Showing by definition that a curve is a 1 dimensional differentiable manifold

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?

• 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 Jan 12, 2021 at 20:43
• There's a typo in the problem. The assumption should be $\gamma'(t)\ne 0$ for all $t$. Jan 12, 2021 at 20:44
• @TedShifrin Thanks for noticing, my bad Jan 12, 2021 at 20:45
• It's still false for the reason Lee Mosher has pointed out, though.
– user239203
Jan 12, 2021 at 20:46
• 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. Jan 12, 2021 at 20:58

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)

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