Jordan–Brouwer separation

The image $K$ is compact and weakly locally contractible, since the sphere is compact and has locally contractible coordinate balls; $S^0$ is discrete. It is proper: otherwise $h$ would be a homeomorphism from $S^{n-1}$ to $S^n$, contradicting the different degree-$n$ homology groups in [F3]. A homeomorphism and its inverse induce mutually inverse chain maps, hence isomorphisms on homology, so that contradiction does not assume invariance of domain. Let $E$ be the standard equator $S^{n-1}\subset S^n$. Its two complementary hemispheres are each homeomorphic to an open $(n)$-ball, by projection to the first $n$ coordinates and the two graphs $t=\pm \sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}$. They are path connected, disjoint, and relatively open and closed in the complement. Thus [F2] gives ${\tilde{H}}_0(S^n\setminus E;\mathbb{Z})=\mathbb{Z}$. Applying [F1] to this explicit equator computes ${\tilde{H}}^{n-1}(S^{n-1};\mathbb{Z})=\mathbb{Z}$, including $n=1$. Transport by $h$ using [F4], then apply [F1] to $K$, to obtain ${\tilde{H}}_0(S^n\setminus K;\mathbb{Z})=\mathbb{Z}$.

A nonempty space with reduced $H_0$ equal to $\mathbb{Z}$ has exactly two path components here. To see this directly from [F2], choose one component $C_0$. The augmentation kernel is freely generated by $e_C-e_{C_0}$ for $C\mathrm{\ne }{C}_0$: subtracting the total coefficient at $C_0$ gives the spanning formula, and comparison of the other coefficients proves independence. There must be at least one such generator because the kernel is nonzero. There cannot be two: their images under an isomorphism to $\mathbb{Z}$ would be nonzero integers $a,b$, and the nonzero relation $b(e_C-e_{C_0})-a(e_{C^{\mathrm{\prime }}}-e_{C_0})$ would map to zero, contradicting injectivity. Hence step 1.1 gives precisely two path components $P,Q$. They are open in the sphere: every point of its open complement has a small path-connected coordinate ball lying in that complement, and this ball is contained in the point's path component. A connected subset cannot meet two members of this open partition, so these path components are also the connected components. Each is closed relative to the complement, whence its sphere boundary is contained in $K$.

We will also use that the complement of any embedded nonempty closed disk $L\cong D^{n-1}$ contained in $K$ is path connected. Such an $L$ is compact, proper and weakly locally contractible, including its boundary (small convex relative neighborhoods contract in the disk). The disk contracts linearly to a point. A point has one singular cochain generator in each nonnegative degree; its positive coboundaries alternate between zero and identity because the alternating face sum is zero or one. Thus its cohomology is $\mathbb{Z}$ in degree zero and zero otherwise, and its reduced cohomology is zero in every degree. By [F4], the same is true of $L$. Alexander duality [F1] gives ${\tilde{H}}_0(S^n\setminus L;\mathbb{Z})=0$. This complement is nonempty because $L\subset K\mathrm{\ne }{S}^n$, and [F2]'s basis-difference description shows it has exactly one path component.

Suppose $x\in K$ were not in the closure of $P$. Choose an open sphere neighborhood $W$ of $x$ disjoint from $P$. In the domain sphere choose a small open spherical cap around $h^{-1}(x)$ whose image $T$ lies in $W$; its complement is a closed $(n-1)$-disk. For $n=1$ choose the singleton cap, leaving the other point. In positive domain dimension, a rotation puts the cap at the north pole, and projection or stereographic coordinates identifies its complementary closed cap with a closed disk. Set $L=K\setminus T$. By step 2.2 the space $S^n\setminus L=P\cup Q\cup T$ is connected. But $P$ is open in it, is nonempty, and is closed in it: its boundary in $S^n$ lies in $K$ by step 2.1 and cannot meet $T\subset W$, so is contained in $L$. Its complement contains $Q$ and is nonempty. This is a separation, a contradiction. Therefore every point of $K$ lies in the closure of $P$; the same proof applies to $Q$. Together with step 2.1 this gives $\mathrm{\partial }P=\mathrm{\partial }Q=K$.

