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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 $Z_1/B_1 \cong \Pi_1(X)_{ab}$.

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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.

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  • $\begingroup$ 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
    Commented Aug 22, 2012 at 5:00
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    $\begingroup$ @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
    Commented Aug 22, 2012 at 5:46

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