By my course, all manifold of dimension 1 is isomorphic to $(0,1), (0,1],[0,1)$ or $\mathbb S^1=\{x^2+y^2=1\mid x,y\in\mathbb R\}$.
I was thinking of a curve in the plan, with a knot. (See picture)
I agree that the gluing (in pink) is possible in $\mathbb R^3$, but if we consider this curve in $\mathbb R^2$, it doesn't look possible to me, otherwise we have to cut the curve like this
and thus, It can't work, no ? Because if it does, we could do the same with the torus, and then, the Torus in $\mathbb R^3$ would be homeomorphic to the sphere $\mathbb S^3=\{x^2+y^2+z^2=1\mid x,y,z\in\mathbb R\}$, which is not the case. Am I right ?
Q1) So if I'm right, don't we have to add those type of curve (i.e. with a knot) in the classification of manifold of dimension 1 ?
Q2) If not, why the torus is not homeomorphic to $\mathbb S^2$ in $\mathbb R^3$ ?