Visualize why the group of isometries of the sphere $S^2$ is $S^3/(\mathbf{Z}/2) ?$ How to visualize
why the group of rotational symmetries (orientation preserving isometries) of the sphere $S^2$ is $$S^3/(\mathbf{Z}/2) ?$$
I meant that because that the rotational symmetry group is $SO(3)=RP^3=S^3/(\mathbf{Z}/2) $. I know this fact, but how to convince ourselves that by visualization that there is a $\mathbf{Z}/2$ mod out?
It may be useful to use the fibration fact that the $S^3/(\mathbf{Z}/2)$ can be a lens space obtained by $S^1$ fiber over $S^2$? But how is that relevant to the isometry group of the sphere $S^2$?
p.s. I certainly know $(3)=^3=^3/(/2)$. My point here is how to visualize the rotational group has a quotient out $(/2)$ part in $^3/(/2)$ for the rotational group? Which path of the loop of the rotational group element gives that $\pi_1(^3/(/2))=(/2)$?
 A: Consider the ball $B$ in $\mathbb{R}^3$ of radius $2\pi$, centered on the origin.
Any point $u\in B$ represents a rotation through an angle $|u|$ radians about the directed axis represented by $u$.  Note there is no ambiguity with $0\in B$ as a rotation through $0$ radians is the identity map - e.g. we do not need an axis.
Note also that all rotations through $2\pi$ radians are the same regardless of axis, so we may quotient the boundary of $B$ to a point, to obtain a sphere $S^3$ and still have every element of $S^3$ representing a well defined rotation.
Any rotation in $SO(3)$ will be represented by precisely two points in $S^3$: a non-trivial rotation will be $\theta$ radians clockwise about some directed axis represented by a unit vector $u$, with $\theta\in [0,2\pi]$, so will be represented by $\theta u\in S^3$.  However it will also be represented by $-(2\pi-\theta)u\in S^3$.
Thus $SO(3)\cong S^3/(\mathbb{Z}/2)$ in a visual way - a rotation is represented by a point in $S^3$, where the distance from the identity is the angle of rotation, and direction from identity is axis of rotation.
An easy way to see the natural fibration $SO(3)\to S^2$ is to define it as as evaluation on a point of $S^2$.  That is fix a point $x\in S^2$ and map a rotation $A\in SO(3)$ to $Ax\in S^2$.
Finally, to visualise the non-trivial element in $\pi_1(SO(3))$ simply pick an axis and rotate about it going form $0$ radians to $2\pi$ radians.  This lifts to a path in $S^3$ going from the origin to the point representing rotations by $2\pi$ radians.  Thus it represents a non-trivial element of $\pi_1(SO(3))$ (to be trivial it would have to lift to a loop).

Let $f:S^3\to SO(3)\to S^2$ denote the composition of the quotient map $S^3\to SO(3)$ and the fibration described above, sending each $A\in SO(3)$ to $Ax$.
The fiber $F_y$ in $SO(3)$ above a point $y\in S^2$ is the set of rotations which map to $y$.  That is $F_y$ is the set of rotations $\{C| Cx=y\}$.
If $Ax=y$ then $F_y=\{AT_\theta |\theta\in[0,2\pi]\}$, where $T_\theta$ denotes a rotation through $\theta$ radians about $x$.
Consider in particular $f^{-1}(x)$.  This is all points in $S^3$ which map to a $T_\theta$.  That is precisely the line through the origin and $x$.  Note this line is actually a circle, as the boundary points of $B$ are identified.  In other words $2\pi x=-2\pi x$ in $S^3$.
Now consider the fiber $F_y$ where $y$ is the point in $S^2$ opposite $x$.  The fiber above $y$ in $SO(3)$ is all $\pi$ radian rotations about an axis perpendicular to $x$. These are precisely the rotations taking $x\mapsto y$.
Thus $f^{-1}(y)$ is the circle about the origin of radius $\pi$, perpendicular to $x$.  Thus we can see that $f^{-1}(x)$ and $f^{-1}(y)$ form a pair of linked circles in $S^3$.

Note that for any points $z\neq w\in S^2$ there are paths in $S^2$ from $x$ to $z$ and from $y$ to $w$ which do not cross. Following these paths gives us isotropies of the fibers from $f^{-1}(x) \to f^{-1}(z)$ and $f^{-1}(y) \to f^{-1}(w)$, which also do not cross.  Thus  any pair of fibers $f^{-1}(z)$ and $f^{-1}(w)$ will be linked in the same way as in the picture above.
A: One way is to think of S3 as the group of unit quaternions.
Then for any q ∊ S3, you can define the map
fq : S3  →  S3
via fq(x)  =  q x q-1.  The purely imaginary unit quaternions form a S2  ⊂  S3. Restricting fq to this S2 can be seen to define a rotation of S2.
Then the assignment
q  →  fq
defines a map S3 → SO(3) that is a surjective Lie group homomorphism whose kernel can be seen to be just {±1}, which is what you wanted to prove.
