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Reparametrization#

Definition: Reparametrization

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^m \to \mathbb{R}^n\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^m \to \mathbb{R}^n\) be real vector functions.

A reparametrization between \(f\) and \(g\) consists of a bijective function \(h_{\mathcal{D}_f \to \mathcal{D}_g}: \mathcal{D}_f \to \mathcal{D}_g\) and its inverse \(h_{\mathcal{D}_g \to \mathcal{D}_f}: \mathcal{D}_g \to \mathcal{D}_f\) with the following properties:

\[\begin{aligned}f & = g \circ h_{\mathcal{D}_f \to \mathcal{D}_g} \\ g & = f \circ h_{\mathcal{D}_g \to \mathcal{D}_f}\end{aligned}\]

The above is the most general definition for reparametrization. However, it is quite common to require that both \(h_{\mathcal{D}_f \to \mathcal{D}_g}: \mathcal{D}_f \to \mathcal{D}_g\) and \(h_{\mathcal{D}_g \to \mathcal{D}_f}: \mathcal{D}_g \to \mathcal{D}_f\) have additional properties such as continuity, total differentiability, etc. In those cases, when we say that a reparametrization has some property, we mean that both \(h_{\mathcal{D}_f \to \mathcal{D}_g}: \mathcal{D}_f \to \mathcal{D}_g\) and \(h_{\mathcal{D}_g \to \mathcal{D}_f}: \mathcal{D}_g \to \mathcal{D}_f\) have this property.

Equivalence#

In general, it is not always possible to find a reparametrization between two arbitrary real vector functions.

Definition: Equivalence of Real Vector Functions

Two real vector functions are equivalent if there exists a reparametrization between them.

Again, this is the most general definition of equivalence for real vector functions and the term "equivalent" is used very broadly here. On its own, equivalence does not guarantee many useful properties because the reparametrization is allowed to be arbitrary. When we want to emphasize the existence of a reparametrization with specific properties such as continuity, total differentiability, etc., we say that the functions are equivalent up to a "PROPERTY" reparametrization.

Theorem: Equivalence \(\implies\) Equal Images

If two real vector functions are equivalent, then they have the same image.

Proof

TODO