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 ?

  • 1
    $\begingroup$ Why is it an immersed submanifold? $\endgroup$ Sep 2, 2021 at 16:49
  • $\begingroup$ what definition of "manifold" are you using? $\endgroup$
    – Kajelad
    Sep 2, 2021 at 16:49
  • $\begingroup$ See wikipedia under "mathematical properties". $\endgroup$ Sep 2, 2021 at 16:49
  • 2
    $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Sep 2, 2021 at 17:03

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

  • 1
    $\begingroup$ Aaah! 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 ! $\endgroup$
    – manifold
    Sep 2, 2021 at 17:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .