Find a homeomorphism of the sphere with itself Question: Let $x$ and $y$ be points on the sphere. Find a homeomorphism of the sphere with itself which takes $x$ to $y$. 
Solution:
The special orthogonal group $SO(n)$ are rotations of $n$ space given by square $n × n$ matrices $M$ with $det(M) = 1$ and $MM^T = I$. 
These are length (and angle) preserving transformations. An orthogonal matrix therefore takes the unit sphere to itself. 
The columns are orthogonal unit vectors. Let $x$ be any point on the unit sphere. Extend $\{x\}$ to an orthonormal basis and make a matrix $R$ from these vectors with $x$ as the first column. 
Then this matrix takes the first standard basis vector $e_1$ to $x$. 
Likewise there is an orthogonal matrix $S$ that takes $e_1$ to $y$. Then $SR^{−1}$ is a transformation that takes $x$ to $y$. 
Linear transformations are continuous. Things in $SO(n)$ are invertible, so give homeomorphisms.

Can anyone if its possible help me break down this solution (or parts of it) in pieces that are easier to understand with a little less advanced group theory, linear algebra etc or more intuitive arguments? I am working hard on rehearsing my group theory and linear algebra but i am still far from understanding a solution like this. Thanks.

edit: This is what is left for me to understand. The rest i think i got know. 
My linear algebra is not good at all. 
Extend $\{x\}$ to an orthonormal basis and make a matrix $R$ from these vectors with $x$ as the first column. 
Then this matrix takes the first standard basis vector $e_1$ to $x$. 
Likewise there is an orthogonal matrix $S$ that takes $e_1$ to $y$. Then $SR^{−1}$ is a transformation that takes $x$ to $y$. 
 A: Find any rotation of $n$-dimensional space that maps $x$ to $y.$ The whole group theoretic machinery is just to ensure the existence of such a rotation. You can get around it with an explicit construction.
Consider the 2-dimensional subspace $P$ (of $n$-dimensional space) that is spanned by $x$ and $y.$ In this 2-dimensional subspace $x$ can be mapped to $y$ by a rotation $\rho$ in the plane (over an angle $\theta$ with $\cos\theta=x.y$).
We extend this planar rotation to an $n$-dimensional rotation $R$ as follows. For any $n$-vector $x$ consider its orthogonal projection $\pi x$ on $P.$ Define $Rx=\rho(\pi x)+(x-\pi x),$ in other words: apply $\rho$ to the component of $x$ parallel to $P,$ and leave the component of $x$ perpendicular to $P$ alone.
It is not difficult to see that $R$ maps the unit sphere to itself. Since $\rho(\pi x)$ is still inside $P,$ it is perpendicular to $x-\pi x$ and therefore Pythagoras applies:
$$\eqalign{
\|Rx\|^2
&=\|\rho(\pi x)+x-\pi x\|^2 \\
&=\|\rho(\pi x)\|^2+\|x-\pi x\|^2\hbox{ (Pythagoras)} \\
&=\|\pi x\|^2+\|x-\pi x\|^2\hbox{ (planar rotations preserve length)} \\
&=\|\pi x+x-\pi x\|^2\hbox{ (Pythagoras)} \\
&=\|x\|^2
}$$
The fact that $R$ is one-to-one and onto can be verified by realising that its inverse can be defined in exactly the same way with the sign of the angle $\theta$ reversed.
Continuity can be verified directly with the $\epsilon-\delta$ definition by considering the components of each vector parallel and perpendicular to $P$ separately.
