# The Closed disc $D$ is a manifold with boundary

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. –  t.b. May 20 '12 at 18:07
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. –  Paul Siegel May 20 '12 at 19:58
@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? –  Jr. May 21 '12 at 2:24
@PaulSiegel What do you mean by "a closed manifold"? –  Jr. May 21 '12 at 2:25
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. –  Paul Siegel May 26 '12 at 17:53