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Is a continuos and injective mapping from circle to a jordan curve is surjective ?

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closed as too localized by Bill Johnson, Yemon Choi, Fernando Muro, Alain Valette, Michael Renardy Jan 5 '13 at 12:33

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That's a cool question for an exam in point set topology! –  Alain Valette Jan 5 '13 at 14:37
    
What percentage of answers in the class was right: 0-20? 21-40? 41-60? 61-80? 81-100? 101-120? –  Włodzimierz Holsztyński Feb 14 at 2:52

1 Answer 1

Yes. Let $f:S^{1}\rightarrow S^{1}$ be continuous and injective. Then by compactness of $S^{1}$, the image $f[S^{1}]$ is homeomorphic to $S^{1}$. In particular, $f[S^{1}]$ is a compact connected subset of $S^{1}$, so $f[S^{1}]$ is either an interval or $S^{1}$ and $f[S^{1}]$ is not an interval since $f[S^{1}]$ is homeomorphic to $S^{1}$. Therefore $f[S^{1}]=S^{1}$.

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