isZero_h1_complement_of_embedding
plain-language theorem explainer
If every embedded arc in the D-sphere has H₁-acyclic complement, then every embedded circle in S^D (D≥1, D≠3) also has H₁-acyclic complement. Cited by anyone running the Hatcher 2B.1 circle reduction or closing the RS spatial-dimension lock. Proof splits the circle into two semicircle images and feeds their complements, plus the two-point endpoint complement, into a Mayer-Vietoris vanishing lemma.
Claim. Let $D\ge 1$ with $D\ne 3$. Assume every topological embedding of the unit interval into $S^D$ has $H_1$-acyclic complement. Then for every continuous embedding $g\colon S^1\hookrightarrow S^D$, the group $H_1(S^D\setminus g(S^1))$ is zero.
background
Classical Alexander duality (Hatcher §2B) controls the homology of sphere complements by codimension. Recognition Science uses the circle case as the linking obstruction that forces spatial dimension three (T8/T9): nontrivial first homology of an embedded-circle complement is possible only when $D=3$.
This file works in the unit-sphere model $\mathrm{Sph},D\subset\mathbb{R}^{D+1}$. The standing hypothesis (arc-complements acyclic) asserts that every topological embedding of the unit interval into $S^D$ has vanishing $H_1$ of the complement. Classically true for all $D$ by compact-support bisection, it is kept as an explicit parameter so the circle case reduces to it.
Two standard semicircle arcs cover $S^1$ and meet only at a pair of antipodal endpoints. Their images under an embedded circle therefore cover the image circle and intersect at exactly two points.
proof idea
Compose the circle embedding $g$ with the two semicircle arc maps. Each composite is an embedding of the unit interval (composition of embeddings). Range identities give: the two semicircle images cover the full image of $g$, and their intersection is exactly the two endpoint images, which are distinct by injectivity of $g$.
The $H_2$ of the complement of those two points vanishes by the already-proved two-point complement lemma (hypothesis $D\ne 3$). Arc-acyclicity supplies $H_1$-vanishing for each semicircle-image complement. The union-complement lemma then yields $H_1$-vanishing for the complement of the union, i.e. of the embedded circle. Rewrite along the cover identity to finish.
why it matters
This is the circle half of Hatcher 2B.1 inside the RS foundation stack. It is the sole upstream input to the high-dimensional vanishing theorem: granting arc-acyclicity, no embedded circle in $S^D$ ($D\ge 1$, $D\ne 3$) has homologically nontrivial complement, so no such circle detects nontrivial linking.
That vanishing is the high-dimensional side of the dimension lock. Linking, and therefore the RS spatial dimension $D=3$ forced along the T8/T9 chain, can occur only in dimension three. The remaining open interface is precisely the arc hypothesis; once arcs are granted, this theorem discharges every embedded circle, so the forcing argument reduces to one classical input.
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