# Why does a circle cut a torus into an annulus?

Let $\phi : S^1 \rightarrow T^2$ be an (topological. Not necessarily smooth) imbedding of the circle in the 2-torus and let $\iota : S^1 \rightarrow T^2, \theta \mapsto (\theta,0)$ be the imbedding into the first factor. How can I show that there is a homeomorphism $\psi : T^2 \rightarrow T^2$ such that $\psi \circ \iota = \phi$? Another way to ask this is "why does $\phi$ cut the torus into an annulus?" Thank you for considering my question.

-
I think you assume that $\phi$ is not contractible in $T^2$ for otherwise, $T^2 \setminus \phi(S^1)$ has two components whereas $T^2 \setminus \iota(S^1)$ has just one and so can't be homeomorphic. But $\psi$ would induce such a homeomorphism. –  Marek Jan 22 '13 at 17:06
Rolfsen, in section 2C of Knots and Links, classifies 1-knots in the torus basically by proving this result. It's not a long proof, but I wouldn't call it trivial either. –  Hew Wolff Jan 23 '13 at 0:16