Problem to proof that $SO(3)$ is a differentiable manifold I and my friend were trying to proof that $SO(3)$ is a Lie Group, and in particular a differentiable manifold.
His attempt was to consider a closed ball in $\mathbb{R}^3$, centered in origin and whose radius is $\pi$.
Then for any point $x$ in this closed ball, we consider the rotation that preserves the axis passing through the origin and $x$ and with a rotation angle given by the oriented lenght of $0$ to $x$.
Antipode points on boundary (which represent the same rotation of angle $\pi$), are identified by a equivalence relation. In this way we obtain that $SO(3)$ is homeomorphic to $\mathbb{R}\rm{P}^3$.
That is my attempt: If we look $SO(3)$ like 3x3 matrices whose columns are unitary orthogonal vectors with determinant equal to 1. 
Then we can conclude that the first column is a choice of one arbitrary vector $w_1 \in \mathbb{S}^2$, the second column is a choice of a vector $w_2 \in \mathbb{S}^2 \cap \{w_1\}^\perp$, which is homeomorphic to $\mathbb{S}^1$, and finally, the third column is
$w_3 = w_1 \times w_2$, to obtain $det = 1$. 
In other words, I mean that $SO(3)$ is homeomorphic to $\mathbb{S}^2 \times \mathbb{S}^1$.
But, if you evaluate the fundamental groups of these spaces, you'll obtain $\pi_1(\mathbb{R}\rm{P}^3) \cong \mathbb{Z}_2$ and $\pi_1(\mathbb{S}^2 \times \mathbb{S}^1) \cong \mathbb{Z}$.
The Question is: what is wrong in my interpretation?
Remark: The first attempt is described in many texts.
 A: What you obtain is that $\text{SO}(3)$ is diffeomorphic (not just homeomorphic) to an $S^1$-bundle over $S^2$, which is true. But you cannot conclude that this bundle is trivial.  In other words, given $w_1$, the space of possible choices for $w_2$ is diffeomorphic to $S^1$, but it turns out that you cannot smoothly choose a family of such diffeomorphisms as a function of $w_1$ (in fact the fundamental group computation you describe proves this). This is a great example of the crucial difference between knowing that two things are isomorphic and knowing a particular isomorphism between them. 
This bundle is closely related to the Hopf fibration, which also exhibits $\text{SU}(2)$, the double cover of $\text{SO}(3)$, as a nontrivial $S^1$-bundle over $S^2$. One way of describing its structure is that $\text{SO}(3)$ can naturally be identified with the unit tangent bundle of $S^2$ (via its natural action on $S^2$). That the unit tangent bundle of $S^2$ is nontrivial is implied, for example, by the hairy ball theorem. 
