Existence of regular projections

Tangencies and double points. The empty link has every direction regular, so assume $C\mathrm{\ne }\mathrm{\varnothing }$. Projection fails to be immersive exactly when $u$ is parallel to a tangent, not orthogonal to it. Both signed unit tangent images are null by [F3], since their sources are finite unions of circles. At a secant parallel to $u$, the differential of $\sigma$ has image spanned by the two projected tangent vectors: differentiating the normalized secant gives their multiples in $u^{\perp }$. Thus a nontransverse double point makes $u$ or $-u$ a critical value of $\sigma$. Those critical values and their antipodes are null by [F3].

Genuine triple incidence. On distinct triples define $s(x,y,z)=(\sigma (x,y),\sigma (x,z))\in S^2\times S^2$. Restrict to the open set where these two directions are not parallel to the tangents at $y,z$ and the projections of those tangents to their common direction plane are independent whenever the directions agree. At a point with $\sigma (x,y)=\sigma (x,z)=u$, the normal differential to the diagonal of $S^2\times S^2$ has the two independent columns ${\pi }_u{f}^{\mathrm{\prime }}(y)\mathrm{/}\mathrm{\mid }f(y)-f(x)\mathrm{\mid }$ and $-{\pi }_u{f}^{\mathrm{\prime }}(z)\mathrm{/}\mathrm{\mid }f(z)-f(x)\mathrm{\mid }$. Hence $s$ is transverse to that diagonal on this open set. Its preimage $T$ is a smooth one-dimensional manifold by [F4], and the direction map $T\to S^2$ has null image by [F3]. If a triple occurs for a direction not excluded in step 1.1, order its three points along their common line and take $x$ to be an extreme point. Then the two secants agree up to the same sign, and the projected tangents at $y,z$ are independent because that direction is a regular secant value for every pair. This triple therefore contributes a direction in the image of $T$ or its antipode. These two images are null.

Compactness away from the diagonal. Outside the null sets of steps 1.1–2.1 the projection is immersive, all double points are transverse, and there are no triples. Its double-point pairs cannot accumulate on the diagonal: near each source point one coordinate of the projected derivative is nonzero, so the projection is injective on a small arc; finitely many such arcs give a neighbourhood of the diagonal containing no double-point pair. The pairs consequently form a closed subset of a compact subset of $C\times C$ away from the diagonal. Transversality makes them isolated, so there are finitely many. The complement of a null set in $S^2$ is dense, proving the assertion. Countable choice is used only in [F3].