Proof:

Since the $\Pi_X$ is a map into an abelian group, its kernel contains the commutator subgroup. To prove the converse suppose that $\gamma$ is a loop based at $x_0$ such that $[\gamma]\in$   Ker$\;\Pi_X$. When considered as a singular one cycle it is a boundary of a singular two chain $\sum n_j\sigma_j$ where $\sigma_j:\Delta_2\longrightarrow X$. Writing the boundary $\partial\sigma_j$ as a sum of its faces

$$\partial\sigma_j = \lambda_j + \mu_j + \nu_j$$

we see that

$$\sum_{j=1}^kn_j\partial\sigma_j = \sum_{j=1}^k n_j(\lambda_j + \mu_j + \nu_j) = \gamma.\eqno(32.8)$$

We proceed as in lemma (32.4). Let $S$ be the set distinct singular one simplicies in the list

$$\lambda_j, \mu_j, \nu_j\quad j = 1, 2, \dots, k. \eqno(32.9)$$

and choose auxiliary paths $\beta_p$ joining $x_0$ and the endpoints $p$ of each of the one simplicies in $S$. The loop $\gamma$ also appears in the list (32.9) but since its ends are both $x_0$ there is no need to take the auxiliary paths $\beta$ in this case.

Figure:

[width=0.7]GKSBook/fig__/fig__.eps

As in lemma (32.4), for each $\theta$ in the list (32.9), we denote by $m_{\theta}$ the sum of the coefficients of $\theta$ in (32.8) so that,

$$m_{\theta} = \left\{\begin{array}{lll} 0 & & \mbox{ if } \theta \... ...mma \\ 1 & & \mbox{ if } \theta = \gamma \\ \end{array} \right. \eqno(32.10)$$

For each two simplex $\sigma_j$ we have the three loops (suppressing the subscript $j$)

$$\beta_{\lambda(0)}*\lambda *\beta^{-1}_{\lambda(1)},\; \beta_{\mu(0)}*\mu *\beta^{-1}_{\mu(1)},\; \beta_{\nu(0)}*\nu *\beta^{-1}_{\nu(1)},\;$$

whose juxtaposition $\eta_j$ is easily seen to be homotopic to the trivial loop. For proving this one uses the equations $\lambda(1) = \mu(0)$, $\mu(1) = \nu(0)$ and $\nu(1) = \lambda(0)$. Corresponding to (32.8) we form the loop

$$\eta_1^{n_1}*\eta_2^{n_2}*\cdots *\eta_k^{n_k}*\gamma^{-1} \eqno(32.11)$$

which is homotopic to $\gamma^{-1}$ since the piece $\eta_1^{n_1}*\eta_2^{n_2}*\cdots *\eta_k^{n_k}$ is a juxtaposition of loops homotopic to the constant loop. On the other hand if we write out the expression (32.11) completely, we see that for each $\theta$ in the list (32.9), the factor $\beta*\theta *\beta^{-1}$ appears, probably in several positions, but the sum of its exponents is $m_{\theta}$. In view of (32.10) and lemma (30.5) we see that the element of $\pi_1(X, x_0)$ represented by (32.11) lies in the commutator subgroup, that is to say, $[\gamma]^{-1}$ lies in the commutator subgroup of $\pi_1(X, x_0)$. The proof is complete.