Orbits of mapping class group on four-punctured sphere

Question

Let $\mathcal{M}_{0,4}$ be the mapping class group of the four-punctured sphere $S_{0,4}$. Denote the simple closed curve around boundary components $b_1$ and $b_2$ by $x$, the one around $b_2$ and $b_3$ by $y$ and the one around $b_3$ and $b_1$ by $z$. I'm interested in the orbit of $x$ under the action of the mapping class group $\mathcal{M}_{0,4}$, in particular, in the question whether $x$ can be taken to $y$.

I understand that if two curves $\alpha$ and $\beta$ have geometric intersection number 1 they are in the same orbit (and similarly for chains of curves). This is a direct consequence of the braid relation between the Dehn twists $T_\alpha$ and $T_\beta$. Since the intersection number for any two of the curves $x$, $y$ and $z$ is 2 there is no pairwise relation at all between the Dehn twists $T_x$, $T_y$ and $T_z$. Yet, the beautiful lantern relation connects all three of them with Dehn twists along the boundary components (and I think this relation essentially determines $\mathcal{M}_{0,4}$).

Can the lantern relation tell me whether $x$ and $y$ lie in the same orbit? If so, how would I find a sequence of Dehn twists mapping one to the other?

Answer 1

If you require each boundary component to be fixed pointwise, the homeomorphism does not exist. Since $x$ separates the boundaries $\{b_1$, $b_2\}$ to the boundaries $\{b_3$, $b_4\}$, the images of $x$ under homeomorphisms that preserving the boundaries still hold this property. But $y$ does not have such a property.

Answer 2

For the existence of this product you can use classification of the surfaces: if you cut $S_{0,4}$ along $x$, you obtain a pair of pants $P_{x}$, similarly if you cut $S_{0,4}$ along $y$ then you obtain another pair of pants $P_{y}$. There exist a homeomorphism which maps $P_{x}$ to $P_{y}$ and $x$ to $y$. So isotopy class of this homeomorphism can be written product of Dehn twists.