Do Diffeomorphism between connected compact subsets of the real line constitute a convex set?

Question

Could anyone please confirm or refute the following proposition:

Let $I_0 = [a_0, b_0], I_1=[a_1, b_1], I_2=[a_2, b_2]$ be connected and compact subsets of $\mathbb{R}$, and the maps $\xi_1:I_0\rightarrow I_1$ and $\xi_2:I_0\rightarrow I_2$ be diffeomorphism. Then the following map $$ \xi_3(t) = \lambda\xi_1(t) + (1-\lambda)\xi_2(t) $$ where $\lambda\in[0, 1]$, is also a diffeomorphism between $I_0$ and $I_3=[a_1+a_2, b_1+b_2]$. It follows that diffeomorphisms from $I_0$ onto compact intervals $[a, b]$ constitute a convex set.

Proof:

1. $\xi:I_0 \rightarrow I_3 $ is a homeomorphism because it is $C^0$ and monotonically increasing.

   In fact, the sum of two monotonically increasing functions is monotonically increasing.

2. $\xi:I_0 \rightarrow I_3 $ is a diffeomorphism because it is $C^1$ and strictly monotonically increasing.

   In fact, the sum of two monotonically increasing functions is monotonically increasing.

Answer 1

A very concrete family of diffeomorphisms constitute a convex cone.

Let $I = [0, T], I_1=[0, T_1], I_2=[0, T_2]$ be connected and compact subsets of $\mathbb{R}$ with $T, T_1, T_2 >0$, and the maps $\xi_1:I\rightarrow I_1$ and $\xi_2:I\rightarrow I_2$ be diffeomorphism. Then the following map $$ \xi_3(t) = \lambda\xi_1(t) + (1-\lambda)\xi_2(t) $$ where $\lambda\in[0, 1]$, is also a diffeomorphism between $I$ and $I_3=[0, T_1+T_2]$. It follows that the set of diffeomorphisms that take the interval $I$ onto another interval which lower limit is zero is a convex cone.