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Let $\gamma :(a,b) \to \mathbb{R}^d$ be a $C^1$ curve and $S = \left \{ x | \exists t\in(a,b) : \gamma(t)=x \right \}$ its image.

Show that if $\gamma$ is injective and $\gamma' (t) \neq \vec{0}, \forall t \in (a,b)$, then $S$ is a one dimensional differentiable manifold in $\mathbb{R}^d$.

Just to clarify the definition we were taught: I need to show that $\forall x \in S$, there exists an open neighborhood $V \subset \mathbb{R}^d$ of $x$ s.t $S \cap V$ is the graph of a differentiable function $f: U \subset \mathbb{R} \to \mathbb{R}^{d-1}$.

I believe this should be a pretty basic exercise but I'm a little lost. Help appreciated.

*EDIT: The above question is false as mentioned, modifying the question this way:

Let $\gamma :(a,b) \to \mathbb{R}^d$ be a $C^1$ curve and $S = \left \{ x | \exists t\in(a+\varepsilon ,b-\varepsilon) : \gamma(t)=x \right \}$ for some $\varepsilon>0$, the image of $\gamma |_{(a+\varepsilon ,b-\varepsilon)}$.

Show that if $\gamma$ is injective and $\gamma' (t) \neq \vec{0}, \forall t \in (a,b)$, then $S$ is a one dimensional differentiable manifold in $\mathbb{R}^d$.

how would we prove it?

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    $\begingroup$ This statement is false. For a discussion and a depiction of a counterexample, see the (somewhat poorly written) entry on wikipedia: en.wikipedia.org/wiki/Submanifold#Immersed_submanifolds $\endgroup$
    – Lee Mosher
    Jan 12, 2021 at 20:43
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    $\begingroup$ There's a typo in the problem. The assumption should be $\gamma'(t)\ne 0$ for all $t$. $\endgroup$ Jan 12, 2021 at 20:44
  • $\begingroup$ @TedShifrin Thanks for noticing, my bad $\endgroup$ Jan 12, 2021 at 20:45
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    $\begingroup$ It's still false for the reason Lee Mosher has pointed out, though. $\endgroup$
    – user239203
    Jan 12, 2021 at 20:46
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    $\begingroup$ Compactness is crucially a part of analysis. But here's the other way: You need to know that $\gamma^{-1}$ is continuous (as a map with domain $\gamma((a,b))$, of course). The point is that you'll need to know that each point of the image has a neighborhood in $\Bbb R^d$ that is just diffeomorphic to an interval in $\Bbb R$. Then you can apply the inverse/implicit function theorem. But yes, I think your instructor messed up. $\endgroup$ Jan 12, 2021 at 20:58

1 Answer 1

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NOTE: we'll assume the function $\gamma$ proper, in order to avoid pathological cases (i.e. the curve getting infinitely close to itself - though without self-intersections). I thank Ted Shifrin for his observations and corrections on this proof.

I'll use the notation $\gamma(t)=(\gamma_1(t),...,\gamma_d(t))$.

For a fixed $t_0$ you can take one of the $d$ axis, say $\hat{x_1}$ (if not, $\hat{x_2}$...) such that $\gamma '(t_0)$ is not perpendicular to $\hat{x_1}$.

Then $\gamma_1'(t_0)\neq 0$ and I can take an interval on $\hat{x_1}$ where it remains $\neq0$. Over this interval, I reparametrise $Im\gamma$ as:

$\alpha(s)=(\gamma \circ \gamma_1^{-1})(s) = (\gamma_1 \circ \gamma_1^{-1},\gamma_2 \circ \gamma_1^{-1},...,\gamma_d \circ \gamma_1^{-1})(s) = (s, \alpha_2(s),...,\alpha_d(s)) $

by using that $\gamma_1' \neq 0 \Rightarrow \gamma_1$ bijective.

And there we have a local graph (if we restrict on $V\cap S$ with $V\subset\mathbb{R}^d$ a suitable open set)

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  • $\begingroup$ For the second line, I can't have that $\gamma'(t_0)$ is perpendicular to every axis. If it was, then $||\gamma'(t_0)||^2=(\gamma'(t_0)\cdot \hat{x_1})^2+...+ (\gamma'(t_0)\cdot \hat{x_d})^2=0$ $\endgroup$
    – rod
    Jan 30, 2021 at 0:26
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    $\begingroup$ Right. The question then is why shrinking to $S$ solves the problem that the proof would otherwise be wrong. $\endgroup$ Jan 30, 2021 at 0:35
  • $\begingroup$ Well, I skipped that because since I restrict $\gamma_1$, I should also restrict $\alpha$ and therefore $Im\alpha$ $\endgroup$
    – rod
    Jan 30, 2021 at 0:40
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    $\begingroup$ No, that isn’t sufficient. Far-away points of the curve can approach your point ... or could before $S$ was introduced. $\endgroup$ Jan 30, 2021 at 0:47
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    $\begingroup$ Think about the figure 8 example that we referred to in the original comments. $\endgroup$ Jan 30, 2021 at 0:50

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