Joining Arcs makes Another Arc

Theorem
Let $T$ be a topological space.

Let $\mathbb I \subseteq \R$ be the closed unit interval $\closedint 0 1$.

Let $a, b, c$ be three distinct points of $T$.

Let $f, g: \mathbb I \to T$ be arcs in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.

Let $h: \mathbb I \to T$ be the mapping given by:


 * $\map h x = \begin{cases}

\map f {2 x} & : x \in \closedint 0 {\dfrac 1 2} \\ \map g {2 x - 1} & : x \in \closedint {\dfrac 1 2} 1 \end{cases}$

Then either:
 * $h$ is an arc in $T$

or
 * There exists some restriction of $h$ which, possibly after reparametrisation, is an arc in $T$.

Proof
From Arc in Topological Space is Path, $f$ and $g$ are also paths in $T$.

So by Joining Paths makes Another Path it follows that $h$ is a path in $T$.

Now if $\Img f \cap \Img g = b$ it can be seen that:
 * $\forall x \in \Img h : x = \begin{cases}

\map f y & \text {for some $y \in \closedint 0 {\dfrac 1 2}$}, \text{ or} \\

\map g z & \text {for some $z \in \closedint {\dfrac 1 2} 1$} \end{cases}$ and it follows that $h$ is an injection, and therefore an arc.

On the other hand, suppose:
 * $\exists y \in \closedint 0 {\dfrac 1 2}: \exists z \in \closedint {\dfrac 1 2} 1: \map f y = \map g z$

such that $\map f y \ne b$.