The Gauss linking integral is unchanged when the stereographic pole moves along a path avoiding both loops

Let $\gamma_1,\gamma_2:\mathbb{R}\to S^3\subset\mathbb{R}^4$ be $C^2$, $1$-periodic loops with disjoint images. Let $P:\mathbb{R}\to S^3$ be a $C^2$ path of poles that misses both loops, $P(\tau)\notin\gamma_1(\mathbb{R})\cup\gamma_2(\mathbb{R})$, and is never antipodal to $P(0)$, i.e. $P(\tau)\neq -P(0)$. Then the Gauss linking integrals after stereographic projection from $P(0)$ and from $P(1)$ agree:

Proof idea: let $N=P(0)$ and $R_\tau=S_{P(\tau)+N}\circ S_{P(\tau)}$, where $S_v$ is the reflection in $v^\perp$. Then $R_\tau\in SO(4)$ depends $C^2$ on $\tau$ and $R_\tau P(\tau)=N$. Stereographic projection is equivariant, $\sigma_{N}(R_\tau y)=R_\tau\sigma_{P(\tau)}(y)$, and the integrand $\det(-N,a,b,c)/|c|^3$ is invariant under $R_\tau$. So the integral from the pole $P(\tau)$ equals the fixed-pole Gauss integral of the loops $\sigma_N\circ R_\tau\circ\gamma_i$. These form a $C^2$ homotopy of disjoint loops in $N^\perp$, so the integral is constant (gaussIntegral_eq_of_homotopy).