I'm studying curves in $\mathbb{R}^3$, and I came up with a question that I don't manage to answer on my own, so I would like your help. Since the definitions can vary from book to book, I'll first give some definitions and conventions that my book uses:
Regular curve
Let $(\gamma,\mathbf{r})$ be a curve
$\mathbf{r}:I\subseteq \mathbb{R} \to \mathbb{R}^3 , I \text{ interval}$
$(\gamma,\mathbf{r})$ is regular iff $\mathbf{r} \in C^1(I)$ and $\mathbf{r}'(t)\neq 0 \ \forall t\in I$.
Intuitively(if I didn't misunderstood) a curve is regular if and only if $\mathbf{r}$ "walks on $\gamma$" in a continous way withouth jumping or coming back or stopping or zig-zagging, etc. Then there is this definition:
Reparametrization
Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular curves.
$\mathbf{r}_1:I_1\subseteq \mathbb{R} \to \mathbb{R}^3 $
$\mathbf{r}_2:I_2\subseteq \mathbb{R} \to \mathbb{R}^3 $
$(\gamma,{\mathbf{r}_2})$ is a reparametrization of $(\gamma,{\mathbf{r}_1})$ if there exists a bijection:
$\varphi:I_2 \to I_1$
$\varphi \in C^1(I_2)$
$\varphi'(t)\neq 0 \ \forall t \in I_2$
Such that $\mathbf{r}_2=\mathbf{r}_1 \circ \varphi$
Clearly the properties of $\varphi$ are imposed by the fact that $\mathbf{r}_2$ has to be regular. Now I can start to come to my question.Let's proceed step by step. I wondered if two regular curves $(\gamma,\mathbf{r}_1)$, $(\gamma,\mathbf{r}_2)$ are always one the reparametrization of the other. And I come up to a negative answer. In fact I can take: $$\mathbf{r}_1(t)=(\cos(t),\sin(t)) \ \ t\in [0,2\pi]$$ $$\mathbf{r}_2(t)=(\cos(t),\sin(t)) \ \ t\in [0,4\pi]$$ And this curves are not one the reparametrization of the others, because if they were they should pass the same number of times on each point of $\gamma$ in fact if $\mathbf{r}_2=\mathbf{r}_1 \circ \varphi$, let $P$ be a point of $\gamma$: $$|\mathbf{r}_2^{-1}(P)|=|(\mathbf{r}_1 \circ \varphi)^{-1}(P)|=|\varphi^{-1}(\mathbf{r}_1^{-1}(P))|=|\mathbf{r}_1^{-1}(P)|$$ Where the last step is justified because $\varphi^{-1}$ is a bijection so it preserves cardinalities. So I tried to solve this problem by imposing another hyphothesis.
Conjecture
Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular curves.
$|\mathbf{r}_1^{-1}(P)|=|\mathbf{r}_2^{-1}(P)|<\aleph_0 \ \forall P \in \gamma$.
$\Rightarrow $ $(\gamma,\mathbf{r}_1)$, $(\gamma,\mathbf{r}_2)$ are one the reparametrization of the other
However I think that I found a counterexample also to this one, even though it's just a sketch:
(The arrows indicate how each parametrizations walks on $\gamma$, the red point is the start point the coincides with the end point, and just to make it clear $\mathbf{r}_1$ and $\mathbf{r}_2$ are injective(except for the start end point))
This is a countexample because if $\mathbf{r}_1$ and $\mathbf{r}_2$ were one the reparametrization of the other the tangent versors should be equal in every point or opposite in every point. When I come to this point I thought that the problem was that the way that the self-intersection was constructed(because not all self-intersection give problems), but I soonely found out that self-intersections can get pretty nasty and sneaky,so I gave up and I came up with this weaker conjecture:
Final Conjecture
Let $(\gamma,\mathbf{r}_1)$ and $(\gamma,{\mathbf{r}_2})$ be two regular simple curves(if they are close suppose in addiction that the starting\end point of the two parametrizations coincide). Then one is the reparametrization of the other.
I would like to know if this Final Conjecture is true, and if possible I would like some insight about the self-intersections problem and if there is some result about this. I'm sorry if it's a long post, but I'm really interested in this topic so I get carried away.
Thank you in advance.
:)
If I understand your hypotheses correctly, I think the answer to your final conjecture is yes. By your notation $(\gamma, r)$, I take it to mean that $\gamma = r(I)$ is the image of $r$. Then a simple curve presumably means one such that $r: I\to \gamma$ is one-to-one. The key point is that if $r$ is regular at $p\in I$ (meaning $r'(p) \neq 0$), then there is are open neighborhoods $U\subset I$ of $p$ and $V\subset \mathbb R^3$ of $r(p)$, and a $C^1$ function $g: U\to V$, such that $g\circ r = id_U$. In other words, near a regular point $p$, $r$ is a local diffeomorphism to its image. This is a consequence of the inverse function theorem; note that assuming that $r$ is simple means there is an inverse function defined on $\gamma$, but regularity means that the inverse is (locally) the restriction of a differentiable function. Equivalently, if the language is familiar, $\gamma(U)$ is an embedded submanifold of $\mathbb R^3$.
Now if $r_1$, $r_2$ are as in your final conjecture, let $h: \gamma\to I_2$ be the inverse of $r_2$. This exists as a set function since $r_2$ is one-to-one. Moreover, the composition $h\circ r_1: I_1 \to I_2$ is $C^1$: to see this it's enough to check at a given point $p\in I_1$, and since $h$ is the restriction of a $C^1$ function $g$ as above, defined near $r_1(p)$, we get that, near $p$, $h\circ r_1 = g\circ r_1$ is the composition of $C^1$ functions, hence is $C^1$. In fact, $\varphi = h\circ r_1$ is a $C^1$ bijection, since it has an inverse of the form $\tilde h\circ r_2$ where $\tilde h$ is the inverse of $r_1$. Moreover $\varphi$ has nonzero derivative everywhere (from the chain rule and using regularity of $r_1$ and $r_2$), and we have $r_2\circ\varphi = r_1$ since $r_2\circ h = id$ on $\gamma$. So $\varphi$ is the desired reparametrization.
Note that in general for a simple regular curve $r$ there need not be a differentiable function $g$ defined on an open neighborhood of all of $\gamma$ such that $g\circ r = id_{I}$. However, such a $g$ does exist if $I$ is compact.