Homeomorphisms of the 3D sphere Given 2 points $x,y\in\,S^2$, how could we define a homeomorphism $f$ of $S^2$ such that $f(x)=y$?
The idea is to show that rotations of the sphere are essentially homeomorphisms, but are there any canonical ways to go about this problem?
 A: The points $x$, $y$ and the center of the sphere (viewing $S^2$ as the subspace $\{x\in\mathbb{R}^3| |x|=1\}\subset\mathbb{R}^2$) lie in a unique plane as long as $x\neq y$. Rotate the sphere, about the normal to this plane through the center of the sphere, until x reaches y. This is your canonical homeomorphism (in fact this is more than a homeomorphism, it is an isometry).
A: If you regard $S^2$ as a Riemannian manifold, then there is a canonical rotation mapping $x$ to $y$, namely the one that transports $x$ to $y$ along a geodesic.
A: I think I can give an answer "canonical" enough for any connected, at least differentiable, $n$-manifold with $n>1$. Let $M$ be this manifold
If $X$ is any $\mathcal{C}^1$ vector field on $M$, let's note $\varphi^X_t$  its flow at time $t$. 
1) Suppose $x,y \in M$ are contained in a open set $U$ such that $\psi : U \longrightarrow \mathbb{D}^n$ is a local chart ($\mathbb{D}^n$ is the open unit ball of $\mathbb{R}^n$). We can now construct a differentiable vector field Y on $\mathbb{D}^n$ that equals $\psi(y) - \psi(x)$ on the segment from $x$ to $y$ and $0$ outside a compact of $\mathbb{D}^n$ (ask me if you don't know the method to do such a construction).
Let's confuse $Y$ with $\psi^*Y$ on $U$. Define $\tilde{Y} = Y$ on $U$ an $0$ outside.
$\varphi_1^Y(x) = y$ and  $\varphi_1^Y \equiv Id$ on $M - U$ 
2) Now took $x$ and $y$ in $M$, since $M$ is a connected manifold, it is path-connected. Take $\gamma : [a,b] \longrightarrow M$ a path from $x$ to $y$ and take $a=a_0 < a_1 < ... < a_m = b$ such that $\gamma([a_i,a_{i+1}])$ lies in a small ball $U_i$ such thtat $\psi_i : U_i \longrightarrow \mathbb{D}^n$ is a local chart. 
Note that we can choose the $U_i$'s small enough so $\bigcup_{i}{U_i} \subset V$ for any neighborhood of $\gamma([a,b])$. 
We can apply 1) to $\gamma(a_i)$ and $\gamma(a_{i+1})$ to get $\varphi_i$ a diffeomorphism taking  $\gamma(a_i)$ to $\gamma(a_{i+1})$ which is the identity on $M- U_i$. Now $ \varphi = \varphi_{m-1} \circ ... \circ \varphi_0$ take $x$ to $y$ leaving $M - V$  pointwise invariant, where $V$ is a neighborhood of $\gamma([a,b])$.
3) Now it is not really difficult to conclude. Take $x_1,...,x_l \in M$ and $y_1,.., y_l \in M$ two collections of disjoints points. Then you can find paths $\gamma_i$ taking $x_i$ to $y_i$ that are pairwise disjoint(that's where we use $dim(M) > 1$). 
Then using a theorem of topology(I guess it's Urysohn or something like that), since the images of the $\gamma_i$ are compact, we can find pairwise disjoint neigborhoods of the $\gamma_i([x_i,y_i])$. Now you can apply 2) to each couple $(x_i,y_i)$, compose the $l$ diffeomorphism you have obtained and the result is proved. 
