Proof of equivalence well defined function

Question

This is the definition:

Definition 1:

A function $f:D\subseteq R\to R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.

Definition 2:

A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] \to R^n$, the points f(a) and f(b) are then called the end points of the arc$

Now comes the theorem:

Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]\subseteq R\to R^n $ Then any smooth parametrisation $h:[c,d]\subseteq R\to R^n$ of C is 1-1 and is equivalent to f.

Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:

It starts as follows:

Proof

The function $\Phi=f^{-1}\circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=f\circ\Phi$.

Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $\Phi$ is well defined.

Question

1) Why is $\Phi$ well defined?

2) Why does it satisfy $h=f\circ\Phi$

Answer 1

1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $f\bigl([a,b]\bigr)\bigl(=h\bigl([a,b]\bigr)\bigr)$ into $\mathbb R$ and because $h\bigl([a,b]\bigr)\subset D_{f^{-1}}$.
2. Because $\Phi= f^{-1}\circ h\implies f\circ\Phi=f\circ(f^{-1}\circ h)=h$.