Mapping Class Group of $S^2$ I try to give a detialed verification why the mapping class group of $2-$sphere is trivial $$\mbox{MCG}(\mathbb{S}^2) =1.$$
I already know that $$\mbox{MCG}(\mathbb{D}^2) = 1 \ \mbox{and} \ \mbox{MCG}(\mathbb{S}^2\setminus\{p\}) = 1.$$
There are different sources that very briefly guide the proof, but mostly depends on the fact that $\mbox{MCG}(\mathbb{S}^2\setminus\{p\}) = 1$
Approach (1) (Wikipedia) Claim that any homeomorphism of $\mathbb{S}^2$ is either isotopic to identity or (restriction of) symmetry around $z=0$ in $\mathbb{R}^3$. But the latter class is not orientation-preserving. So if the claimed fact holds, the result is immediate.
(https://en.wikipedia.org/wiki/Mapping_class_group_of_a_surface#The_mapping_class_groups_of_the_sphere_and_the_torus)
But how to verify the claim in this case ? It does nor sound like a trivial fact
Approach (2) Claim that the key is to isotope the image of circle to itself. (clearly not a trivial fact. Even quite involving facts to prove.) The tools (claimed) to be used is Jordan curve theorem and Alexander trick (like one use in $\mathbb{D}^2$ case).
I more understand how appraoch (1) works, but not sure what it takes to show the claim. 
Can anyone help pointing out how to verify it ? Or even other more hands on appraoch to see $\mathbb{MCG}(\mathbb{S}^2) = 1$
 A: Let us take it for granted that $MCG(R^2)=\{1\}$, i.e. every orientation-preserving homeomorphism $R^2\to R^2$ is isotopic to the identity. 
Lemma. For any two points $p_1, p_2\in S^2$ there is a homeomorphism $\phi$ isotopic to the identity such that $f(p_1)=p_2$. 
Proof. Pick a great circle $C\subset S^2$ through the points $p_1, p_2$. The stabilizer of $C$ in $SO(3)$ contains a subgroup $G\cong SO(2)$ acting transitively on $C$. Since $SO(2)$ is connected, you get a rotation  $\phi\in  G$ sending $p_1$ to $p_2$ and isotopic to the identity through rotations. (I will leave you as an exercise to write down the detail which are just linear algebra.) qed 
Theorem. $MCG(S^2)=1$. 
Proof. For $f\in Homeo^+(S^2)$, we pick a point $p\in S^2$ (the north pole) and find $\phi$ as in the lemma such that $f_1=\phi\circ f$ satisfies $f_1(p)=p$. 
Since $S^2-\{p\}\cong R^2$, we can regard $S^2$ as the 1-point compactification of $R^2$ so that $\infty$ corresponds to $p$. 
Let $f_2$ denote the restriction of $f_1$ to $R^2$. Of course, $f_2$ is still orientation-preserving. By our assumption, $f_2$ is isotopic to $id$, let $F: [0,1]\times R^2\to R^2$ be this isotopy, where $F(0,z)=f_2(z)$, $F(1,z)=z$ for all $z\in R^2$. 
For each $t$, the map $F_t(z)=F(t,z)$ is a self-homeomorphism of $R^2$. By applying the invariance of domain theorem to the injective continuous map
$$
(t,z)\mapsto (t, F(t,z)), [0,1]\times R^2\to [0,1] \times R^2,
$$
we see that the family of inverses
$$
[0, 1]\times R^2\to R^2, (t,w)\mapsto F_t^{-1}(w)
$$
is also continuous. 
Extend $F$ to a map 
$$
G : [0,1]\times R^2 \cup \{\infty\} \to R^2 \cup \{\infty\}
$$
by $G(t,\infty)=\infty$. We need to show that $G$ is continuous. 
(Most likely, the authors of the book overlooked this issue.) 
Suppose, that $G$ is not continuous. Since $F$ is continuous, discontinuity of $G$ means that there is a sequence $t_n\to t\in [0,1]$ and a sequence $z_n\to \infty$ such that $F(t_n,z_n)$ converges to some $w\in R^2$. (You have to pass to subsequences few times to achieve this.) But applying continuity of the family of inverse maps (see above), we then obtain that $z_n$ converges to $F_t^{-1}(w)$, which is a contradiction. 
Thus, $G$ is continuous and yields an isotopy from $f_1$ to $id$. Since isotopy is an equivalence relation, it follows that $f$ is isotopic to $id$ as well. qed
