# Joining Arcs makes Another Arc

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## Theorem

Let $T$ be a topological space.

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

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:

- $h \left({x}\right) = \begin{cases} f \left({2 x}\right) & : x \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\ g \left({2 x - 1}\right) & : x \in \left[{\dfrac 1 2 \,.\,.\, 1}\right] \end{cases}$

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

$\blacksquare$