Let $S$ be a surface (possibly with boundary, and punctures), and let $\alpha,\beta$ be two simple closed curves on $S$ which intersect once. If $a,b$ denote the isotopy classes of $\alpha,\beta$, respectively, then why is the subgroup of $\text{Mod}(S)$ generated by $T_a,T_b$ isomorphic to the braid group $B_3$? [Here, $T_a$ is the Dehn twist around $\alpha$, and $T_b$ the Dehn twist around $\beta$.] I understand why the relation $T_aT_bT_a=T_bT_aT_b$ holds, but why is this the only relation?

If it makes any difference, I am reading the "Primer on Mapping Class Groups" by Farb and Margalit (available here); they claim this is true, but give no proof. The relevant section in that PDF is 3.5, specifically pages 91--94 (in the PDF).

Thanks for any help, Steve