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](http://www.math.uchicago.edu/~margalit/mcg/mcgv50.pdf)); 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



Subgroup of mapping class group generated by two Dehn twists
A regular neighborhood of these two curves is a 1-holed torus, so you're question is equivalent to asking why the mapping class group $M\_{1,1}$ of a 1-holed torus is isomorphic to the 3-strand braid group $B\_3$. The key observation is that you can construct a homomorphism $f : M\_{1,1} \rightarrow B\_3$ as follows. Let $i : \Sigma\_{1,1} \rightarrow \Sigma\_{1,1}$ be the hyperelliptic involution. This is not an element of $M\_{1,1}$ since it rotates the boundary component by $\pi$, but it lies in a degree 2 extension $M\_{1,1}'$ of $M\_{1,1}$. In fact, $i \in M\_{1,1}'$ lies in the center of $M\_{1,1}'$. It follows that $M\_{1,1} \cong M\_{1,1}' / \langle i \rangle$ acts (modulo homotopy) on the quotient $\Sigma\_{1,1}/i$, which is a disc with $3$ punctures. Since $B\_3$ is the mapping class group of a $3$-punctured disc, this give you a homomorphism $f : M\_{1,1} \rightarrow B\_3$. The dehn twists $T\_{\alpha}$ and $T\_{\beta}$ go to the standard generators of $B\_3$, so the proof is completed by observing that they satisfy the braid relation.


