0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.