Proof:

Let $\lambda$ be a singular one cycle say $\lambda = \sum {n_j}\gamma_j,$ where $n_j\in \mathbb{Z}$ and $\gamma_j:[0, 1]\longrightarrow X$. Since $\partial\lambda = 0,$

$$\sum_{j=1}^k n_j(\gamma_j(1) - \gamma_j(0)) = 0. \eqno(32.7)$$

The idea is to complete each of the paths $\gamma_j$ into a loop at $x_0$ by means of paths joining $x_0$ to the ends $\gamma_j(0)$ and $\gamma_j(1)$. The only non-trivial part is the book-keeping which has to be done carefully. Let $S$ denote the set of endpoints

$$S = \{\gamma_j(1),\;\gamma_j(0)/\;j = 1, 2 \dots, k\}.$$

For each $p\in S$, if $m_p$ denotes the sum of the coefficients of $p$ in (32.7) then $m_p$ must be zero. Taking a path $\beta_p$ in $X$ joining $x_0$ and $p\in S$ we construct for each $j$ a loop $\eta_j$ in $X$ based at $x_0$ namely,

$$\eta_j = \beta_{\gamma_j(0)}*\gamma_j * \beta^{-1}_{\gamma_j(1)}.$$

Finally

$$\Pi_X(\eta_1^{n_1}*\eta_2^{n_2}*\cdots *\eta_k^{n_k}) = \sum_{j=1... ...amma_j -\sum_{j=1}^k n_j(\beta_{\gamma_j(1)} - \beta_{\gamma_j(0)}) = \lambda$$

since

$$\sum_{j=1}^k n_j(\beta_{\gamma_j(1)} - \beta_{\gamma_j(0)}) = \sum_{p\in S} m_p\beta_p = 0.$$