 Parallel transport around a spherical triangle

For the positively oriented boundary of a spherical triangle on the unit sphere, parallel transport is rotation by its angle excess modulo $2\pi$. The tangent is parallel along each geodesic side and turns by $\pi -\alpha$ at a corner of angle $\alpha$. The three corner turns sum to $3\pi -\alpha -\beta -\gamma$, and combining them with parallel transport returns the initial tangent. Hence the transport angle is the negative of that sum modulo $2\pi$, giving the formula. The spherical excess formula identifies the angle with enclosed area; reversing traversal changes its sign.