Half-twist (Dehn twist) on braids represented as elements of the mapping class group on an n-punctures disk Let $\mathbb{D}_n$ denote a disk with $n$ marked points or punctures. The mapping class group of $\mathbb{D}_n$ is isomorphic to the braid group $B_n$.
Elements of the mapping class group of $\mathbb{D}_n$ can be represented by curve diagrams, as below (taken from Ordering Braids by Dehornoy et al.):

Later in the chapter, they claim that “a half-twist on $e_i$ corresponds to $\sigma_i$“.
Consider the diagrams in the above image. Clearly, these all correspond to some braids on 3 strands (since there are 3 punctures). Further, the arc $e_3$ corresponds to $\sigma_3$ in $B_3$, but what is $\sigma_3$ in $B_3$? This is not one of the standard generating elements of $B_3$, since $B_3$ is generated by only $\sigma_1$ and $\sigma_2$. A similar question can be asked about the arc $e_0$.
Additionally, what would the half-twist on the arcs $e_0$ or $e_3$ look like? I am confused about these since they connect to the boundary.
 A: There's not too much going on here. The quoted statement

... a half-twist on $e_i$ corresponds to $\sigma_i$ ...

is only valid for a restricted range of values of the index $i$, namely
$$1 \le i \le n-1
$$
So in your diagram, that statement is valid only for $i=1,2$. I suspect that this restricted range is mentioned somewhere in Dehornoy's text, at least in the fine print (it would be a mistake to leave this out).
The reason for this restriction, as you have cottoned on to, is that the half-twists $e_0$ and $e_n$ are not defined in $\mathbb D_n$.

Now, let's say that you were interested and willing to move into a somewhat different realm, as follows.
Suppose you take the quotient space of the closed disc by collapsing its boundary circle to a single point. You get a space homeomorphic to the $2$-sphere, with $n+1$ special points on it: the point you get from collapsing the boundary circle; and the (images under the quotient map of the) original $n$ special points of $\mathbb D_n$.
Having taken this quotient, now the half-twists around $e_0$ and $e_n$ do indeed make sense, and they correspond to elements $\sigma_0$ and $\sigma_n$ which, together with the original $\sigma_1,...,\sigma_{n-1}$, generate a group called the $n+1$ strand braid group of the sphere. In fact braid groups on general surfaces are highly interesting objects of study.
