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:

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

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

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

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

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

Similarly, $$h \left({\left[{\tfrac 1 2 \, . \, . \, 1}\right]}\right)$$ 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 $$h \left({1}\right) = g \left({1}\right) = c$$.