I'm reading the book A Primer of Mapping Class Group by Benson Farb and Dan Margalit. In order to show that the mapping class group of a surface $S$ is finitely presentable. They make use of an abstract simplicial complex called the arc-complex $\mathcal{A}(S)$. The vertices of this complex are the free isotopy classes of proper, essential, and simple arcs.
I'm trying to understand why Dehn twists about boundary components act trivially on $\mathcal{A}(S)$
The word proper means that for an arc $\alpha: [0,1] \rightarrow S$, we have that the $\{0,1\}$ map to either a marked point or a point on the boundary and the interior $(0,1)$ must not map to a marked point or to the boundary.
The word essential here means that $\alpha$ is not homotopic to a marked point. The homotopy can be unbased and must consist of proper arcs (otherwise any arc would be homotopic to a marked point).
I've been thinking about the arc complex of the annulus $A= S^{1} \times [0,1]$. I believe that the arc complex to the annulus should include vertices which correspond to curves such as the bold arcs in the image below. The bold arc on the left thought of as the arc that sits on $(0,t)$ and the one on the right can be thought of as image curve of the Dehn twist $T_{\gamma}$ about $\gamma$.
I'm wondering why a Dehn twist about say the inner boundary circle does not take vertex associated to curve on the left to the vertex associated to the curve on the right.
I'm sorry for how poorly worded this question is.....
The Dehn twist does take the curve on the left to the curve on the right, the curves are just (properly) isotopic! The reason is that you are allowed to slide the endpoints as long as they stay on the boundary. For example, even without a Dehn twist you could isotope the original curve to the other curve by dragging the endpoint in the inner circle counter clockwise.