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This question is partially answered here: How to see the real projective plane is a Möbius band glued to a disk?

In that answer it is shown that for a very specific embadding $f: D \rightarrow P$, space $P - int(D)$ is homeomorphic to Mobious strip and it can be (rigorously) proved constructively.

I want to show that that is the case for every embedding.

I would like to somehow use Jordan curve theorem and Schoenflies theorem but I must just stuck on how to prove that for a non specific embedding of a disk $D$.

Anyone any ideas?

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I am hesitant to use the Jordan curve theorem here but perhaps it is possible. Instead what I would use is the important fact that the space $Emb(D^n , M)$ is equivalent to $Fr(M)$, the total space of the fiber bundle over $M$ where the fiber over $x$ is the space of bases of the tangent bundle. Here $M$ is n-dimensional. This equivalence just comes from pushing forward the standard basis along the derivative.

For a nonorientable connected manifold, this space is path connected because the base is path connected and there exists a path in the total space that connects any two points in a specific fiber (this is equivalent to being nonorientable).

From this we conclude that any two embeddings of the disk are isotopic. A very nice result called isotopy extension then implies that there is a diffeomorphism of the manifold taking one of these disks to the other. This implies that as pairs (the manifold and the image of the embedded disk), they are exactly the same topologically (as well as smoothly). This implies that removing the interior gives homeomorphic (as well as diffeomorphic) manifolds.

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  • $\begingroup$ Thanks for the reply - although I thought this could be answered in more elementary terms. I have not yet learned fiber bundles. :( $\endgroup$
    – espasios
    Nov 10 '20 at 8:35
  • $\begingroup$ @espasios Perhaps it can, but questions about embeddings are just very difficult. $\endgroup$ Nov 10 '20 at 12:24

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