To be more precise, as sets, $S\subset M$, and S has structure such that it is a manifold on its own right but not a submanifold of the manifold $M$?
Is there a case such that $S$ is a manifold on its own right but not a submanifold?
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differential-geometry
manifolds
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0Might be mistaken, but have you heard about [irrational winding of a torus](https://en.wikipedia.org/wiki/Linear_flow_on_the_torus#Irrational_winding_of_a_torus)? – 2017-02-24
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1Or take the mapping $(-\pi/2,\pi/4)\to\Bbb R^2$ given by $t\rightsquigarrow \big(\cos(2t)\cos t,\cos(2t)\sin t\big)$, whose image is ... a figure 6. :) – 2017-02-24
1 Answers
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Take $M = \mathbb R^2$. Take $S$ to be the image of $i: \mathbb R \to M$ given by $i(t) = (t^2, t^3)$. Then $S$ has a cusp at $(0,0)$, so it does not inherit a manifold structure from $M$. However, $S = \mathbb R$ as a set, which has an obvious manifold structure of its own.
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0Right: In the smooth category. In the topological category it would be a wild sphere in $R^3$. – 2017-02-24
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0How is $S=\mathbb{R}$ as a set? should not S be a curve in $R^2$?, its y component equals set R, no doubt. – 2017-02-25
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0And S is not a manifold, right? – 2017-02-25
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0@jack $S = \mathbb R$ as a set because $S$ is the image of an injective function of $\mathbb R$. – 2017-02-25