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I just want to check if my intuition is right:

Let $\phi:U\subset\mathbb{R}^2\rightarrow\mathbb{R}^3$ be a differentiable, one-to-one, map from an open set $U$, such that at each point $q\in U$ the derivative $f'(q):\mathbb{R}^2\rightarrow\mathbb{R}^3$ is also one-to-one (that is, $\phi$ is an immersion). Then $\phi$ is a homeomorphism onto its image, correct? Because if I apply the local form of immersions, for each $q\in U$ I can guarantee a neighborhood $V\subset U$ such that $\phi|_V$ is a homeomorphism. Because of the unicity of the inverse $\phi^{-1}|_{\phi(V)}=\left(\phi|_V \right)^{-1}$, hence $\phi^{-1} $ is continuous at $q$. Since $q$ is arbitrary we prove the statement.

Also, many books start the definition of surfaces by defining parametrizations as maps $\phi:U\subset\mathbb{R}^2\rightarrow\mathbb{R}^3$ such that: (i) $\phi$ is differentiable, (ii) $\phi$ is homeomorphism onto its image, (iii) $\phi$ is an immersion. However, if the above is true, it suffices to change the second property to (ii)': $\phi$ is one-to-one, correct?

Arctic Char
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Fernando Nazario
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  • In general the image of an injective immersion doesn't need to be even a topological manifold. See the Chapter 6 of J. Lawson's notes. I doubt that the situation is different for maps $\mathbb R^2\to \mathbb R^3$, although I don't have a counterexample. – Paweł Czyż Sep 21 '21 at 14:21
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    $\phi$ doesn't need to be an homeomorphism, by example take the figure eight for here and add some height to make it a surface – Masacroso Sep 21 '21 at 14:25
  • Where does the demonstration fail? Also, in do Carmos' Differential Geometry of Curves and Surfaces, proposition 4 chap 2-2 states: if $p\in S$ is a point in a regular surface $S$ and $\phi:U\subset\mathbb{R}^2\rightarrow\mathbb{R}^3$ with $p\in\phi(U)$ such that conditions (i) and (iii) holds, then $\phi^{-1}$ is continuous, but the proof doesn't use the fact that $S$ is a regular surface, only the local form of immersions – Fernando Nazario Sep 21 '21 at 16:24
  • The error in your argument is that $\phi(V)$ need not be open in $\phi(U)$, so testing continuity of the inverse on $\phi(V)$ does not suffice to make it globally continuous (think this through in the figure eight example to see what's going on). do Carmo uses the fact that $S$ is a regular surface to get a parametrization, from which he then uncorks a similar argument, but where you actually have open subsets of the image. – Thorgott Sep 21 '21 at 19:42

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