Homeomorphisms of the 2-sphere $S^2$ fixing a set of points. I was reading the book by Benson Farb and Dan Margalit titled A Primer on Mapping Class Groups. In chapter 2, Proposition 2.3, it is given that 

The action given by $$\text{Mod}(S_{0,3}) \longrightarrow \Sigma_3 $$ is an isomorphism.

They begin with the fact that the given action is a surjective homomorphism. My question is how?
To be precise what I mean to say is that, since we know that the elements of Mod$(S_{0,3})$ fix the set of punctures, we can proceed along that line. But, given any non-trivial permutation how do I know that there is an orientation preserving homeomorphism.
Then, the question boils down to the following,

Consider a set of n points on the 2-sphere ($S^2$), can every permutation of these n points be extended to an orientation preserving homeomorphism?

Thanks in advance.
 A: This is true not just for $S^2$, but for any manifold of dimension at least $2$.  That is, 

For any $n\geq 2$, and any $n$-manifold $M$, any permuation of a finite subset $\{x_1, x_2,..., x_k\}\subseteq M$ extends to an orientation preserving diffeomorphism of $M$.

In fact, one can choose this diffeomorphism to be isotopic to the identity - that is what the below proof actually gives.  Further, we really do need $n\geq 2$: if $a<b<c\in \mathbb{R}$, there is a no homeomorphism of $\mathbb{R}$ which maps $a$ to $b$, $b$ to $c$ and $c$ to $a$.
I'll prove the above claim via a series of propositions.
Proposition 1: Suppose $M^n$ is a manifold with $n\geq 2$.  Choose distinct points $x,y, z_1,..., z_k\in M$.  Then there is an orientation preserving diffeomorphism $f:M\rightarrow M$ for which $f(x) = y$, but $f(z_i) = z_i$ for all $1\leq i\leq k$.
Proof:  Because $n\geq 2$, $M\setminus\{z_1,..., z_k\}$ is path connected.  Let $\gamma:[0,1]\rightarrow M$ be a simple regular curve with $\gamma(0) = x$ and $\gamma(1) = y$.  Let $U$ denote a tubular neighborhood of $\gamma$, chosen small enough so that $U\cap\{z_1,..., z_k\} = \emptyset$.  Now, create a vector field $X$ extending $\gamma'$ which is supported in $U$.  Flowing for an appropriate amount of time gives the desired $f$. $\square$
Now, let $X\subseteq M$ be any finite set.  Given a permutation $\sigma$ of $X$, we will say $f$ is an extension of $\sigma$ if $f:M\rightarrow M$ is an orientation preserving diffeomorphism and $f|_X = \sigma|_X$.
Proposition 2:  If $\sigma:X\rightarrow X$ is a transposition, there is an extension $g$ of $\sigma$.
Proof:  Suppose $\sigma(x_i) = x_j$ and $\sigma(x_j) = x_i$ for $x_i\neq x_j$, with all other $x_k\in X$ being fixed.  Applying the lemma once with $x = x_i, y = x_j$ and with $\{z_1,..., z_k\} = X\setminus \{x_i, x_j\}$, we get a diffeomorphism $f_1$ for which $f_1(x_i) = x_j$, but $f_1(x_k) = x_k$ for all other $k$.  However, there is no reason that $f_1(x_j) = x_i$.  If $f_1(x_j) \neq x_i$, then we use the lemma again with $x = f_1(x_j)$, $y = x_i$ and $\{z_1,..., z_{k+1}\} = X\setminus\{x_i\}$ to get a diffeomorphism $f_2$ with $f_2 f_1(x_j) = x_i$ and $f_2(x_k) = x_k$ for any other $k$.  Then $g = f_2\circ f_1$ is the desired diffeomorphism of $M$.  $\square$
Proposition 3:  Suppose $\sigma_1, \sigma_2$ are both permutations of $X$ with extensions $f_1$ and $f_2$. Then $\sigma_1 \sigma_2$ is extended by $f_1\circ f_2$.  Further, $\sigma_1^{-1}$ is extended by $f_1^{-1}$.
Proof:  Let $x_i\in X$.  For the first statement, $f_2(x_i) = \sigma_2(x_i)\in X$ and so $f_1 (f_2(x_i)) = \sigma_1(f_2(x_i)) = \sigma_1(\sigma_2(x_i))$.
For the second statement, note that $x_i = \sigma_1(\sigma_1^{-1}(x_i)) = f_1(\sigma_1^{-1}(x_i)$, and also that $x_i = f_1(f_1^{-1}(x_i))$.  Since $f_1$ is injecive, $\sigma_1^{-1}(x_i) = f_1^{-1}(x_i)$.$\square$
Proposition 3 essentially claims the following:  The set $\{\sigma \in S_k: \exists \text{ extension }f\}$ is a subgroup of the symmetric group on $k$ letters, $S_k$.  Proposition 2 then claims that this set contains all transpositions.  Since the symmetric group is generated by transpositions, the claim above now follows.
A: Surjectivity follows from a theorem in complex analysis: Let a,b,c,x,y,z be any six points on the Reimann sphere.  Then, there exists a Möbius transformation taking a to x, b to y and c to z. 
The image of these points determine a Möbius transformation completely.  So, the answer to the more general question is No.  If a map takes 0 to 0, 1 to 1 and infinity to infinity, then the map has to be the identity map.  
