Zeroth-homology of a complex of $n$ connected components I am new at Algebraic topology and am reading Basic Concepts of Algebraic Topology of Croom.
I have a question. In Theorem 2.4/page 25, it states that if $K$ is a complex with $n$ connected components, then $H_0(K)$ is isomorphic to $\mathbb{Z}^n$.
Anyway, in the proof, Croom showed that "Applying this result to each connected component $K_1,..., K_n$ 
 of $K$,
there is a vertex $a_i$ 
 of $K_i$
 such that any $0$-cycle on $K$
 is homologous to a $0$-chain
of the form $\sum h_i \langle a_i \rangle$ where $h_i$ is a integer and $\langle a_i \rangle$ denotes the $0$-cycle that maps $(a_i)$ to 1 and other $0$-simplicies to $0$.
Hence it suffices to show that the representation here is unique, which means if we have $\sum (g_i- h_i) \langle a_i \rangle=\partial (c)$, then $g_i=h_i$. This part is clearly trivial to Croom, but I do not understand. 
Can you clarify this part for me? Thank u

The boundary of a $1$-chain is a linear combination of
boundaries of $1$-simplex. A $1$-simplex in $K$ has endpoints in the
same component of $K$, so in the same $K_i$. If $\partial(c)=\sum_v r_v
\langle v\rangle$ then the sum of the $r_v$
over the vertices $v$ in the same $K_i$ is zero. In your example this sum
is $g_i-h_i$.
