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?
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.