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Pardon me for repeating a question, but I am not able to justify one aspect of this proof that $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic, and I was hoping that someone could clarify it for me.

Claim: $SO(3)$ is diffeomorphic to $\mathbb{R}P^3$.

Proof. Let $D^3$ be the closed unit ball centered at the origin in $\mathbb{R}^3$. Map each nonzero $x \in D^3$ to the matrix in $SO(3)$ corresponding to clockwise rotation by $\pi|x|$ of $\mathbb{R}^3$ around the directed axis determined by $x$. Also map the origin to the identity matrix.

This is a surjective map because every nonidentity element of $SO(3)$ corresponds to a clockwise rotation of $\mathbb{R}^3$ about some directed axis passing through the origin. It is injective on the interior of $D^3$, and two-to-one on the boundary of $D^3$ with antipodal points mapping to the same element of $SO(3)$.

This map is also continuous, so we obtain a continuous bijection from $D^3 / \sim$ to $SO(3)$, where $\sim$ is the equivalence relation given by identifying antipodal points on the boundary of $D^3$. Since $\mathbb{R}P^3 = D^3 / \sim$ is compact and $SO(3)$ is Hausdorff, this mapping is in fact a homeomorphism.

This map can also be shown to be a diffeomorphism. $\tag*{$\blacksquare$}$

The problem is I am unable to properly justify that it is a diffeomorphism. If $f : D^3 \to SO(3)$ is the map under discussion, then we have $$ f(x) = A(x/|x|, \pi|x|) \quad \text{ for all nonzero } x \in D^3, $$ where $A(v,\theta)$ is the matrix in $SO(3)$ corresponding to clockwise rotation of $\mathbb{R}^3$ by $\theta$ about the directed axis determined by the unit vector $v \in \mathbb{R}^3$. If $R_x(\theta)$, $R_y(\theta)$ and $R_z(\theta)$ are the clockwise rotations by $\theta$ about the $x$-, $y$- and $z$-axes, respectively, then we have $$ A(v,\theta) = R_z(\alpha) R_y(\beta)^T R_z(\theta) R_y(\beta) R_z(\alpha)^T, $$ where $v = (\cos \alpha \sin \beta, \sin \alpha \sin \beta, \cos \beta)$ in spherical coordinates. This is taken from @Randall's answer to this question.

With this description, it is easy to see that $f$ is continuous, because the entries of $A(v,\theta)$ are continuous (and, in fact, smooth) functions of $\theta$, $\alpha$ and $\beta$ (and hence of $v$ too). Thus, $f$ is a smooth map when we view $SO(3) \subset M_3(\mathbb{R}) \underset{\text{diff.}}{\cong} \mathbb{R}^{3^2}$, and in particular it is continuous.

Let $g : \mathbb{R}P^3 \to SO(3)$ be the map induced by $f$ on the quotient space. To show that it is a diffeomorphism, it only remains to show that $g^{-1}$ is smooth.

The only way I know to show that a map between smooth manifolds is smooth is to use the definition: take smooth charts on both sides and compose to get a map between subsets of Euclidean spaces, and then check that this is smooth. Here, I have a problem. I know that $SO(3)$ is a smooth manifold because I have shown that the $O(3)$ is the inverse image of a regular value, and $SO(3)$ is a connected component of $O(3)$. I don't have an explicit description of a smooth atlas on $SO(3)$ and so I am unable to show that $g^{-1}$ is smooth.

Can someone help me complete this last step of the problem?

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  • $\begingroup$ Try extending this map to all of $\Bbb R^3$ by the sane formula and showing that on the ball of radius less than $2\pi$, it is a submersion. $\endgroup$ – user98602 Nov 11 '18 at 3:50
  • $\begingroup$ @MikeMiller how would that help, can you elaborate? $\endgroup$ – Brahadeesh Nov 11 '18 at 4:38

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