm kind of loss here! A question: What is the dimension of $D$ as a manifold?2012-06-23
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    @Mercy I guess the following: As a real manifold with bondary, it has dimension 2, I don't know the answer for the complex case.2012-06-23
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    Do you mean the open disk or the closed one?2012-06-23
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    @Mercy the closed one2012-06-24

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