Alexander duality for the standard equator

The equator is nonempty, since it contains $((1,0,\dots ,0),0)$, and proper, since it misses the two poles $p_{\pm }=(0,\pm 1)$. It is closed and bounded in Euclidean space, hence compact by [F5]. It is weakly locally contractible. For $n=1$ it consists of two isolated points. For $n>1$, at a point of $E$ some coordinate $u_j$ is nonzero; restrict to its fixed sign and solve $u_j=\pm \sqrt{1-{\sum }_{k\mathrm{\ne }j}{u}_k^2}$. Projection to the remaining coordinates is a homeomorphism onto an open subset of ${\mathbb{R}}^{n-1}$. Inside any prescribed neighborhood a small ball in these coordinates contracts within it by straight segments. Thus all hypotheses of [F1] hold.

Put $U_{\pm }=\{(u,t)\in S^n:\pm t>0\}$. They are disjoint open sets whose union is the complement of $E$. Projection $(u,t)\mapsto u$ identifies each with the open unit ball $B^n$, with inverse $u\mapsto (u,\pm \sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2})$. These formulas are continuous inverses. Contract $B^n$ by $u\mapsto (1-s)u$, which stays in the ball for all $0\le s\le 1$. Transporting this contraction gives a contraction of $U_{\pm }$ to $p_{\pm }$. No path joins the two hemispheres: its last coordinate would change from positive to negative and take zero by [F6]. Hence these are exactly the two path components.

By [F2], $H_0(S^n\setminus E;\mathbb{Z})$ is $\mathbb{Z}[p_{+}]\oplus \mathbb{Z}[p_{-}]$. Its augmentation sends $(a,b)$ to $a+b$, so its kernel consists of $(a,-a)$ and is generated by $[p_{+}]-[p_{-}]$. Thus reduced $H_0$ is $\mathbb{Z}$. In every positive degree the singular chain complex splits as the direct sum of the two hemisphere complexes. To see this on generators, a simplex cannot meet both hemispheres: restrict its last-coordinate function to a straight segment between two preimages of opposite signs and apply [F6]. Each simplex therefore lies in exactly one hemisphere, and every face stays there, proving the direct-sum assertion degreewise and for the differential. Kernels and images in this two-summand complex split componentwise. Both hemisphere homology groups vanish in positive degree by [F4] and step 1.2: the point complex has one generator in each degree with boundary alternating between identity and zero. Therefore all positive reduced homology of the complement is zero. Negative reduced groups are zero by the convention in [F1].

With AC supplied by [F3], apply [F1] to the hypotheses verified in step 1.1. It identifies each group computed in step 2.1 with ${\tilde{H}}^{n-i-1}(E;\mathbb{Z})$, proving the displayed formula. In particular this also computes the equator's reduced integral cohomology: it is $\mathbb{Z}$ exactly in degree $n-1$. The duality image of $[p_{+}]-[p_{-}]$ is a generator there; its sign is the one determined by the fixed ambient orientation and the duality construction. No additional sign convention on an independently chosen equator generator is silently asserted.

At $n=1$, the two hemispheres are open semicircles, and the equator is $S^0$; its reduced degree-zero cohomology is the quotient of ${\mathbb{Z}}^2$ by constant pairs, as in [F1], so the same duality conclusion holds. The excluded case $n=0$ would have empty equator and does not satisfy [F1]'s nonempty hypothesis. Neither hemisphere is empty: its specified pole is its contraction endpoint. The contraction at time zero is the identity and at time one is the pole map. Zero homology degrees and zero input classes were accounted for in step 2.1; degenerate simplices stay in their hemisphere just as other simplices do. Every chart and contraction here is explicit, and only [F1]'s atlas/UCT and controlled-extension choices invoke AC.