$SO(3)$ homeomorphic to $\mathbb{R}P^3$ I'm doing some topological base-exercises, but I can't come up with this problem (That I suppose should be quite trivial):
$SO(3)$ is homeomorphic to $\mathbb{R}P^3$.
Any hints?
thank you in Advance!
 A: Aside from the clean indirect approach by Max on the comments, there is a nice way to visualize the homeomorphism. 
Just consider the map
$$f: D^3 \to SO(3)$$
given by $f(x)=$ Rotation (as per the right-hand rule) around $x$ with angle $\pi \Vert x \Vert$. This can be verified to be continuous (you can take matrices, for instance). Now, notice that two points in the boundary of $D^3$ are taken to the same rotation. Therefore, this induces a map
$$f: \mathbb{R}P^3 \to SO(3),$$
which is bijective. Since it is the induced map on the quotient of a continuous map, it is continuous. Since it is a bijective continuous map on a compact set with Hausdorff codomain, it is a homeomorphism.
A: I searched on internet proofs of the heomeomorphism $\mathbb{RP}^3 \cong SO(3)$ but the only proof I managed to find is the one posted here that uses the homeomorphism between $\mathbb{RP}^3$ and the closed 3-dimensional unit disk with antipodal border points identified.
This is another proof that I came up with :
We will need to use the non-commutative field of quaternions $\mathbb{H}$.
Let $\mathbb{H}^* := \mathbb{H} \setminus \{0\}$
Let $q \in \mathbb{H}^*$, I define the following map
$$\begin{split} f_q \; : \; \mathbb{H} &\to \mathbb{H} \\
v &\mapsto f_q(v) = qvq^{-1} \end{split}$$
Clearly I can see $\mathbb{H}$ as a four-dimensional vector space over $\mathbb{R}$, in this way it's clear that $f_q$ is linear, I also have that $\forall v \in \mathbb{H}$
$$|f_q(v)| = |q v q^{-1}| = |q| \cdot |v| \cdot |q|^{-1} = |v|$$
Therefore $f_q$ is an isometry, therefore is injective, therefore is also surjective (since $\mathbb{H}$ is finite dimensional).
Let $\alpha \in \mathbb{R}$, then
$$f_q(\alpha) = q \alpha q^{-1} = \alpha$$
So in particular $\mathbb{R}$ is invariant under $f_q$, so in particular if $V := \{ v \in \mathbb{H} \; : \; Re(v) = 0\}$
Then also $V$ is invariant under $f_q$.
Now let $q \in \mathbb{H}^*$, I define
$\phi_q := f_q\big|_V$
Since $V$ is a 3-dimensional vector space over $\mathbb{R}$ I can identify $V$ with $\mathbb{R}^3$, since $f_q$ is an isometry also $\phi_q$ is, so is well defined the following map
$$\begin{split} \Psi \; : \; \mathbb{H}^* &\to O(3) \\ q &\mapsto \phi_q \end{split}$$
Now I prove that $\Psi$ is  continuos.
Let $p,q \in \mathbb{H}^*$, let $v \in V \; : \; |v| \leq 1$, then
$$|\phi_q(v) - \phi_p(v)| = |qvq^{-1} - pvp^{-1}| \leq |q| \cdot |v| \cdot |q^{-1} - p^{-1}| + |q - p| \cdot |v| \cdot |p|^{-1} = 2|v| |q - p| \cdot |p|^{-1} \leq 2 |p - q| \cdot |p|^{-1}$$
Taking the sup when $|v| \leq 1$ I find
$$||\phi_q - \phi_p|| \leq 2 |p - q| \cdot |p|^{-1}$$
So $\Psi$ Is continuos.
$\mathbb{H}^{*}$ is connected therefore $Im(\Psi)$ has to lie in a connected component of $O(3)$, notice that $SO(3)$ is a connected component of $O(3)$ and $\Psi(1) = Id \in SO(3)$ so $Im(\Psi) \subset SO(3)$
Is not hard to prove that $\Psi$ is also a group homomorphism ($\phi_{qp}(v) = q(pvp^{-1})q^{-1} = \phi_q(\phi_p(v))$ )
Therefore to prove that $\Psi$ is surjective (meaning that $Im(\Psi) = SO(3)$ ) is enough to prove that a set of generators of $SO(3)$ lies in $Im(\Psi)$, this is not hard to prove but I will skip it.
Now I want to show that $\Psi(q) = \Psi(p)$ if and only if there exist a $\lambda \in \mathbb{R}$ such that $q = \lambda p$.
$\impliedby$ is trivial so I only prove $\implies$, to do it I use a lemma :

Lemma :
Let $a \in \mathbb{H}$ be such that
$ab = ba \; \forall b \in \mathbb{H}$, then $a \in \mathbb{R}$

Now let $\Psi(p) = \Psi(q)$, let $b \in \mathbb{H}$, then there exist a $\alpha \in \mathbb{R}$ and a $v \in V$ such that $\alpha + v = b$, then
$$qbq^{-1} = q(\alpha + v)q^{-1} = \alpha + \phi_q(v) = \alpha + \phi_p(v) = pbp^{-1}$$
So I just proved that
$$qbq^{-1} = pbp^{-1} \; \forall b \in \mathbb{H}$$
Which implies that
$$(p^{-1}q)b = b(p^{-1}q) \; \forall b \in H$$
Which by the lemma implies that exists $\lambda \in \mathbb{R}$ such that
$p^{-1}q = \lambda$, so $q = \lambda p$
Now I can consider the map $\pi$ from $\mathbb{R}^4 \setminus \{0\}$ to $\mathbb{RP}^3$ and the map $\Psi$ from $\mathbb{H}^*$ to $SO(3)$
I identify $H^*$ with $\mathbb{R}^4 \setminus \{0\}$ and I have that $\Psi(p) = \Psi(q) \iff \pi(p) = \pi(q)$, therefore by the foundamental theorem of homeomorphism ( the version for topological spaces ) I know that there exists a continuos map $F \; : \; \mathbb{RP}^3 \to SO(3)$ such that $F(\pi(p)) = \Psi(p) \; \forall p \in \mathbb{H}^{*}$
I also know that $F$ is bijective and since $\mathbb{RP}^3$ is compact and $SO(3) \subset \mathbb{R}^9$ is Hausdorff I know that $F$ is closed, therefore $F$ is a homeomorphism and finally
$\mathbb{RP}^3 \cong SO(3)$
Something interesting is that, once again identifying $\mathbb{H}^*$ with $\mathbb{R}^4 \setminus \{ 0\}$, I can define a group structure on $\mathbb{RP}^3$ by letting $[p] \cdot [q] = [p\cdot q]$ (basically $\mathbb{RP}^3 \cong \mathbb{H}^* / \mathbb{R}^* $ ) With this group structure $F$ is also an isomorphism of groups.
