Definition of the sum of paths $\gamma_1:[a_1,b_1]\rightarrow \Omega$ and $\gamma_2:[a_2,b_2]\rightarrow \Omega$ Throught our lecture notes and textbox in complex analysis the sum of paths is given by

Given paths $\gamma_1:[a_1,b_1]\rightarrow \Omega$ and $\gamma_2:[a_2,b_2]\rightarrow \Omega$ such that $\gamma_1(b_1)=\gamma_2(a_2)$ we defined the path $\gamma_1+\gamma_2:[a_1,b_1,+b_2-a_2]\rightarrow \Omega$ by
$$
(\gamma_1+\gamma_2)(t)=
\begin{cases}
 \gamma_1(t)&\text{if}\, t\in[a_1,b_1]\\
 \gamma_2(t-b_1+a_2)&\text{if}\, t\in [b_1,b_1+b_2-a_2] \\
\end{cases}
$$

What is the reason for this? Why not just define the sums of path $\gamma_1+\gamma_2:[a_1,b_2]\rightarrow \Omega$ by
$$
(\gamma_1+\gamma_2)(t)=
\begin{cases}
 \gamma_1(t)&\text{if}\, t\in[a_1,b_1]\\
 \gamma_2(t)&\text{if}\, t\in [a_2,b_2] \\
\end{cases}
$$
 A: The suggested solution has two issues:

*

*If $[a_1,b_1]$ and $[a_2,b_2]$ are disjoint, then the resulting function has a disconnected domain.  This would not be a path since the domain for a path must be connected (an interval).


*If $[a_1,b_1]$ and $[a_2,b_2]$ overlap, then the definition is not well-defined because it is unclear if $(\gamma_1+\gamma_2)(t)$ would be $\gamma_1(t)$ or $\gamma_2(t)$.
While the original formula seems complicated, it is doing a very simple geometric operation.  It is moving the interval $[a_2,b_2]$ so that it starts at $b_1$ and then gluing them together.  Let's see why:

*

*The length of $[a_1,b_1]$ is $b_1-a_1$, and, similarly, the length of $[a_2,b_2]$ is $b_2-a_2$.  On the other hand, the length of the given interval for the sum is $(b_1+b_2-a_2)-a_1=(b_1-a_1)+(b_2-a_2)$.  In other words, the length of the given interval is the sum of the original intervals.


*What the piecewise formula is doing is traveling along $\gamma_1$ for $[a_1,b_1]$ and then traveling along $\gamma_2$ for the rest of the time.  What looks messy here is that the interval $[a_2,b_2]$ has been shifted so that it comes right after $b_1$, by adding $b_1-a_2$ to the interval to get
$[a_2+(b_1-a_2),b_2+(b_1-a_2)]=[b_1,b_1+b_2-a_2]$.
A: What is actually more common, is to take "paths" into $\Omega$ as continuous functions $f: [0,1] \to \Omega$, so that all paths are starting from $f(0)$ and "travelling" to $f(1)$ in a fixed amount of "time" $1$. Then the "product" (or composition) of two paths $f,g: [0,1] \to \Omega$ is defined differently:
$$f\ast g(t) = \begin{cases}
                    f(2t) & 0 \le t \le \frac12\\
                    g(2t-1) & \frac12 \le t \le 1\\
\end{cases}$$
(it's usually not called "sum" because that suggests commutativity, but here order matters).
$f \ast g: [0,1]\to \Omega$ is then the same geometrically as your definition: follow the path $f$ first (In twice the speed, to fit into $[0,1]$) and then follow the path $g$ (ditto), which must start at the end point of $f$ for continuity reasons.
In some cases it can be handy to allow for arbitrary closed intervals $[a,b]$ as the domain, but it doesn't matter really because we can always  reparametrise the same function to have domain $[0,1]$ instead. And for purposes like defining a homotopy group of equivalence classes (modulo homotopy) for loops (paths with end = begin), it's handy to have such a common domain, see my note on the definition of this so-called first homotopy group, e.g. In your case we need some other reparametrisations but ones that preserve the "length" of the domain, which is maybe more with a view towards integration over such paths.
But geometrically it's all the same. First one path then the other.
