Let $\gamma_1$ be a curve in $\mathbb{R}^k$, defined on $[a, b]$; let $\phi$ be a continuous 1-1 mapping of $[c, d]$ onto $[a, b]$, such that $\phi(c) = a$; and define $\gamma_2(s) = \gamma_1(\phi(s))$. Prove that $\gamma_2$ is an arc, a closed curve, or a rectifiable curve if and only if the same is true of $\gamma_1$. Prove that $\gamma_2$ and $\gamma_1$ have the same length.
My attempt:
(1) Claim: $\gamma_2$ is an arc $\iff$ $\gamma_1$ is an arc. Proof: ($\Rightarrow$) first suppose $\gamma_2$ is one-to-one. We need to show that, $\gamma_1$ is also 1-1, i.e. $\forall x,x^\prime \in [a,b],$ If $\gamma_1(x)=\gamma_1(x^\prime)$, then $x=x^\prime$. Since $\phi$ is bijective and $x,x^\prime \in [a,b]$, $\exists ! u,v \in [c,d]$ such that $x=\phi(u)$ and $x^\prime=\phi(v)$, respectively. So $\gamma_1(\phi(u))=\gamma_1(\phi(v))$. $\gamma_2$ is one-to-one, i.e. $\forall s,s^\prime \in [c,d],$ If $\gamma_2(s)=\gamma_1(\phi(s))= \gamma_1(\phi(s^\prime)) =\gamma_1(s^\prime)$, then $s=s^\prime$. Since $u,v \in [c,d]$, $\gamma_1(\phi(u))=\gamma_1(\phi(v)) \Rightarrow u=v \Rightarrow x=x^\prime$. Thus, $\phi(u)=\phi(v)=x=x^\prime$.
($\Leftarrow$) conversely suppose $\gamma_1$ is 1-1. We need to show that, $\gamma_2$ is also 1-1, i.e. $\forall x,x^\prime \in [c,d],$ If $\gamma_2(x)=\gamma_2(x^\prime)$, then $x=x^\prime$. $\gamma_2(x)=\gamma_2(x^\prime)=\gamma_1(\phi(x))=\gamma_1(\phi(x^\prime))$. Since $\phi(x), \phi(x^\prime) \in [a,b]$ and $\gamma_1$ is one-to-one, we have $\phi(x)=\phi(x^\prime)$. Since $\phi$ Is bijective, $x=x^\prime$. Thus, $\forall x,x^\prime \in [c,d], \gamma_2(x)=\gamma_2(x^\prime) \Rightarrow x=x^\prime$. Is this proof correct?
(2) Claim: $\gamma_2$ is closed curve $\iff$ $\gamma_1$ is closed curve. Proof: we’ll first show, If $\phi$ is continuous, bijective and $\phi(c)=a$ , then $\phi(d)=b$. Proof: Assume towards contradiction, i.e. $\phi(d) \neq b$, $a \lt \phi(d) \lt b$. Since $\phi$ is bijective, $\exists ! x\in(c,d)$ such that $\phi(x)=b$. Since $\phi$ is continuous on $[c,d]$, it’s also continuous on $[c,x]$. By theorem 4.23, $\phi(c)=a \lt \phi (d) \lt \phi(x)=b, \exists y \in (c,x)$ such that $\phi(y)=\phi(d)$. By injectivity property of $\phi$, $\phi(y)=\phi(d) \Rightarrow y=d$. Which is a contradiction, since $y \lt d$. Thus our initial assumption must be wrong. This completes the proof.
($\Rightarrow$) suppose $\gamma_2$ is closed. So, $\gamma_2(c)=\gamma_2(d)$. $\gamma_2(c)= \gamma_1(\phi(c)) = \gamma_1(a)$, by hypothesis of the problem. $\gamma_2(d)= \gamma_1(\phi(d))= \gamma_1(b)$, by above claim. Thus, $\gamma_1(a)=\gamma_1(b)$, $\gamma_1$ is closed. ($\Leftarrow$) conversely suppose $\gamma_1$ is closed, i.e. $\gamma_1(a)= \gamma_1(b)$. So, $\gamma_2(c)= \gamma_1(\phi(c))= \gamma_1(a)$ and $\gamma_2(d)= \gamma_1(\phi(d))= \gamma_1(b)$. Thus, $\gamma_2(c)= \gamma_2(d)$. Is this proof correct?
