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Project from the north pole $N=(0,0,1)$ onto the equatorial plane, identifying that plane with the complex plane. The line from $N$ to $P=(X,Y,Z)$ meets the plane inThis is stereographic projection. Its inverse isRotation through $\theta$ about the $z$-axis changes $X+iY$ to $e^{i\theta }(X+iY)$ and preserves $Z$, so it becomes the Möbius transformation $u\mapsto e^{i\theta }u$.

Let $B$ be the rotation matrix supplied in the question. It sends the $z$-axis to the $x$-axis, and therefore $R_x(\theta )=B{R}_z(\theta )B^{-1}$. If $b=\varphi B{\varphi }^{-1}$ is its given Möbius transformation, thenwhich is again a Möbius transformation. Such $z$- and $x$-axis rotations can send any chosen point of the sphere to the south pole: first rotate its horizontal projection onto the $y$-axis, then rotate in the $yz$-plane. This supplies the rotations needed below.

The antipodal stereographic coordinate relation is $\varphi (-P)=-1/\overline{\varphi (P)}$, as follows immediately from the inverse formula. Fix the cross-ratio conventionWe claim the required ordering isFor the antipodal cross-ratio and spherical distance identity, rotate $P$ to the south pole. Its coordinate becomes zero and its antipode becomes infinity. A point at angular spherical distance $d$ from the south pole has vertical coordinate $-\cos d$ and horizontal magnitude $\sin d$, so its new projected coordinate $v^{\prime }$ has modulus $\sin d/(1+\cos d)=\tan (d/2)$. By Möbius invariance of the cross-ratio, the displayed expression equalsFor finite nonsingular coordinates the same ratio is $-|u-v{|}^2/|1+u\bar{v}{|}^2$. Zero or infinite antipodal coordinates are interpreted by limits. When $Q=-P$, $d=\pi$ and the ratio has the extended value infinity; the finite real formula applies for $0<d<\pi$.

Project from the north pole $N=(0,0,1)$ along the line through a sphere point onto the equatorial plane $Z=0$, identified with $\mathbb{C}$. Parameterizing this line givesThe inverse stereographic projection isNegating the three sphere coordinates consequently sends $z$ to $-1/\bar{z}$, with $0,\infty$ interchanged. This proves the antipodal stereographic coordinate relation.

A nonidentity Möbius transformation $T(z)=(az+b)/(cz+d)$ has $ad-bc\ne 0$. For $c\ne 0$, its fixed points are the roots of $c{z}^2+(d-a)z-b=0$, giving one or two distinct points. If $c=0$ and $a\ne d$, there is one finite root and the fixed point infinity. If $c=0$, $a=d$, and $b\ne 0$, it is a translation and only infinity is fixed. The remaining case is the identity, already excluded. Thus the claim includes fixed points at infinity, not just finite solutions of the quadratic.

A nontrivial rotation of the sphere fixes exactly the two poles on its axis, which are antipodal. It therefore gives exactly two fixed stereographic points with $z_2=-1/{\bar{z}}_1$. Here the nonzero rotation angle is understood modulo $2\pi$; an angle that is a whole multiple of $2\pi$ would be the identity.

Conversely, use a sphere rotation whose Möbius map $S$ sends the two antipodal fixed points to $0,\infty$. The conjugate $ST{S}^{-1}$ fixes both, hence equals $z\mapsto az$ for some $a\ne 0$. If $|a|=1$, write $a=e^{i\theta }$; the inverse stereographic formula shows that this rotates $X+iY$ while leaving $Z$ fixed. Conjugating back by the rotation $S$ gives a sphere rotation. If $|a|<1$, $a^{n}z\to 0$ for every finite $z$. If $|a|>1$, $a^{n}z\to \infty$ for every $z\ne 0$, in the sphere topology. Pulling these conclusions back proves the antipodal fixed-point classification of Möbius transformations: either the map is a sphere rotation or one fixed point attracts every point except the other fixed point.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.