The coeffcients of a generator of $H_0(X)$ sum to $\pm 1$? I'm reading Theorem 4.14 (p. 70) of Rotman's Intro to Algebraic Topology. He proves that if $X$ is a nonempty path connected space, then $H_0(X)\simeq\mathbb{Z}$, and if $x_0,x_1\in X$, then $cls(x_0)=cls(x_1)$ is a generator of $H_0(X)$. Here $cls(x)$ is the homology class of $x$ in $H_0(X)$. 
He writes

If $cls(\gamma)$ is a generator of $H_0(X)$, where $\gamma=\sum m_ix_i$, then $\sum m_i=\pm 1$. 

Why is this? I don't follow this step in his proof.

$\DeclareMathOperator{\cls}{cls}$Let $\epsilon : C_0(X) \to \mathbb{Z}$ be defined by $\epsilon(\sum m_i x_i) = \sum m_i$. This is clearly a homomorphism, and it vanishes on boundaries (easy check), so it yields $\epsilon : H_0(X) \to \mathbb{Z}$. Suppose that $\cls(\gamma)$ is a generator of $H_0(X)$ and let $n = \epsilon(\cls(\gamma))$. Let $x \in X$ be some point. Then since $\cls(\gamma)$ is a generator, there is some $k \in \mathbb{Z}$ such that
$$\cls(x) = k \cls(\gamma) \Rightarrow \epsilon(\cls(x)) = k \epsilon(\cls(\gamma)) \Rightarrow 1 = kn$$
The only invertible elements of $\mathbb{Z}$ are $\pm 1$, therefore $n = \sum m_i = \pm 1$.