(3) claim: $\wedge(\gamma_1) \lt \infty \iff \wedge(\gamma_2) \lt \infty$ and $\gamma_2$ & $\gamma_1$ have the same length. While solving this problem, I first thought that, I should show $\wedge(\gamma_2) \leq \wedge(\gamma_1) \lt \infty$ and $\wedge(\gamma_1) \leq \wedge(\gamma_2) \lt \infty$ so that $\wedge(\gamma_1)=\wedge(\gamma_2)$. But I end up doing the following. Proof: ($\Rightarrow$) first suppose $\wedge(\gamma_1) \lt \infty$. $\wedge(\gamma_1)= \sup \wedge(P,\gamma_1)$; $\wedge(P,\gamma_1)= \sum_{i=1}^n | \gamma_1 (x_i) – \gamma_1 (x_{i-1})|$. $P=\{x_0, ….. , x_n\}$. Since $\phi$ is bijective, $\exists ! y_i \in [c,d]$ such that $\phi(y_i)=x_i, \forall i \in J_n \cup \{0\}$. It’s clear that $\phi(y_0)=x_0=a$ and $\phi(y_n)=x_n=b$. To each partition $P=\{x_0, ….. , x_n\}$ of $[a,b]$ corresponds a partition $Q=\{y_0, … , y_n\}$ of $[c,d]$, so that $x_i=\phi(y_i)$. All partition of $[c,d]$ are obtained in this way. So $\wedge(Q,\gamma_2)= \sum_{i = 1}^n | \gamma_2(y_i)-\gamma_2(y_{i-1})| = \sum_{i = 1}^n | \gamma_1(\phi(y_i))-\gamma_2(\phi(y_{i-1}))| = \sum_{i = 1}^n | \gamma_1(x_i)-\gamma_1(x_{i-1})| = \wedge(P, \gamma_1)$. Thus $\wedge(\gamma_1)=\sup \wedge(P,\gamma_1)= \sup \wedge(Q,\gamma_2) = \wedge(\gamma_2) \lt \infty$. Hence, $\gamma_1$ and $\gamma_2$ have the same length. [Que: My last step, to be more specific, second equality from left side, is not satisfactory to me. It’s not rigorous. Can you make it rigorous?]. ($\Leftarrow$) conversely suppose $\wedge (\gamma_2) \lt \infty$. $\wedge (\gamma_2) = \sup \wedge (Q,\gamma_2)$; $\wedge (Q,\gamma_2) = \sum_{i=1}^n | \gamma_2 (y_i) – \gamma_2 (y_{i-1})|$. $Q=\{y_0, ….. , y_n\}$. To each partition $Q=\{y_0, ….. , y_n\}$ of $[c,d]$ corresponds a partition $P=\{x_0, ….. , x_n\}$ of $[a,b]$, so that $x_i = \phi(y_i)$. All partition of $[a,b]$ are obtained in this way. $\wedge(P,\gamma_1)= \sum_{i = 1}^n | \gamma_1(x_i)-\gamma_1(x_{i-1})| = \sum_{i = 1}^n | \gamma_1(\phi(y_i))-\gamma_2(\phi(y_{i-1}))| = \sum_{i = 1}^n | \gamma_2(y_i)-\gamma_2(y_{i-1})| = \wedge(Q, \gamma_2)$. Thus, $\wedge(\gamma_2)=\sup \wedge(Q,\gamma_2)= \sup \wedge(P,\gamma_1) = \wedge(\gamma_1) \lt \infty$. Hence, $\gamma_1$ and $\gamma_2$ have the same length.[Que: same question as above]. Is this proof correct?
Best Answer
We know that $\phi$ has a continuous 1-1 inverse $\varphi$, and that the composition of one-to-one functions is one-to-one. Hence, since $\gamma_1 = \gamma_2 (\varphi(x))$ we see that $\gamma_1$ and $\gamma_2$ are both arcs (one-to-one) if either is. Since necessarily $\phi(d) = b$, we see that $\gamma_1(a) = \gamma_1(b)$ if and only if $\gamma_2 (c) = \gamma_2 (d)$. Hence both are closed curves if either is. Finally, since $\phi$ and $\varphi$ establish a one-to-one correspondence between partitions $\{s_i\}$ of $[a, b]$ and $\{t_i\}$ of $[c, d]$ such that $\sum | \gamma_1(s_i) - \gamma_1 (s_{i-1})| = \sum | \gamma_2(t_i) - \gamma_2 (t_{i-1})|$, it follows that the two curves have the same length.