Zeroth-homology of a complex of $n$ connected components

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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

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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 $g_i-h_i$.