Joining Arcs makes Another Arc

Theorem
Let $T$ be a topological space.

Let $I \subseteq \R$ be the unit interval $\left[{0 \,.\,.\, 1}\right]$.

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

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


 * $h \left({x}\right) = \begin{cases}

f \left({2x}\right) & : x \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\ g \left({2x - 1}\right) & : x \in \left[{\dfrac 1 2 \,.\,.\, 1}\right] \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 $\operatorname{Im} \left({f}\right) \cap \operatorname{Im} \left({g}\right) = b$ it can be seen that:
 * $\displaystyle \forall x \in \operatorname{Im} \left({h}\right): x = \begin{cases}

f \left({y}\right) & \text{for some } y \in \left[{0 \,.\,.\, \dfrac 1 2}\right], \text{ or} \\

g \left({z}\right) & \text{for some } z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right] \\ \end{cases}$ and it follows that $h$ is an injection, and therefore an arc.

On the other hand, suppose:
 * $\exists y \in \left[{0 .. \dfrac 1 2}\right]: \exists z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]: f \left({y}\right) = g \left({z}\right)$

such that $f \left({y}\right) \ne b$.