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A set $S \subseteq \mathbb R^{n+k}$ is an $n$-surface iff there is an open set $U$ in $\mathbb R^{n+k}$ and a smooth map $f:U \to \mathbb R^{k}$ such that $S=f^{-1}(\{0\})$ and rank $Df(a)=k , \forall a \in S $ . I am looking for some statement of the form " smooth ... injective ... (may be injective , full rank) image of a (compact or connected may be) $n$-surface is again a surface (or atleast a smooth manifold) " .. Can you please give a reference or some links ( possibly with proofs) of any such statement available ? Also I am looking for such similar statement about when the (smooth) image of a smooth manifold is again a manifold . Please help . Thanks in advance

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  • $\begingroup$ An embedding $f: M \rightarrow \mathbb{R}^k$ will map $M$ to a diffeomorphic submanifold. It means that $f$ is a homeomorphism on its image with an everywhere injective derivative. See en.wikipedia.org/wiki/Embedding $\endgroup$ – Olivier Mar 12 '17 at 6:32
  • $\begingroup$ @Olivier : and "embedding" here means ? $\endgroup$ – user228169 Mar 12 '17 at 6:33
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An example of a classical statement is one in which you encounter during your first course in general topology, namely; ''If $f: X \to Y$ is a continuous bijection with $X$ compact and $Y$ Hausdorff, then $f$ is a homeomorphism." Also look here: How to use the corollary (manifold version rank theorem) to prove the quoted theorem?

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  • $\begingroup$ Sorry .. but what that point-set topology statement has anything to do with smooth images of smooth manifolds and surfaces ? $\endgroup$ – user228169 Mar 12 '17 at 13:17
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    $\begingroup$ Certainly if you have a smooth map then it is continuous and certainly if you have a smooth manifold, then it is a topological manifold. The difference in the two being a smooth manifold has differentiable structure. Also, I think the link I attached is perfect for your question, as it add two more sources that weren't mention. Now, tell me what is the problem? $\endgroup$ – Faraad Armwood Mar 12 '17 at 13:27

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