Is a continuos and injective mapping from circle to a jordan curve is surjective ?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
0
1
|
|||||||
|
closed as too localized by Bill Johnson, Yemon Choi, Fernando Muro, Alain Valette, Michael Renardy Jan 5 at 12:33 |
|
4
|
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 $f[S^{1}]=S^{1}$. |
||
|
|
