Proof:

We shall denote the ends of $\Delta_1$ by $a$ and $b$. If $\sigma:\Delta_1\longrightarrow X$ is a singular one simplex then $\partial _1\sigma = \sigma(b) - \sigma(a)$ which is obviously in ker$\;\epsilon$ and we conclude that $B_0(X)\subset$   ker$\;\epsilon$. To prove the reverse inclusion, let $\sigma$ be an arbitrary element of ker$\;\epsilon$ given by (31.1). That is, the coefficients satisfy $c_1 + c_2 + \dots + c_k = 0$. Pick any point $p \in X$ and for each $j$ let $\sigma_j:\Delta_1\longrightarrow X$ be a path in $X$ joining $p$ and $p_j$. We claim that $\sigma$ is the boundary of the one chain $\tau = c_1\sigma_1 + c_2\sigma_2 + \dots + c_k\sigma_k$.

$$\partial _1\tau = c_1(\sigma_1(b) - \sigma_1(a)) + c_2(\sigma_2(b) - \sigma_2(a)) + \dots + c_k(\sigma_k(b) - \sigma_k(a))$$

$$= (c_1p_1 + c_2p_2 + \dots + c_kp_k) - (c_1 + c_2 + \dots + c_k)p = \sigma.$$

The last part follows from the fundamental theorem on group homomorphisms.