# Is a figure eight a manifold

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 ?

• Why is it an immersed submanifold? Sep 2, 2021 at 16:49
• what definition of "manifold" are you using? Sep 2, 2021 at 16:49
• See wikipedia under "mathematical properties". Sep 2, 2021 at 16:49
• In my opinion the concept of an immersed submanifold leads to misunderstandings as in your question. See my answer to math.stackexchange.com/q/3899905. Sep 2, 2021 at 17:03

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 $$(-\pi,\pi)$$).