# Joining Paths makes Another Path

## Theorem

Let $T$ be a topological space.

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

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

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

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

Then $h$ is a path in $T$.

## Proof

First we see that $h$ is well-defined, because on $\closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set{\dfrac 1 2}$ we have $\map f 1 = b = \map g 0$.

Now $\mathbin {h} {\restriction_{\closedint 0 {\frac 1 2}}} \mathop = f \circ k$ where $k: \closedint 0 {\dfrac 1 2} \to \closedint 0 1$ is given by $\map k x = 2x$.

So by Continuity of Composite Mapping, $\mathbin {h} {\restriction_{\closedint 0 {\frac 1 2}}}$ is continuous.

Similarly, $\mathbin {h} {\restriction_{\closedint {\frac 1 2} 1}}$ is continuous.

By Continuity from Union of Restrictions, it follows that $h$ is continuous.

Finally, $\map h 0 = \map f 0 = a$ and $\map h 1 = \map g 1 = c$.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 4$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $6.3$: Path-connectedness: Lemma $6.3.3$