$\phi$ is a homeomorphism from the 2-sphere to itself which represents an element of $PMCG(S^2,A)$ (we also denote it by $\phi$), where $A$ is a finite set of $S^2$. $\gamma$ is an essential Jordan curve (i.e. both components $F$, $G$ of $S^2-\gamma$ contain at least two points of $A$). Suppose $\phi$ fixes the homotopy class of $\gamma$. Then it seems that $\phi$ can be generated by those Dehn twists along curves which do not intersect $\gamma$ and the Dehn twists along $\gamma$. Is that true?
For the case $\# A = 4$, the question can be stated simply as follows: $PMCG(S^2,A)=\langle D\_\gamma,D\_\theta \rangle$; if $\phi \in PMCG$ fixes the homotopy class of $\gamma$, can we conclude $\phi \in \langle D\_\gamma \rangle$?
I am sure that this question is not too hard, but I really don't know how to solve it.



Homeomorphism of the punctured sphere which fixes an essential Jordan curve
To prove this you just need to put together a few simple known results. 


First, you need that the subgroup of $PMCG(S^2,A)$ that stabilizes the homotopy class of $\gamma$, a subgroup I shall denote $Stab(\gamma)$, can be written as a short exact sequence
$$1 \to \langle D(\gamma)\rangle \hookrightarrow Stab(\gamma) \to PMCG(F) \times PMCG(G)
$$
The projections onto $PMCG(F)$ and $PMCG(G)$ are just the restriction homomorphisms.


Second, you need that $PMCG(F)$, $PMCG(G)$ are each generated by Dehn twists.


