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On p.33 of Sergei Matveev's "Lectures on Algebraic Topology", the relative chain group of a pair $(K,L)$, where $L \subset K$ is a subcomplex, is defined as the free abelian group on simplices with interiors in $K \setminus L$. How is this different from just the chains on $K \setminus L$? How is this the same notion as $C_n(K)/C_n(L)$, i.e. chains in $K$ modded out by chains in $L$, which is another definition I have seen for the relative chain groups?

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The interior of the boundary of a simplex with interior in $K/L$ need not lie in $K/L$. As a result, the boundary map on the relative chain group $(K,L)$ (in your notation) has to be defined such that the (homological) boundary of a simplex includes only the pieces of the (geometric) boundary that don't lie in $L$. In this manner, we get the same thing as $C_*(K)/C_*(L)$.

Strictly speaking, $K/L$ is not a simplicial complex, so it doesn't make sense to consider simplicial chains on $K/L$. It does make sense to consider singular chains. In that case $H_*(K,L)$ and $H_*(K-L)$ are generally different. (As a very simple example, take the index to be zero, and consider the case where $L$ disconnects the space $K$.)

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Ah yes, I'd convinced myself that the interior of the boundary of a simplex with interior in K/L would itself have to be in K/L (to be more precise: the closure of K/L, but it is still false, of course). The disconnecting example furnishes an obvious counterexample. Thanks! –  Tony Dec 31 '10 at 4:16
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