Is a figure eight a manifold

Question

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 ?

Answer 1

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