What is the "reverse" path in the case of a piecewise defined curve?

Question

Consider a curve $\gamma(t): [a,b] \to \mathbb{R}^n$. The curve $$-\gamma(t)=\gamma(a+b-t) \,\,\,\,\,\,\,\,\,\,\,\,\,\, t \in [a,b]$$

is called the "reverse" curve (or path) of $\gamma(t)$.

This definition is clear, but how is the "reverse" path defined in the following case?

Take two regular curves $\gamma_1:[a,b] \to \mathbb{R}^n$ and $\gamma_2:[c,d] \to \mathbb{R}^n$ with $\gamma_1(b)=\gamma_2(c)$ and define $\gamma:[a,d] \to \mathbb{R}^n$ as $$\gamma(t)=\begin{cases} \gamma_1(t) & t\in [a,b] \\ \gamma_2(t) &t \in [c,d] \end{cases}$$

Now, what is $-\gamma(t)$?

I think there are two possibilities:

1. $$-\gamma(t)=\begin{cases} \gamma_1(a+b-t) & t\in [a,b] \\ \gamma_2(a+b-t) &t \in [c,d] \end{cases}$$
2. $$-\gamma(t)=\begin{cases} \gamma_1(a+b-t) & t\in [a,b] \\ \gamma_2(c+d-t) &t \in [c,d] \end{cases}$$

Answer 1

You want to map $a$ to $\gamma_2 (d)$, $b$ and $c$ to $\gamma_2 (c) = \gamma_1 (b)$ and $d$ to $\gamma_1 (a)$. Looking for a linear map $\varphi(t) = At + B$ taking $a$ and $b$ to $d$ and respectively $c$ we find $\begin{cases} Aa + B = d \\ Ab + B = c \end{cases}$, whence $A = \frac {d-c} {a-b}$ and $B = \frac {ac-bd} {a-b}$, so that $\varphi (t) = \frac {d-c} {a-b}t + \frac {ac-bd} {a-b}$.

The reverted curve is then $-\gamma : [a,b] \cup [c,d] \to \Bbb R^n$ given by

$$-\gamma (t) = \begin{cases} \gamma_2 \big( \varphi (t) \big), & t \in [a,b] \\ \gamma_1 \big( \varphi^{-1} (t) \big), & t \in [c,d] \end{cases} .$$

Answer 2

A picture would really help here - the intuitive idea behind this is a bit like how, when looking at compositions of functions (or even group elements), $(fg)^{-1}=g^{-1}f^{-1}$.

Here, you're dealing with a concatenation of paths, which is an analog of composition (well, almost) in this setting. You're travelling along $\gamma_1$ first, and then along $\gamma_2$. With this in mind, to reverse your path, you should retrace your steps along $\gamma_2$, and then along $\gamma_1$.

Once such way of doing this would be:

$$-\gamma (t) = \begin{cases} \gamma_2 (a+d-t), & t \in [a,a+d-c] \\ \gamma_1 (a+d-t), & t \in [d-b+a,d] \end{cases} .$$