homology with compact supports In one of the exercises in McDuff and Salamon, they mention homology with compact supports.  I know how to define *co*homology with compact supports, but I can't picture the homology version.  How do I say that a chain has compact support?  If I use singular chains, don't they all have compact support anyway?
Google isn't a big help here, so any references would really help me out. 
Also, I've added some basic sub-questions that would also help me out tremendously.  This must all be pretty simple, but my background in algebraic topology is weak and it completely baffles me!


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*McDuff-Salamon go on to state that for the open annulus $(1/2 < r < 1)$ in the plane, the first compact homology group is generated by the arc $\theta = 0$, $1/2 < r < 1$, which I can understand with hindsight: this is just the generator of the homology rel. the boundary and most likely there will be an isomorphism between the compact and relative homologies, just like there is one between the compact and relative co-homologies.

*They also implicitly use an isomorphism between the compact homology and compact cohomology in certain dimensions.  Should I just use this as the definition for the compact homology?  I.e.  $H_{k, c} = (H^k_c)^\ast$?  
 A: I think what McDuff-Salamon call homology with compact support is more commonly known as homology of infinite chains. The chains are formal infinite sums of singular simplices that are locally finite in the sense that each compact subset meets only finitely many singular simplices. The boundary is defined in the usual way.
Note that the usual singular homology are with compact support: the cycles have compact images. By contrast, the usual singular cohomology do not have compact support as a cocycle may take nonzero value on a sequence of cycles that run off to infinity. There is a book by Massey, "Homology and cohomology theory. An approach based on Alexander-Spanier cochains", where these matters are discussed in a very general setting. 
A: To elaborate on Ekedahl's comment:
It will be easiest to describe things for a triangulated space, so I can work with simplicial chains and cochains.  (But my space could be infinitely triangulated; e.g. think of Escher's famous picture of the infinitely triangulated hyperbolic plane.  I will also assume that my triangulation is locally finite.)
As you observed, usual chains have compact support.  This is why you can pair them with
cochains (which have arbitrary support).  Borel--Moore chains can have unbounded support.  These can be paired with not with arbitrary cochains, but only with compactly supported ones.  
So Borel--Moore homology is the "homology analogue" of compactly supported cohomology.
(But the support conditions are reversed, since homology is dual to cohomology.)
One can often interpret Borel--Moore homology as relative homology.  E.g. if $M$ is a compact manifold
with boundary $\partial M$, then the Borel--Moore homology of $M\setminus \partial M$ (the interior of $M$) is the same as the homology of the pair $(M,\partial M)$.
A: See http://eom.springer.de/H/h047870.htm
A: For any manifold (compact or not), compactly supported cohomology is Poincare dual to  (ordinary) homology, via capping with the fundamental class, which is an infinite chain (i.e the sum of all the top simplices in a triangulation).  Likewise, (ordinary) cohomology is Poincare dual to homology with locally finite infinite chains. In notation, $ H^{n-k}_{comp}(X)\cong H_{k}(X)$ and $H^{n-k}(X)\cong H_{k, inf}(X) $. 
