Rotations by π form a projective plane; the Cayley transform covers the rest

Part 1: description of $C$. By the normal form of the previous problem, $\theta(U)=\pi$ means $\operatorname{tr}U=-1$, i.e. $U\sim1\oplus R_\pi=\operatorname{diag}(1,-1,-1)$. So $U$ fixes a unit vector $n$ and acts as $-1$ on $n^\perp$: $Ux=2(n\cdot x)n-x$, i.e. $U=2nn^T-I$. Conversely such matrices are orthogonal with determinant $(+1)(-1)(-1)=1$ and trace $2-3=-1$.

Part 1: topology. The map $\phi:S^2\to C$, $n\mapsto2nn^T-I$, is continuous and onto. It satisfies $\phi(n)=\phi(m)$ iff $nn^T=mm^T$ iff $m=\pm n$. So $\phi$ induces a continuous bijection $S^2/\{\pm1\}=\mathbb{RP}^2\to C$. As $\mathbb{RP}^2$ is compact and $C$ is Hausdorff, this is a homeomorphism.

Part 2: well defined. If $K$ is skew, then $I-K$ is invertible: if $(I-K)x=0$ then $x^TKx=x^Tx$, but $x^TKx=0$ for skew $K$, so $x=0$. Put $U=(I+K)(I-K)^{-1}$. Since $I+K$ and $(I-K)^{-1}$ commute (both are functions of $K$),
$$U^TU=(I+K^T)^{-1}\ldots$$
more directly: $U^T=(I-K^T)^{-1}(I+K^T)=(I+K)^{-1}(I-K)$, and so $U^TU=(I+K)^{-1}(I-K)(I+K)(I-K)^{-1}=I$ (again using commutation of functions of $K$). Moreover $\det U=\det(I+K)/\det(I-K)=1$ because $\det(I+K)=\det\bigl((I+K)^T\bigr)=\det(I-K)$. So $U\in SO(3)$.

Part 2: avoids $C$. If $Ux=-x$ with $x\ne0$, put $y=(I-K)^{-1}x\ne0$; then $(I+K)y=-(I-K)y$, so $2y=0$, a contradiction. So $-1$ is not an eigenvalue of $U$. But every element of $C$ has the eigenvalue $-1$ (it is $-1$ on $n^\perp$). Hence the image lies in $SO(3)\setminus C$.

Part 2: onto. Let $U\in SO(3)$ without eigenvalue $-1$, so $I+U$ is invertible. Define $K=(U-I)(U+I)^{-1}$. Then, using $U^T=U^{-1}$,
$$K^T=(U^T+I)^{-1}(U^T-I)=(U^{-1}+I)^{-1}(U^{-1}-I)=(I+U)^{-1}(I-U)=-K,$$
where in the last step we multiplied numerator and denominator by $U$ and used that $(U-I)$ and $(U+I)^{-1}$ commute. So $K$ is skew, and solving $K(U+I)=U-I$ for $U$ gives $U=(I+K)(I-K)^{-1}$: the Cayley transform of $K$ is $U$.

Part 2: injective. From $U=(I+K)(I-K)^{-1}$ we get $U(I-K)=I+K$, hence $K(U+I)=U-I$, which determines $K=(U-I)(U+I)^{-1}$ uniquely.

Remark. Since $SO(3)\setminus C$ is the image of the *whole vector space* $\mathrm{Skew}_3\cong\mathbb{R}^3$ under a homeomorphism, $SO(3)=C\cup(\text{open ball})$: this is the cell decomposition of $SO(3)\cong\mathbb{RP}^3$ as the one-point-more... namely $\mathbb{RP}^3=\mathbb{RP}^2\cup\mathbb{R}^3$.

Lesson. The "cut locus" of a Lie group chart is the set where the chart fails to cover; here it is exactly the rotations by $\pi$.