I am given two equivalent $C^1$ paths $\alpha : [a,b] \to \mathbb R^n $ and $\beta : [c,d] \to \mathbb R^n$ such that they have different orienation. That means that there exists $\varphi : [a,b] \to [c,d]$ which is $C^1$ diffeomorphism, such that $\alpha = \beta \circ \varphi$ and $\varphi'(t)<0, t \in [a,b]$
How do i show that $\alpha(a) = \beta(d)$ and $\alpha(b) = \beta(c)$. Thought about using Bolzano-Weierstrass but i am not sure how.
I sort of get that $\alpha(a) \in \{\beta(c), \beta(d)\}$ since $\varphi$ is homeomorphism, and so it maps boundary to boundary, but i cant show that it is equal to $\beta(d)$
Since $ \varphi'(t) < 0 $ for all $t \in [a, b]$ this means that $\varphi(t)$ is monotonically decreasing.
Since $\varphi: [a, b] \to [c, d]$ this can only mean $\varphi(a) = d$ and $\varphi(b) = c$ because $c < d$.
Therefore $$\alpha(a) = (\beta \circ \varphi)(a) = \beta(\varphi(a)) = \beta(d)$$ and $$\alpha(b) = (\beta \circ \varphi)(b) = \beta(\varphi(b)) = \beta(c).$$