Property of diffeomorphism between closed intervals of $\mathbb R$?

Question

Let $f:[a, b]\longrightarrow [c, d]$ be a diffeomorphism between closed intervals $[a, b]$ and $[c, d]$ of $\mathbb R$ such that $f(a)=c$. How to show $f(b)=d$? Can we weaken the assumptions on $f$?

Thanks.

Answer 1

From the definition of diffeomorphism, combined with the formula $f'(t) = \frac{1}{(f^{-1})'(f(t))}$, it follows that $f$ is a bijection and that $f(t) \ne 0$ for all $t \in [a,b]$.

Consider $t \in (a,b)$. If $f(t) > 0$ then by applying the definition of the derivative we may find $\delta>0$ such that $(t-\delta,t+\delta) \subset (a,b)$, such that if $s \in (t-\delta,t)$ then $f(s)<f(t)$, and such that if $s \in (t,t+\delta)$ then $f(s)>f(t)$; it follows that $f(t) \in (c,d)$. If $f(t)<0$ then a similar argument shows also that $f(t) \in (c,d)$.

Thus, the only values of $t \in [a,b]$ whose image $f(t)$ is one of the endpoints $c,d$ are the endpoint values $t \in \{a,b\}$. Since $f(a)=c$ and $f$ is one-to-one, the only possibility is $f(b)=d$.

Answer 2

Your function $f$ is injective and continuous, hence strictly monotonic (see Continuous injective map is strictly monotonic). Hence, as $f(a)=c$, $f$ is strictly increasing, and it follows that $f(b)=d$. We have only used the continuity of $f$ (and the fact that it is a bijection from $[a,b]$ to $[c,d]$, of course).

Answer 3

If an endpoint is mapped to a non-endpoint, we get a contradiction: removing the endpoint, we still have a connected space, while removing its image gives a two-component space. (Removing one point from domain and its image from image set, we still have a homeomorphism.)