Let $S$ be image of function of $f : (-\pi,\pi) \rightarrow \mathbb{R}^2$ defined $f(t) = (\sin 2t, \sin t)$. Which is figure-eight. Now it is not manifold because there is self-intersection. But it is immersed submanifold of dimension 1. Which implies that it is a smooth manifold. Isn't this contradiction. where am I going wrong ?
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1Why is it an immersed submanifold? – Mark Saving Sep 02 '21 at 16:49
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what definition of "manifold" are you using? – Kajelad Sep 02 '21 at 16:49
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See wikipedia under "mathematical properties". – Dietrich Burde Sep 02 '21 at 16:49
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2In my opinion the concept of an immersed submanifold leads to misunderstandings as in your question. See my answer to https://math.stackexchange.com/q/3899905. – Paul Frost Sep 02 '21 at 17:03
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The figure-eight, with the standard topology inherited from $\mathbb{R}^2$, is not a manifold because in the crossing point there is no neighborhood homeomorphic to some Euclidean space.
However the figure-eight IS a manifold with the topology induced by the immersion $f$, because in this topology there is a neighborhood of the crossing point that is homeomorphic to an open interval in the real line (the topology induced by $f$ imply that the figure-eigth is homeomorphic to $(-\pi,\pi)$).
Masacroso
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1Aaah! Thanks so much. I had this fixed idea that self-intersection => Not a manifold. Makes sense. immersed submanifold has flexibility of what kind of topology we can impose unlike embedded submanifold. This example will make me remember that ! – manifold Sep 02 '21 at 17:19