Prob. 19, Chap. 6, in Baby Rudin: Relation between two curves in $\mathbb{R}^k$

Question

Here is Prob. 19, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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.

Here is the link to a Math SE post of mine where I've included all the relevant definitions.

Prob. 18, Chap. 6, in Baby Rudin: Analysis of Some Curves in the Plane

My Attempt:

As $\phi$ is a continuous, bijective mapping of the compact set $[c,d]$ onto $[a, b]$, so $\phi$ has an inverse $\phi^{-1}$, which is a bijective, continuous mapping of $[a, b]$ onto $[c, d]$, by Theorem 4.17 in Rudin.

And, as $\gamma_2 = \gamma_1 \circ \phi$, so we have $\gamma_1 = \gamma_2 \circ \phi^{-1}$.

Now suppose that $\gamma_1$ is an arc; this means that $\gamma_1$ is 1-1, and since $\phi$ too is 1-1, so is the composite $\gamma_1 \circ \phi$; that is, $\gamma_2$ is an arc.

Conversely, suppose that $\gamma_2$ is an arc; this means that $\gamma_2$ is 1-1, and as $\phi^{-1}$ too is 1-1, so is the composite $\gamma_2 \circ \phi^{-1}$; that is, $\gamma_1$ is an arc.

Thus we have shown that $\gamma_1$ is an arc if and only if $\gamma_2$ is an arc.

Am I right?

Now suppose that $\gamma_1$ is a closed curve. Then $\gamma_1(a) = \gamma_1(b)$. And as $\gamma_2(c) = \gamma_1(\phi(c)) = \gamma_1(a) = \gamma_1(b)$.

What next?

Conversely, suppose that $\gamma_2$ is a closed curve. Then $\gamma_2(c) = \gamma_2(d)$; that is, $\gamma_1(\phi(c)) = \gamma_1(\phi(d))$.

What next?

How to show that $\gamma_1$ is rectifiable if and only if $\gamma_2$ is rectifiable?

And, how to show that $\gamma_1$ and $\gamma_2$ have the same length?

Answer 1

The argument for the one-to-one (arc) part could be written in a shorter way: the composition of one-to-one functions is one-to-one. Note that in this part, the continuity of $\phi^{-1}$ is not needed (but yes, it would be needed later).

For the "closed curve" part, you have already observed that $\phi$ is a homeomorphism (continuous bijection with continuous inverse); note that by assumption $\phi(c)=a$ one must have $\phi(d)=b$. (This can be done using properties of homeomorphisms.) Thus we see that $\gamma_1(a)=\gamma_1(b)$ if and only if $\gamma_2(c)=\gamma_2(d)$.

Finally for the "same length part", observe that $\phi$ and its inverse $\phi^{-1}$ gives a one-to-one correspondence between partitions $\{s_i\}$ of $[a,b]$ and $\{t_i\}$ such that $$ \sum|\gamma_1(s_i)-\gamma_1(s_{i-1})|=\sum|\gamma_2(t_i)-\gamma_2(t_{i-1})|. $$ Hence the two curves must have the same length by definition.