$\mathbb{R} \mathbb{P}^2$ is homeomorphic to $\mathcal{S} \setminus \{I_3\}$ Let $\mathcal{S} = \{ B \in SO(3); B^T = B\}$. Define $\varphi: \mathbb{R}\mathbb{P}^2 \to SO(3)$ by
$$
     \varphi(l) = \text{the rotation by $\pi$ about the line $l \subset \mathbb{R}^3$}.
 $$
Show that $\varphi$ maps $\mathbb{R} \mathbb{P}^2$ homeomorphically onto $\mathcal{S} \setminus \{I_3\}$.
Now, the rotation about the line $l$ is the following matrix:
$$\begin{bmatrix} 1& 0& 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta &\cos\theta \end{bmatrix}$$
So, $\varphi$ takes any line to this matrix. First of all, I don't understand how exactly it takes a line and turns it into this matrix, I mean what thing of the line should $\varphi$ take to give us this matrix? And then, if I understand this part, it's easy to show that it's a bijection, but I have problem with the continuity part.
 A: Given a line $\ell$, choose a unit length vector $v_1 \in \ell$.  Complete $\{v_1\}$ to an orthonormal basis $\{v_1,v_2,v_3\}$.  Consider the $3\times 3$ matrix $V$ whose columns are $v_1,v_2,v_3$.  Because the $\{v_i\}$ are an orthonormal basis, $\det V \in \{\pm 1\}$.  If it is negative one, switch $v_2$ and $v_3$.  Thus, $V\in SO(3)$.
Now, consider the matrix $\phi(\ell):=V\operatorname{diag}(1,-1,-1) V^{-1}$.  A simple calculation reveals that $\phi(\ell) v_1 = v_1$, while $\phi(\ell) v_j = -v_j$ for $j\in \{2,3\}$.  Linearity implies that $\phi(\ell)$ is given geometrically by fixing the line $\ell = \operatorname{span}\{v_1\}$ and acting by a rotation by angle $\pi$ on the orthogonal complement to $\operatorname{span}\{v_1\}$.  It follows that $\phi(\ell)$ is independent of the choices made (we made two in defining $V$:  Picking $\pm v_1$, and the choice of completed basis).
Note that it also follows from this that the map $\phi$ described above is the same as the one in your post.
Now, can you show that $\phi(\ell) \in S$?
As far as showing $\phi$ is a bijection, the key is to note that an element of $S$ has eigenvalues $1,-1,-1$, and is characterized by its $1$-eigenspace.  Can you prove this?
For continuity, you can argue as follows.  Suppose $\ell_1$ and $\ell_2$ are close.  We need to show that $\phi(\ell_1)$ and $\phi(\ell_2)$ are close.  Because $\ell_1$ and $\ell_2$ are close, we can pick vectors $v_1\in \ell_1,w_1\in \ell_2$ which are close.  We can complete these to orthonormal basis which are close, which means the Vs we get are close.  It then follows that $\phi(\ell_1)$ and $\phi(\ell_2)$ are close.  How you make this rigourous depends on your definitions of the topologies on $\mathbb{R}P^2$ and $SO(3)$.
A: Here is an explicit description of your map $\varphi$:
Let ${\mathbf n}$ be a unit vector in ${\mathbb R}^3$. The map
$$
\sigma_{{\mathbf n}}({\mathbf x})= {\mathbf x} - (2 {\mathbf x} \cdot  {\mathbf n}) {\mathbf n}
$$
defines the reflection in the hyperplane in ${\mathbb R}^3$ orthogonal to ${\mathbf n}$. (There is a good chance that you have seen this formula in your linear algebra class. The same formula workds in all dimensions.) Therefore, the map
$$
R_{{\mathbf n}} ({\mathbf x})= - \sigma_{{\mathbf n}}({\mathbf x})= - {\mathbf x} + (2 {\mathbf x} \cdot  {\mathbf n}) {\mathbf n}
$$
defines the rotation by the angle $\pi$ around the line spanned by the vector ${\mathbf n}$. From this, you see continuity of the map
$$
\Phi: {\mathbf n} \mapsto R_{{\mathbf n}}, S^2\to O(3). 
$$
Surjectivity of this map to the subset $\Sigma$ of non-identity symmetric orthogonal matrices is a nice exercise in linear algebra. Moreover,
$$
\Phi({\mathbf n})= \Phi({\mathbf n}') \iff {\mathbf n}=\pm {\mathbf n}'.
$$
Now, you use some point-set topology (the definition of the quotient topology) to check that the map $\Phi$ descends to a continuous bijection
$$
\varphi: RP^2\to \Sigma,
$$
such that $\Phi= \varphi \circ q$, where $q: S^2\to RP^2$ is the quotient map. Lastly, use Hausdorffness of $\Sigma$ and compactness of $RP^2$ to conclude that $\varphi$ is a homeomorphism.
