# On a special case of Alexander duality

Let $S^n$ be the $n$-dimensional sphere and let $K\subseteq S^n$ be a compact, locally contractible subspace of real codimension $\geq 2$. Applying Alexander duality we find that

$$\tilde{H}_{i}(S^n-K)\simeq \tilde{H}^{n-i-1}(K)$$ where $\tilde{H}$ denotes the reduced homology (cohomology) with coefficients in $\mathbf{Z}$ and $0\leq i\leq n-1$. In particular, taking $i=0$ we find that $$\tilde{H}_{0}(S^n-K)\simeq \tilde{H}^{n-1}(K)=0.$$ Thus $S^n-K$ is connected and therefore path connected (since $S^n-K$ is a euclidian open set and therefore locally path connected).

Q: Is there a simple (low-tech and/or geometrical proof) that $S^n-K$ is path connected ?

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If everything is smooth, you get a simple proof that the complement is path connected from transversality -- generically a path in the sphere does not intersect the set $K$. If the co-dimension was at least $3$, you could similarly deduce that the complement is simply connected. These techniques are well written up in Guillemin and Pollack's "Differential Topology" text. – Ryan Budney Aug 2 '11 at 16:42
Or Milnor's "topology from a differentiable viewpoint", which is more fun to read than G&P. – Igor Rivin Aug 2 '11 at 16:55
What do you mean by "subspace of real codimension >= 2?" Does it mean that dim(K)<= n-2, where 'dim' is the covering dimension? – t22 Aug 2 '11 at 17:36
If you don't assume smooth (or something weaker, like Lipschitz or "tame"=tubular neighborhood), you should not expect an easy proof because closely related results are difficult or false. The Jordan curve theorem is related to codimension 1. The double suspension theorem shows that a circle in a large sphere need not have simply connected complement, so general position arguments are tricky for topological manifolds. en.wikipedia.org/wiki/Double_suspension_theorem – Ben Wieland Aug 2 '11 at 18:14

Since you're asking for a geometric proof, let's just work with manifolds. Then the general claim is that if $N\subset M$ is a codimension 2 submanifold of a connected manifold, then $M\setminus N$ is connected. Iterating the claim, we can reduce to the case when $N$ is also connected.

By the tubular neighborhood theorem, we reduce$^{\dagger}$ to showing that the (total space of the) normal bundle of $N$ in $M$ minus the zero section is connected. I.e., we reduce to the case of showing that a vector bundle of rank $\geq 2$ minus its zero section is connected.

This is clearly true, but I'll include one possible proof for completeness' sake. Take two points in the vector bundle $\mathcal{E}$ over $N$ which don't lie in the zero section. Take a smooth simple path in $N$ between their projections, and then take a tubular neighborhood of this path. This neighborhood is contractible and $\mathcal{E}$ is a vector bundle over it, so it's a trivial vector bundle. Therefore, we reduce to showing that $\mathbb{R}^n$ minus a codimension 2 linear subspace is connected, which is unobjectionable.

$^{\dagger}$ Here's how you do the reduction. Take two points in $M\setminus N$ and a path between them in $M$. If this path doesn't go through $N$, we're certainly in good shape. Otherwise, choose two points on the path in the neighborhood, one "before" the path crosses through $N$, one "after" (i.e., after it has finished passing through $N$). Since the tubular neighborhood is diffeomorphic to the normal bundle of $N$ in $M$, we can choose a path in this tubular neighborhood which doesn't cross $N$, and modifying our original path by this procedure, we get the result.

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Thanks Moosbrugger for the nice argument. – Hugo Chapdelaine Aug 3 '11 at 19:18
I want to add a more elementary argument: Suppose $x,y\in M\setminus N$.Take any path $p\colon I\to M$ with $p(0)=x$ and $p(1)=y$. We can firstly perturb it into a path such that $\{\,t\in I\,\vert\,p(t)\in N\}$ is finite (Hint: partition $I$ so that the image of each piece is contained in a coordinate neighborhood of $(M,N)$), then a further perturbation clears all such $t\in I$. – Frank Science Feb 23 at 18:30

The proof that $S^n \setminus K$ is path connected follows directly from general position if $K$ is, say, a finite simplicial complex of codimension two (I don't know what "codimension two" means in your general formulation).

If we take two points $x,y$ not in $K$, and we take a generic path in $S^n$ connecting them, then general position taken with respect to each simplex of $K$ guarantees that a generic path won't intersect $K$.

Addendum: a more general fact is true: if $K \subset S^n$ is a simplicial complex of codimension $m$, then general position implies that the complement $S^n \setminus K$ is $(m-2)$-connected (the proof is similar).

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If you don't make some kind of assumption, like that $K$ is a subcomplex of a subdivision of the usual triangulation of the sphere, the addendum is not true, as the double suspension theorem gives a pathological embedding of the circle in a large sphere. – Ben Wieland Aug 20 '11 at 2:37
Yes, I'm fully aware of that. I should have specified that K is a embedded in a locally flat way. – John Klein Aug 21 '11 at 1:54