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Colatitude is the angle from the positive polar axis, usually denoted $\theta$ with $0\le \theta \le \pi$. On a sphere it complements latitude $\lambda$: $\theta =\pi /2-\lambda$. Thus the north pole, equator and south pole have colatitudes zero, $\pi /2$ and $\pi$.

With longitude $\varphi$, the standard spherical coordinates are $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$. The surface element is $r^2\sin \theta d\theta d\varphi$, which shows why an integral over the sphere must include the sine weight. Some disciplines exchange the names of $\theta$ and $\varphi$, so formulas should accompany symbol conventions.

Relative angular velocities add as vectors: ${\mathbf{\omega }}_{B/C}={\mathbf{\omega }}_{B/A}+{\mathbf{\omega }}_{A/C}$. Geographic pole latitudes and longitudes must first be converted into Cartesian vector directions.

Longitude is an angular coordinate around a chosen polar axis, measured from a reference half-plane. On Earth the conventional reference is the Greenwich meridian. In a Cartesian coordinate convention it is $\varphi =atan2(y,x)$, with a specified range such as $(-\pi ,\pi ]$.

Together with latitude and a radius it describes a point using spherical coordinates. Longitude is undefined at the poles because every meridian meets there. Its numerical jump at the chosen meridian cut is a coordinate discontinuity, not a discontinuity of the sphere itself. On an oblate reference ellipsoid, geographic latitude needs an additional geodetic convention.

Let $\hat{x}$ point to the equatorial meridian at zero longitude, $\hat{y}$ to the equator at $90^{\circ }$ E, and $\hat{z}$ to the north pole. Use a right-hand positive Euler angular velocity vector. Relative rotations add as vectors, not as geographic coordinates:ConsequentlyHere $\lambda ={\tan }^{-1}(12/5)$ is the longitude of the positive Euler pole, and its latitude $\varphi$ is zero. The antipodal pole with the opposite angular-velocity sign is an equivalent convention; mixing a pole reversal with an unchanged sign would give the wrong motion.

At latitude $\varphi$ and longitude $\lambda$, $\mathbf{r}=R(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$. Motion is ${\mathbf{v}}_{B/C}={\mathbf{\omega }}_{B/C}\times \mathbf{r}$. A meridian boundary has horizontal normal ${\hat{e}}_{\lambda }=(-\sin \lambda ,\cos \lambda ,0)$, so its across-boundary velocity isFor a transform fault this must vanish along the whole meridian, not only at an isolated point. Thus $5\cos \lambda +12\sin \lambda =0$. Select the solution in the western hemisphere:The other solution is the antipodal meridian in the eastern hemisphere. Geometrically, the transform meridian is in a great-circle plane perpendicular to the Euler angular velocity vector. Its outgoing tangent direction at either pole, expressed in the fixed $x,y$ axes, is ${\hat{t}}_F=(12,-5)/13$.

For an open cover $X=U\cup V$, the Mayer–Vietoris theorem gives the long exact sequenceIt ends in $H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$. To prove it, let $C_{\ast }^{U,V}(X)$ be the subcomplex generated by singular simplices lying wholly in $U$ or wholly in $V$. There is a short exact sequence of chain complexesExactness follows from the singular-simplex bases: the intersection of the two chain subgroups consists exactly of simplices lying in both sets. The small-chain locality theorem for singular homology says the inclusion of this small-chain complex into $C_{\ast }(X)$ induces an isomorphism on homology; barycentric subdivision is its standard justification. The long exact sequence of a short exact sequence of chain complexes now proves the theorem. For a homological cycle represented as $a+b$, the connecting map is $[a+b]\mapsto [\partial a]$, since $\partial a=-\partial b$ lies in the intersection. This also identifies the displayed maps and their signs.

For the glued solid tori, use open collar enlargements of the two halves. Each retracts to a circle; the overlap retracts to their boundary torus. Thus the halves have $H_0=H_1=\mathbb{Z}$ and higher homology zero, while the overlap has $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$. In the overlap choose longitude and meridian generators. The meridian bounds a disk in each solid torus and the longitude generates each first homology, so the map on $H_1$ isIts kernel is the meridian copy of $\mathbb{Z}$ and its cokernel is $\mathbb{Z}$. The map on $H_0$ is the injective map $z\mapsto (z,-z)$. Exactness therefore gives $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}$. The preceding segment is $0\to H_3(X)\to H_2(U\cap V)\to 0$, giving $H_3(X)=\mathbb{Z}$. Connectedness gives $H_0(X)=\mathbb{Z}$, and all higher groups vanish. HenceAs a geometric check, the identity boundary gluing identifies the product with $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.