The Stacks project

Proof. If $n = 1$, then $K$ is a direct summand in $D(R)$ of a bounded complex $P^\bullet $ whose terms are finite free $R$-modules and whose differentials are zero. Thus it suffices to show any morphism $f : P^\bullet \to K_1$ in $D(R)$ with $H^ i(f) = 0$ for all $i$ is zero. Since $P^\bullet $ is a finite direct sum $P^\bullet = \bigoplus R[m_ j]$ it suffices to show any morphism $g : R[m] \to K_1$ with $H^{-m}(g) = 0$ in $D(R)$ is zero. This follows from the fact that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(R[-m], K) = H^ m(K)$.

For $n > 1$ we proceed by induction on $n$. Namely, we know that $K$ is a summand in $D(R)$ of an object $P$ which sits in a distinguished triangle

with $P' \in \langle R \rangle _1$ and $P'' \in \langle R \rangle _{n - 1}$. As above we may replace $K$ by $P$ and assume that we have

in $D(R)$ with $f_ j$ zero on cohomology. By the case $n = 1$ the composition $f_1 \circ i$ is zero. Hence by Derived Categories, Lemma 13.4.2 we can find a morphism $h : P'' \to K_1$ such that $f_1 = h \circ p$. Observe that $f_2 \circ h$ is zero on cohomology. Hence by induction we find that $f_ n \circ \ldots \circ f_2 \circ h = 0$ which implies $f_ n \circ \ldots \circ f_1 = f_ n \circ \ldots \circ f_2 \circ h \circ p = 0$ as desired. $\square$