I was wondering if anyone could offer some intuition for why [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality) holds. Of course, the proof is easy enough to check, and it is also easy to work out many examples by hand. However, I don't have any feeling for why it is true.


To give you an example of what I am looking for, when I think of Poincare duality I think of the picture in terms of triangulations and dual triangulations. Is there any picture like that for Alexander duality? Is there at least maybe some kind of obvious bilinear pairing between the two sides of it or something?



Intuition behind Alexander duality
Let $M$ be a closed orientable $n$-manifold containing the compact set $X$. 
Given an $n-q-1$-cocyle on $X$ (I am choosing this degree just to match with the notation
of the Wikipedia article to which you linked), we extend it to some small open neighbourhood $U$ of $X$.
By Lefschetz--Poincare duality on the open manifold $U$, we can convert this $n-q-1$-cocylce
into a Borel--Moore cycle (i.e. a locally-finite cycle made up of infinitely many simplices)
on $U$ of degree $q+1$. Throwing away those simplices lying in $U \setminus X$,
we obtain a usual (i.e. finitely supported) cycle giving a class in $H\_{q+1}(U,U\setminus X) = H\_{q+1}(M,M\setminus X)$ (the isomorphism holding via excision).
Alexander duality for an arbitrary manifold then states that
the map $H^{n-q-1}(X) \to H\_{q+1}(M,M \setminus X)$ is an isomorphism. (If $X$ is very pathological, then we should be careful in how define the left-hand side, to be sure
that every cochain actually extends to some neighbourhood of $X$.)


Now if $M = S^{n+1}$, then $H^i(S^{n+1})$ is almost always zero, and so we may use the boundary map for the long exact sequence of a pair to
identify $H\_{q+1}(S^{n+1}, S^{n+1}\setminus X)$ with $H\_{q}(S^{n+1}\setminus X)$ modulo worrying about reduced vs. usual 
homology/cohomology (to deal with the fact that $H^i(S^{n+1})$ is *non-zero* at the extremal
points $i = 0$ or $n$). 


So, in short: we take a cocycle on $X$, expand it slightly to a cocyle on $U$,
represent this by a Borel--Moore cycle of the appropriate degree, throw away those simplices lying entirely outside $X$, so that it is now a chain with boundary lying outside $X$, and finally *take this boundary*, which is now a cycle
in $S^{n+1} \setminus X$. 


(I found [these notes of Jesper Moller](http://www.math.ku.dk/~moller/f03/algtop/notes/homology.pdf) helpful in understanding the general structure of Alexander duality.)


One last thing: it might help to think this through in the case of a circle embedded in $S^2$. We should thicken the circle up slightly to an embedded strip. If we then take our cohomology class to be the generator of $H^1(S^1)$, the corresponding Borel--Moore cycle is just a longitudinal ray of the strip (i.e. if the strip is $S^1 \times I$, where $I$ is an open
interval, then the Borel--More cycle is just $\{\text{point}\} \times I$).


If we cut $I$ down to a closed subinterval $I'$ and then take its boundary, we get a pair
of points, which you can see intuitively will lie one in each of the components
of the complement of the $S^1$ in $S^2$. 


More rigorously, Alexander duality will show that these two points generate the reduced $H^0$ of the complement of the $S^1$, and this is how Alexander duality proves the Jordan curve theorem. Hopefully the above sketch supplies some geometric intuition to this argument.


