Is a continuos and injective mapping from circle to a jordan curve is surjective ?
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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 $f[S^{1}]=S^{1}$. 

