# Proving That Two Paths of Different Lengths Are Adjoined

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 $P_2$, $\overline{P_2P_3}$?

• Sorry, why did you write $\vec{f}\!\left(^{1}\!/_{2}\right) = \vec{f_1}\!\!\left(^{1}\!/_{2}\right)$? We have that $\vec{f}\!\left(^{1}\!/_{2}\right) = \vec{f_1}\!\!\left(1\right)$ by definition of $f$. – Mankind Mar 2 '15 at 20:15
• Ah, thanks for pointing that out! I'll edit my question to fix that. – RandomDSdevel Mar 2 '15 at 22:10
• @HowDoIMath: OK, it's been fixed! – RandomDSdevel Mar 2 '15 at 22:12
• @HowDoIMath: Is the reason you asked because you're working on an answer for me? – RandomDSdevel Mar 5 '15 at 18:10
• I don't think I understand what the issue is. For instance, I don't understand why you write $\vec{f}\!\left(^{1}\!/_{2}\right) = \vec{f_1}\!\!\left(1\right) \neq \vec{f_2}\!\!\left(0\right)\,$ Is this a typo? The way I see it, the paths $f_1$ and $f_2$ are being chosen/constructed, such that $f_1(1)=f_2(0)$, so that the second path starts where the first path left off. This is also the reason why the function $f$ is well-defined at the point $t=1/2$. Does this clear anything up? @RandomDSdevel – Mankind Mar 7 '15 at 15:34
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