My goal is to proof that every simple closed curve is homeomorphic to unit circle. I also have nice theorem for it.

But, I cannot do it before I find continuous surjection.

Let $f([0,1])\subseteq\mathbb{R}^{2}$ be the simple closed curve ($f:[0,1]\rightarrow \mathbb{R}^{2}$).

How I can find continuous surjection $g:f([0,1])\rightarrow S^{1}$?

  • 1
    $\begingroup$ Write a map in the other direction. $\endgroup$
    – user98602
    Nov 21, 2015 at 20:31
  • $\begingroup$ So I should take advantage of the mapping from $[0,1]$ to $S^1$? If that is $h$, then we have $t\mapsto (\cos (2\pi t),\sin (2\pi t))$. Now we should find such $g$ that $g \circ h = f$? $\endgroup$
    – Zzz
    Nov 21, 2015 at 21:34

2 Answers 2


Both are homeomorphic to the quotient space $[0,1] / R$ where $R$ is the equivalence relation that only identifies $0$ and $1$ and no other points. For the circle you can use the map induced by the standard parametrisation, and $f$ has a similar role. And two spaces homeomorphic to the same space are homeomorphic.


Use Jordan-Schoenflies Curve Theorem.


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