Lemma 15.126.2. Let $R$ be a ring. Let $F^\bullet $ be a bounded above complex of free $R$-modules. Given pairs $(n_ i, f_ i)$, $i = 1, \ldots , N$ with $n_ i \in \mathbf{Z}$ and $f_ i \in F^{n_ i}$ there exists a subcomplex $G^\bullet \subset F^\bullet $ containing all $f_ i$ which is bounded and consists of finite free $R$-modules.

**Proof.**
By descending induction on $a = \min (n_ i; i = 1, \ldots , N)$. If $F^ n = 0$ for $n \geq a$, then the result is true with $G^\bullet $ equal to the zero complex. In general, after renumbering we may assume there exists an $1 \leq r \leq N$ such that $n_1 = \ldots = n_ r = a$ and $n_ i > a$ for $i > r$. Choose a basis $b_ j, j \in J$ for $F^ a$. We can choose a finite subset $J' \subset J$ such that $f_ i \in \bigoplus _{j \in J'} Rb_ j$ for $i = 1, \ldots , r$. Choose a basis $c_ k, k \in K$ for $F^{a + 1}$. We can choose a finite subset $K' \subset K$ such that $\text{d}_ F^ a(b_ j) \in \bigoplus _{k \in K'} Rc_ k$ for $j \in J'$. Then we can apply the induction hypothesis to find a subcomplex $H^\bullet \subset F^\bullet $ containing $c_ k \in F^{a + 1}$ for $k \in K'$ and $f_ i \in F^{n_ i}$ for $i > r$. Take $G^\bullet $ equal to $H^\bullet $ in degrees $> a$ and equal to $\bigoplus _{j \in J'} Rb_ j$ in degree $a$.
$\square$

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