Proposition

Let $S$ be the set of all singular $n$-cubes, and $\mathbb{Z}$ the integers. An $n$-chain is a function $f: S \rightarrow \mathbb{Z}$ such that $f(c)$ for all but finitely many $c$. Define $f + g$ and $nf$ by $(f + g)(c) = f(c) + g(c)$ and $nf(c) = n \cdot f(c)$. show that $f + g$ and $nf$ are $n$-chains if $f$ and $g$ are. If $c \in S$, let $c$ also denote the function $f$ such that $f(c) = 1$ and $f(c’) = 0$ for $c’ \ne c$. Show that every $n$-chain $f$ can be written $a_1c_1 + \cdots + a_kc_k$ for some integers $a_1, \cdots, a_k$ and singular $n$-cubes $c_1, \cdots, c_k$.

Solution

$f(c) = 0$ for all but finitely many $c$, and $g(c) = 0$ for all but finitely many $c$. Therefore, $(f + g)(c) = 0$ for all but finitely many $c$.

$nf(c) \ne 0$ only if $f(c) \ne 0$. Thus $nf(c) = 0$ for all but finitely many $c$.

Therefore, $f + g$ and $nf$ are both $n$-chains.

Let $c_1, \cdots, c_n \in S$ be the all the elements in $S$ such that $f(c) \ne 0$. Let $a_i = f(c_i)$ for each $i$. Let $g = a_1c_1 + \cdots + a_nc_n$. (Note that $c_i$ here means the function $c_i \mapsto 1$.) For all $i$, $f(c_i) = a_i = g(c_i)$. For all $c \in S \setminus \{ c_1, \cdots, c_n \}$, $f(c) = 0 = g(c)$. Thus $f = a_1c_1 + \cdots + a_nc_n$.