In the section on 'Adjoining Paths' of its 'Topology' book's page on 'Path Connectedness,' WikiBooks shows that, for any topological space $X$ with members $a$, $b$, and $c$, the following…:
$$\forall a, b, c \in X: \left( \exists \vec{f_1}\colon \left[0, 1\right] \to X\right) \wedge \left(\exists \vec{f_2}\colon \left[0, 1\right] \to X\right) \Rightarrow \exists \vec{f}\!\left(t\right) = \begin{cases}\vec{f_1}\!\!\left(2t\right)\!, & t \in \left[0, ^{1}\!\!/_{2}\right] \\ \vec{f_2}\!\left(2t-1\right)\!, & t \in \left[^{1}\!/_{2}, 1\right]\end{cases}$$
…holds true. It appears, however, that this only works when $\vec{f_1}\!\!\left(\right)$ and $\vec{f_2}\!\left(\right)$ describe paths of equal length. In particular, $\vec{f}\!\left(^{1}\!/_{2}\right) = \vec{f_1}\!\!\left(1\right) \neq \vec{f_2}\!\!\left(0\right)\,$ and $\,\vec{f}\!\left(1\right) \neq \vec{f_2}\!\!\left(1\right)\,$ when, for example, the path described by $\vec{f_1}$ is the path which results from visiting each of the points on the Euclidean line segment $\overline{P_1P_2}$ in turn, the path described by $\vec{f_2}$ is the path which results from visiting each of the points on the Euclidean line segment $\overline{P_2P_3}$ in turn, and $\overline{P_1P_2} = 2\overline{P_2P_3}$. Since joining these two line segments' path functions as described above does not work since they are not congruent, how would I go about writing a piecewise path function to describe the path formed by tracing out first $\overline{P_1P_2}$ and then, continuing on from $P_2$, $\overline{P_2P_3}$?