3
$\begingroup$

Let $X$ be a connected CW-complex. It's easy to see that the 1-cycles (of the cellular chain complex) $\sigma=\sum_i \varepsilon_i\sigma_i\;(\varepsilon_i=\pm 1)$ are linear combinations of 1-cells $\sigma_i$ that (considered as paths with direction given by $\varepsilon_i$) can be composed such that the composition is a loop.

Question: Is there also a nice description of the 1-boundaries $B_1$ that allows to show $Z_1/B_1\cong \Pi_1(X)_{ab}$ ?

According to the cellular boundary formula (cf. Hatcher, p. 140), the boundary operator of a 2-cell is the linear combination of whose 1-cells that form the boundary of the 2-cell, weighted with some mapping degree of the attaching map. But from this description I find it impossible to see $Z_1/B_1 \cong \Pi_1(X)_{ab}$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

It's easy to write a presentation of $\pi_1$ using the attaching maps of the $2$-cells, assuming that you have a single $0$-cell, the base point. (The 1-skeleton is homotopic to a wedge of circles.) You have one generator for each 1-cell, and one relation for each 2-cell. Abelianizing the relations you obtain the boundary operator of the cellular chain complex, hence the isomorphism you sought.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What happens if there are several 0-cells ? I know that $X$ is always homotopy equivalent to a complex with one 0-cell, but isomorphism in question holds for any CW-structure. $\endgroup$
    – Ralph
    Aug 22, 2012 at 5:00
  • 3
    $\begingroup$ @Ralph: If you want to allow more vertices, just pick a maximal subtree $T$ in 1-skeleton and work "modulo" $T$. Then generators will be 1-cells connecting vertices in $T$, relators are products of generators and their inverses, corresponding to boundaries of 2-cells, etc. However, this assumes that you know how to compute $\pi_1$ using Van Kampen theorem. $\endgroup$
    – Misha
    Aug 22, 2012 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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