End points of different parametrizations of the same curve

Question

I was reading about parametrized curves. The definition is in terms of a parametrized curve is in terms of the parametrization itself rather than the curve as a subset.

Precisely, A parametrized curve is a function $z(t)$ which maps a closed interval $[a, b] \subset \mathbb{R}$ to the complex plane. Let us assume that the function is bijective, differentiable with derivative not zero anywhere. Now if we have a different parametrization say $s$ with the same image $z([a, b])$ then what can we say about the end points of $s$? Can we say that $s(a)$ one of $z(a)$ or $z(b)$? It seems to me intuitively that this has to be the case but I am unable to prove. Assume that $s$ is also a nice parametrization from $[a, b]$.

I am interested in this question because this will have implications about integration of functions along different parametrizations of the 'same' curve (as a set).

Please help.

Answer 1

Let $z$ and $s$ be two continuous and bijective maps from $[a, b]$ to the same set $A \subset \Bbb C$ and consider the function $f = s^{-1} \circ z$:

• $f$ is a bijective function from the interval $[a, b]$ to itself.
• $f$ is continuous (see for example Proving the inverse of a continuous function is also continuous).

As a consequence, $f$ is strictly monotone (see for example Continuous injective map is strictly monotonic).

It follows that indeed $z$ and $s$ have the same endpoints (as you suspected). Either $z(a) = s(a)$ and $z(b)=s(b)$ (if $f$ is increasing), or $z(a) = s(b)$ and $z(b)=s(a)$ (if $f$ is decreasing).