Joining Paths makes Another Path

Theorem

Let $T$ be a topological space.

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

Let $f, g: I \to \R$ be paths 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 $h$ is a path in $T$.

Proof

First we see that $h$ is well-defined, because on $\left[{0 . . \dfrac 1 2}\right] \cap \left[{\dfrac 1 2 . . 1}\right] = \left\{{\dfrac 1 2}\right\}$ we have $f \left({1}\right) = b = g \left({0}\right)$.

Now $h \left({\left[{0 . . \dfrac 1 2}\right]}\right) = f \circ k$ where $k: \left[{0 . . \dfrac 1 2}\right] \to \left[{0 . . 1}\right]$ is given by $k \left({x}\right) = 2x$.

So by Continuity of Composite Mapping, $h \left({\left[{0 . . \dfrac 1 2}\right]}\right)$ is continuous.

Similarly, $h \left({\left[{\dfrac 1 2 . . 1}\right]}\right)$ is continuous.

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

Finally, $h \left({0}\right) = f \left({0}\right) = a$ and $h \left({1}\right) = g \left({1}\right) = c$.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 4$