For the Euclidean version, start with a separate embedding $g:S^{n-1}\to {\mathbb{R}}^n$ and write $K_{\mathbb{R}}=g(S^{n-1})$. Let $\mathrm{\infty }=N=(0,\dots ,0,1)$ and identify $S^n\setminus \{N\}$ with ${\mathbb{R}}^n$ by ${\sigma }_N(u,t)=\frac{u}{1-t},{\sigma }_N^{-1}(v)=(\frac{2v}{1+\mathrm{\mid }v{\mathrm{\mid }}^2},\frac{\mathrm{\mid }v{\mathrm{\mid }}^2-1}{1+\mathrm{\mid }v{\mathrm{\mid }}^2})\mathrm{.}$ Direct substitution gives both inverse identities and norm one, while the denominators are nonzero on their domains, so these are continuous mutual inverses. The composite ${\sigma }_N^{-1}g$ is a spherical embedding with image $K^{\ast }={\sigma }_N^{-1}(K_{\mathbb{R}})$; by construction $K^{\ast }$ avoids $\mathrm{\infty }$. Apply steps 1.1–3.1 to this embedding, and denote by $P_{\mathrm{\infty }}$ the component of $S^n\setminus K^{\ast }$ containing $\mathrm{\infty }$, and by $Q$ the other. If $n\ge 2$, deleting $\mathrm{\infty }$ from $P_{\mathrm{\infty }}$ leaves a path-connected space. For two remaining points, start with a path between them in $P_{\mathrm{\infty }}$. Choose a small closed coordinate ball about $\mathrm{\infty }$ contained in $P_{\mathrm{\infty }}$ and avoiding the endpoints. If the path meets a still smaller closed concentric ball, let the first and last meeting times be the minimum and maximum of its closed preimage. Both meeting points are on that smaller ball's sphere, since the endpoints are outside it. Replace the intervening path segment by a path on that sphere. This sphere is path connected for $n\ge 2$: normalized straight segments join non-antipodal points, and for antipodal points insert a third unit vector not on their line. The replacement avoids $\mathrm{\infty }$; the portions before the first and after the last meeting also avoid it. If there were no meeting, use the original path. This proves the assertion. Under ${\sigma }_N$, the two Euclidean complementary components are ${\sigma }_N(P_{\mathrm{\infty }}\setminus \{\mathrm{\infty }\})$ and ${\sigma }_N(Q)$.

The component $P_{\mathrm{\infty }}$ contains a neighborhood of $\mathrm{\infty }$, whose stereographic image contains all points of sufficiently large norm: this follows directly from the inverse formula in step 4.1, whose last coordinate tends to $1$ as the norm tends to infinity. Thus ${\sigma }_N(P_{\mathrm{\infty }}\setminus \{\mathrm{\infty }\})$ is unbounded. The other component lies outside this neighborhood and its stereographic image is bounded by the same formula. The sphere boundaries of both components are $K^{\ast }$ by step 3.1. Since $K^{\ast }$ avoids $\mathrm{\infty }$, the homeomorphism ${\sigma }_N$ sends their finite-point boundaries to $K_{\mathbb{R}}$. For $n=1$, order the two Euclidean image points as $a<b$. A continuous path in the line cannot cross either missing point, by [F6], and each of the three indicated intervals is convex and therefore path connected. These are exactly the three Euclidean components. The middle interval has boundary $\{a,b\}$; the outer intervals have boundaries $\{a\}$ and $\{b\}$ respectively. In the sphere they join through infinity, as consistent with the spherical statement.

This proves all the asserted cases. An embedding has nonempty image, and properness was proved in step 1.1 rather than assumed. The coefficient group used for counting is $\mathbb{Z}$, so the zero coefficient ring cannot conceal the number of components. The case $n=1$ is explicitly separated where punctured coordinate spheres cease to be path connected; $n=0$ is outside this theorem's statement. No smoothness, local flatness or ball-closure conclusion was used. Homology computations retain degenerate simplices. The only AC use is [F5]'s inherited duality assumption; the component count, finite cap choice and single-path detour use no additional choice.