Inclusions of submflds. and diffeomorphisms.

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The discussion centers on the inclusion map of submanifolds in differential geometry. It establishes that if S is a submanifold of M with the weak topology, the inclusion map is a homeomorphism. However, if S is closed in M, the inclusion may not be a diffeomorphism. The distinction between immersed and embedded submanifolds is crucial, as the inclusion is an injective immersion for immersed submanifolds but requires the implicit function theorem to confirm diffeomorphism for embedded submanifolds.

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WWGD
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Hi, everyone:

I am a little confused about the issue of the inclusion map on submanifolds.

AFAIK, if S is a submanifold of M , then if we give S the weak (or initial) topology

of the inclusion, then the inclusion map is a homeomorphism.(his is the way I understand,

of making the inclusion map of a subspace into a space, into a homeomorphism. If

S is open in M, with this initial topology, then I think inclusion is also a diffeomorphism.)


Question: under what conditions on S is the inclusion map a diffeomorphism?.

I think that if S is closed in the topology of M, or at least not open in M , and

given the weak topology, then the inclusion may not be a diffeomorphism.

Is this correct?.

And when is a (topological) subspace S of M a submanifold, other than when S

is open in M, i.e., S is open as a subset of M?



Thanks.
 
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You have to distinguish between immersed and embedded submanifolds (see also http://en.wikipedia.org/wiki/Submanifold" ). For immersed submanifolds the inclusion is an injective immersion, but not necessarily a homeomorphism onto its image (i.e. an embedding). If it is an embedding then the submanifold is an emdedded submanifold and the inclusion is also a diffeomorphism onto its image, this is not completely trivial and requires the implicit function theorem.

And when is a (topological) subspace S of M a submanifold, other than when S

is open in M, i.e., S is open as a subset of M?

If one can find charts of M such that S looks locally like R^k in R^n.
 
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