I'm very stuck on Exercise 1.35 (4) in Kristopher Tapp's Differential Geometry of Curves and Surfaces which reads:
Show that two parametrized simple closed curves have the same trace if and only if they are equivalent (that is, one is a reparametrization of the other).
Note that this text defines curves to be smooth and closed curves to be regular. Further, it is meant to only have the prerequisites of multivariable calculus, linear algebra, and real analysis (not necessarily including multivariable content), does not introduce the idea of a diffeomorphism until talking about surfaces, and does not define the derivative of a curve as a linear function between tangent spaces (but rather as another curve, differeniated componentwise). I am looking for a solution that reflects this.
The "if" direction is clear from the definitions, but I'm stuck on the "only if". I've tried doing it directly by supposing two simple closed curves $\boldsymbol\gamma: [a, b] \to \mathbb R^n$ and $\boldsymbol\beta: [c, d] \to \mathbb R^n$ have the same trace and constructing a bijection $\phi: [a, b] \to [c, d]$ such that $\boldsymbol\gamma = \boldsymbol\beta \circ \phi$, but I get stuck trying to show that $\phi$ is smooth, let alone that its derivative is never zero and that its derivatives all match at $a$ and $b$.
Update: I've been reviewing the differential geometry exercises I was working on when I wrote this problem and made some progress, but I'm still not quite sure how to put everything together. Here's what I have so far.
Let $\boldsymbol\gamma: [a, b] \to \mathbb R^n, \boldsymbol\beta: [c, d] \to \mathbb R^n$ be two parametrized simple closed curves with the same trace $\Gamma$ and components $\boldsymbol\gamma(t) = (\gamma_1(t), \ldots, \gamma_n(t)), \boldsymbol\beta(t) = (\beta_1(t), \ldots, \beta_n(t))$, and assume without loss of generality that $\boldsymbol\gamma(a) = \boldsymbol\beta(c)$. For every $s \in [c, d]$, there is a $k(s) \in \{1, \ldots, n\}$ such that $\beta_{k(s)}'(s) \neq 0$, and hence there are neighbourhoods $U_s$ of $s$ in $[c, d]$ and $V_s$ of $\beta_{k(s)}(s)$ in $\mathbb R$ such that $\beta_{k(s)}$ has a smooth local inverse $\psi_s: V_s \to U_s$. I then want to define $\phi(t) = \psi_s(\gamma_{k(s)}(t))$, where $\boldsymbol\beta(s) = \boldsymbol\gamma(t)$, but I'm getting caught up in writing out the details showing that this is well-defined (and I'm not sure it is well-defined for $t \in \{a, b\}$). Once I have that, it is clear that (a) $\phi$ is smooth, (b) $\phi$ has nonzero derivative, and (c) all derivatives of $\phi$ match at $a$ and $b$.
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$Let $\Gamma_{0}$ denote $\Gamma$ with the point $\gamma(a) = \gamma(b) = \beta(c) = \beta(d)$ removed.
Each of $\gamma:(a, b) \to \Gamma_{0}$ and $\beta:(c, d) \to \Gamma_{0}$ is a diffeomorphism, so $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is a diffeomorphism (as smooth as the less smooth of the two paths),[*] and extends to the endpoints uniquely by continuity. The extended overlap map is a reparametrization.
For the starred point, if you need to go lower-level than this the important point is that $\gamma_{*}:T(a, b) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point because $\gamma$ is regular, and similarly $\beta_{*}:T(c, d) \to T\Gamma$ is an isomorphism of one-dimensional vector spaces at each point. Consequently, $( \beta^{-1}\circ\gamma)_{*} = (\beta^{-1})_{*}\circ\gamma_{*}:T(a, b) \to T(c, d)$ is an isomorphism of one-dimensional spaces at each point.
Added: We know the function $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is bijective and satisfies $\gamma = \beta \circ (\beta^{-1} \circ \gamma)$. As indicated by the comments, we need to show $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.
Proposition: If $\gamma(t_{0}) = \beta(\tau_{0})$ for some $t_{0}$ in $(a, b)$ and $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$ in $(c, d)$, then $\gamma'(t_{0})$ and $\beta'(\tau_{0})$ are proportional, i.e., each is a (non-zero) multiple of the other.
Proof: Pick an orthonormal basis $(e_{j})_{j=1}^{n-1}$ for the hyperplane orthogonal to $\gamma'(t_{0})$ and define the smooth mapping $F:\Reals^{n} \to \Reals^{n}$ by $$ F(u_{1}, \dots, u_{n-1}, t) = u_{1}e_{1} + \cdots + u_{n-1}e_{n-1} + \gamma(t). $$ Since $Df = [e_{1}\ \cdots\ e_{n-1}\ \gamma'(t)]$ is non-singular at $T_{0} := (0, \dots, 0, t_{0})$, the inverse function theorem guarantees there is a neighborhood $U$ of $T_{0}$ in which $F$ is smoothly invertible. Write $V = F(U)$. By construction, the functions $\phi_{j} := u_{j} \circ F^{-1}$ are smooth, vanish simultaneously on the trace of $\gamma$ and nowhere else, and have linearly independent gradients in $V$. That is, the trace of $\gamma$ intersected with $V$ is precisely the zero set of $\Phi := (\phi_{1}, \dots, \phi_{n-1})$, and the kernel of this mapping, $\ker D\Phi(T_{0})$, is one-dimensional. Since $\beta$ has the same trace as $\gamma$, the non-zero vector $\beta'(\tau_{0})$ lies in this same one-dimensional subspace. That is, $\beta'(\tau_{0})$ and $\gamma'(t_{0})$ are proportional. This completes the proof of the proposition.
Write $\gamma = (\gamma_{k})_{k=1}^{n}$ and $\beta = (\beta_{k})_{k=1}^{n}$, and note that $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$ for each $k$. Let $t_{0}$ be an arbitrary point of $(a, b)$ and put $\tau_{0} = \beta^{-1} \circ \gamma(t_{0})$. Because $\beta$ is regular, there exists an index $k$ such that $\beta_{k}'(\tau_{0}) \neq 0$. The preceding proposition guarantees $\gamma_{k}'(t_{0}) \neq 0$. By the one-variable inverse function theorem, there exists a $\delta > 0$ such that $\beta_{k}$ is smoothly invertible in $(\tau_{0} - \delta, \tau_{0} + \delta)$. Consequently, $\beta_{k}^{-1} \circ \gamma_{k}$ is smoothly invertible in some neighborhood of $t_{0}$ as a composition of smoothly invertible functions. But $\beta^{-1} \circ \gamma = \beta_{k}^{-1} \circ \gamma_{k}$ (!) in this neighborhood by applying $\beta_{k}^{-1}$ to both sides of $\gamma_{k} = \beta_{k} \circ (\beta^{-1} \circ \gamma)$. Since $t_{0}$ was arbitrary, $\beta^{-1} \circ \gamma$ is locally smoothly invertible at each point.
In particular, the preceding paragraph shows that the bijection $\beta^{-1} \circ \gamma:(a, b) \to (c, d)$ is continuous. A continuous bijection whose domain is an interval is strictly monotone (by the intermediate value theorem), and a strictly monotone function between bounded intervals extends uniquely by continuity to the endpoints. In summary, we have shown that $\beta^{-1} \circ \gamma:[a, b] \to [c, d]$ is smooth and has smooth inverse.