# Proving that the equivalence of paths is an equivalence relation.

The definition of equivalent paths is as follows :

Two paths $$f: [a,b] \rightarrow \mathbb{R^n}$$ and $$g: [c,d] \rightarrow \mathbb{R^n}$$ are equivalent if there exist a $$C^{1}$$ bijection $$\phi: [a,b]\rightarrow [c,d]$$ such that $$\phi'(t) > 0$$ for all $$t \in [a,b]$$ and $$f = g \circ \phi.$$

And I was proving that the equivalence of paths is an equivalence relation.

And for proving the symmetric condition I know that since $$\phi$$ is a bijection then it has an inverse call it $$\phi^{-1}$$, but I am not sure why it is also a $$C^{1}$$, could anyone clarify this for me please?

For proving transitivity I called the first $$C^{1}$$ bijection due to the equivalence of $$f$$ & $$g$$, $$\phi_{1}$$ and the second $$C^{1}$$ bijection due to the equivalence of $$g$$ & $$h$$, $$\phi_{2}$$, but I am not sure if their composition is also a $$C^{1}$$- bijection, could anyone explain this for me please?

## 1 Answer

In general, the inverse of a differentiable bijection $$\phi$$ doesn't have to be bijective; take$$\begin{array}{ccc}[0,1]&\longrightarrow&[0,1]\\x&\mapsto&x^2,\end{array}$$for instance. But it is true if $$\phi'$$ has no zeros, which is the case here.

Also, the composition of two bijections whose derivatives are always greater than $$0$$ is also a bijection whose derivativ is always greater than $$0$$; just apply the chain rule here.