Lemma 86.3.4. Let $R$ be a ring. Let $M^\bullet$ be a complex of modules over $R$ with $M^ i = 0$ for $i > 0$ and $M^0$ a projective $R$-module. Let $K^\bullet$ be a second complex.

1. If $K^ i = 0$ for $i \leq -2$, then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(M^\bullet , K^\bullet )$,

2. If $K^ i = 0$ for $i \leq -3$ and $\alpha \in \mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet )$ composed with $K^\bullet \to K^{-2}[2]$ comes from an $R$-module map $a : M^{-2} \to K^{-2}$ with $a \circ d_ M^{-3} = 0$, then $\alpha$ can be represented by a map of complexes $a^\bullet : M^\bullet \to K^\bullet$ with $a^{-2} = a$.

3. In (2) for any second map of complexes $(a')^\bullet : M^\bullet \to K^\bullet$ representing $\alpha$ with $a = (a')^{-2}$ there exist $h' : M^0 \to K^{-1}$ and $h : M^{-1} \to K^{-2}$ such that

$h \circ d_ M^{-2} = 0, \quad (a')^{-1} = a^{-1} + d_ K^{-2} \circ h + h' \circ d_ M^{-1},\quad (a')^0 = a^0 + d_ K^{-1} \circ h'$

Proof. Set $F^0 = M^0$. Choose a free $R$-module $F^{-1}$ and a surjection $F^{-1} \to M^{-1}$. Choose a free $R$-module $F^{-2}$ and a surjection $F^{-2} \to M^{-2} \times _{M^{-1}} F^{-1}$. Continuing in this way we obtain a quasi-isomorphism $p^\bullet : F^\bullet \to M^\bullet$ which is termwise surjective and with $F^ i$ free for all $i$.

Proof of (1). By Derived Categories, Lemma 13.19.8 we have

$\mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(F^\bullet , K^\bullet )$

If $K^ i = 0$ for $i \leq -2$, then any morphism of complexes $F^\bullet \to K^\bullet$ factors through $p^\bullet$. Similarly, any homotopy $\{ h^ i : F^ i \to K^{i - 1}\}$ factors through $p^\bullet$. Thus (1) holds.

Proof of (2). Choose $b^\bullet : F^\bullet \to K^\bullet$ representing $\alpha$. The composition of $\alpha$ with $K^\bullet \to K^{-2}[2]$ is represented by $b^{-2} : F^{-2} \to K^{-2}$. As this is homotopic to $a \circ p^{-2} : F^{-2} \to M^{-2} \to K^{-2}$, there is a map $h : F^{-1} \to K^{-2}$ such that $b^{-2} = a \circ p^{-2} + h \circ d_ F^{-2}$. Adjusting $b^\bullet$ by $h$ viewed as a homotopy from $F^\bullet$ to $K^\bullet$, we find that $b^{-2} = a \circ p^{-2}$. Hence $b^{-2}$ factors through $p^{-2}$. Since $F^0 = M^0$ the kernel of $p^{-2}$ surjects onto the kernel of $p^{-1}$ (for example because the kernel of $p^\bullet$ is an acyclic complex or by a diagram chase). Hence $b^{-1}$ necessarily factors through $p^{-1}$ as well and we see that (2) holds for these factorizations and $a^0 = b^0$.

Proof of (3) is omitted. Hint: There is a homotopy between $a^\bullet \circ p^\bullet$ and $(a')^\bullet \circ p^\bullet$ and we argue as before that this homotopy factors through $p^\bullet$. $\square$

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