**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 $F^ i$ projective 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). If (2)(b) holds, then $a^\bullet $ is homotopic to a map of complexes $(a')^\bullet : M^\bullet \to K^\bullet $ which is zero in degree $-1$. On the other hand, let $N \to I^\bullet $ be an injective resolution. We have

\[ \mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , N) = \mathop{\mathrm{Hom}}\nolimits _{D(R)}(K^\bullet , I^\bullet [1]) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(K^\bullet , I^\bullet [1]) \]

by Derived Categories, Lemma 13.18.8. Let $b^\bullet : K^\bullet \to I^\bullet [1]$ be a map of complexes. Since $K^1 = 0$ the map $b^0 : K^0 \to I^1$ maps into the kernel of $I^1 \to I^2$ which is the image of $I^0 \to I^1$. Since $K^0$ is projective we can lift $b^0$ to a map $h : K^0 \to I^0$. Thus we see that $b^\bullet $ is homotopic to a map of complexes $(b')^\bullet $ with $(b')^0 = 0$. Since $K^ i = 0$ for $i \not\in [-1, 0]$ it follows that $(b')^\bullet \circ (a')^\bullet = 0$ as a map of complexes. Hence the map $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M^\bullet , N)$ is zero. In this way we see that (2)(b) implies (2)(a). Conversely, assume (2)(a). We see that the canonical element in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , K^{-1})$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M^\bullet , K^{-1})$. Using (1) we see immediately that we get a map $h^0$ as in (2)(b).

Proof of (3). 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 (3) holds for these factorizations and $a^0 = b^0$.

Proof of (4) 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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