Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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 $K$.)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.