Diffeomorphism between intervals implies derivative is either positive everywhere or negative everywhere

Question

Let $\varphi: [c,d] \rightarrow [a,b]$ be a diffeomorphism between two closed intervals in $\mathbb{R}$. I am following a proof that claims that since $\varphi$ is a diffeomorphism either $\varphi' > 0$ everywhere or $\varphi' < 0$ everywhere.

So far I have the following: assume there exists $x_0$ and $x_1$ such that $\varphi'(x_0) > 0$ and $\varphi'(x_1) < 0$. Then there must be some $x, y \in [a,b]$ such that $\varphi(x) = \varphi(y)$ which contradicts injectivity. But I'm having trouble making this argument solid. Any advice?

Answer 1

Darboux's theorem says that a derivative has the intermediate value property. More specifically, if $\varphi$ is differentiable on an interval, and on that interval we have two points $x_0, x_1$ such that $\varphi'(x_0)>0$ and $\varphi'(x_1)<0$, then there must be an $x_2$ between $x_0$ and $x_1$ such that $\varphi'(x_2) = 0$.

At this point (to be precise, at $\varphi(x_2)$) the inverse of $\varphi$, if it exists, cannot be differentiable. But that contradicts the definition of a diffeomorphism.

Answer 2

Here is an approach without using Darboux's theorem. We only need the following well-known facts:

1. Each homeomorphism $\varphi: [c,d] \rightarrow [a,b]$ maps boundary points to boundary points. That is, either (a) $\varphi(c) = a$ and $\varphi(d) = b$ or (b) $\varphi(c) = b$ and $\varphi(d) = a$.

2. In case $(a)$, $\varphi$ is strictly increasing; in case (b), $\varphi$ is strictly decreasing. This is a consequence of the ordinary IVT.

3. If $\varphi$ is differentiable, then $\varphi' \ge 0$ in case (a) and $\varphi' \le 0$ in case (b).

If $\varphi$ is a diffeomorphism, then it cannot happen that $\varphi'(t) = 0$ for some $t \in [c,d]$. This follows from the chain rule since $1 = (\varphi^{-1} \circ \varphi)'(t) = (\varphi^{-1})'(\varphi(t)) \cdot \varphi'(t)$.

We conclude that either $\varphi' > 0$ or $\varphi' < 0$.