This section is dual to Section 13.18. We give definitions and state results, but we do not reprove the lemmas.
Definition 13.19.1. Let $\mathcal{A}$ be an abelian category. Let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. An projective resolution of $A$ is a complex $P^\bullet $ together with a map $P^0 \to A$ such that:
We have $P^ n = 0$ for $n > 0$.
Each $P^ n$ is an projective object of $\mathcal{A}$.
The map $P^0 \to A$ induces an isomorphism $\mathop{\mathrm{Coker}}(d^{-1}) \to A$.
We have $H^ i(P^\bullet ) = 0$ for $i < 0$.
Hence $P^\bullet \to A[0]$ is a quasi-isomorphism. In other words the complex
is acyclic. Let $K^\bullet $ be a complex in $\mathcal{A}$. An projective resolution of $K^\bullet $ is a complex $P^\bullet $ together with a map $\alpha : P^\bullet \to K^\bullet $ of complexes such that
We have $P^ n = 0$ for $n \gg 0$, i.e., $P^\bullet $ is bounded above.
Each $P^ n$ is an projective object of $\mathcal{A}$.
The map $\alpha : P^\bullet \to K^\bullet $ is a quasi-isomorphism.
Lemma 13.19.2. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be a complex of $\mathcal{A}$.
If $K^\bullet $ has a projective resolution then $H^ n(K^\bullet ) = 0$ for $n \gg 0$.
If $H^ n(K^\bullet ) = 0$ for $n \gg 0$ then there exists a quasi-isomorphism $L^\bullet \to K^\bullet $ with $L^\bullet $ bounded above.
Proof. Dual to Lemma 13.18.2. $\square$
Lemma 13.19.3. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough projectives.
Any object of $\mathcal{A}$ has a projective resolution.
If $H^ n(K^\bullet ) = 0$ for all $n \gg 0$ then $K^\bullet $ has a projective resolution.
If $K^\bullet $ is a complex with $K^ n = 0$ for $n > a$, then there exists a projective resolution $\alpha : P^\bullet \to K^\bullet $ with $P^ n = 0$ for $n > a$ such that each $\alpha ^ n : P^ n \to K^ n$ is surjective.
Proof. Dual to Lemma 13.18.3. $\square$
Lemma 13.19.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be an acyclic complex. Let $P^\bullet $ be bounded above and consisting of projective objects. Any morphism $P^\bullet \to K^\bullet $ is homotopic to zero.
Proof. Dual to Lemma 13.18.4. $\square$
Remark 13.19.5. Let $\mathcal{A}$ be an abelian category. Suppose that $\alpha : K^\bullet \to L^\bullet $ is a quasi-isomorphism of complexes. Let $P^\bullet $ be a bounded above complex of projectives. Then is an isomorphism. This is dual to Remark 13.18.5.
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet ) \]
Lemma 13.19.6. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram where $P^\bullet $ is bounded above and consists of projective objects, and $\alpha $ is a quasi-isomorphism.
There exists a map of complexes $\beta $ making the diagram commute up to homotopy.
If $\alpha $ is surjective in every degree then we can find a $\beta $ which makes the diagram commute.
\[ \xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_\beta } \]
Proof. Dual to Lemma 13.18.6. $\square$
Lemma 13.19.7. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram where $P^\bullet $ is bounded above and consists of projective objects, and $\alpha $ is a quasi-isomorphism. Any two morphisms $\beta _1, \beta _2$ making the diagram commute up to homotopy are homotopic.
\[ \xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_{\beta _ i} } \]
Proof. Dual to Lemma 13.18.7. $\square$
Lemma 13.19.8. Let $\mathcal{A}$ be an abelian category. Let $P^\bullet $ be bounded above complex consisting of projective objects. Let $L^\bullet \in K(\mathcal{A})$. Then
\[ \mathop{\mathrm{Mor}}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet ) = \mathop{\mathrm{Mor}}\nolimits _{D(\mathcal{A})}(P^\bullet , L^\bullet ). \]
Proof. Dual to Lemma 13.18.8. $\square$
Lemma 13.19.9. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough projectives. For any short exact sequence $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ of $\text{Comp}^{+}(\mathcal{A})$ there exists a commutative diagram in $\text{Comp}^{+}(\mathcal{A})$ where the vertical arrows are projective resolutions and the rows are short exact sequences of complexes. In fact, given any projective resolution $P^\bullet \to C^\bullet $ we may assume $P_3^\bullet = P^\bullet $.
\[ \xymatrix{ 0 \ar[r] & P_1^\bullet \ar[r] \ar[d] & P_2^\bullet \ar[r] \ar[d] & P_3^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 } \]
Proof. Dual to Lemma 13.18.9. $\square$
Lemma 13.19.10. Let $\mathcal{A}$ be an abelian category. Let $P^\bullet $, $K^\bullet $ be complexes. Let $n \in \mathbf{Z}$. Assume that
$P^\bullet $ is a bounded complex consisting of projective objects,
$P^ i = 0$ for $i < n$, and
$H^ i(K^\bullet ) = 0$ for $i \geq n$.
Then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(P^\bullet , K^\bullet ) = 0$.
Proof. The first equality follows from Lemma 13.19.8. Note that there is a distinguished triangle
by Remark 13.12.4. Hence, by Lemma 13.4.2 it suffices to prove $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , \tau _{\leq n - 1}K^\bullet ) = 0$ and $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , \tau _{\geq n} K^\bullet ) = 0$. The first vanishing is trivial and the second is Lemma 13.19.4. $\square$
Lemma 13.19.11. Let $\mathcal{A}$ be an abelian category. Let $\beta : P^\bullet \to L^\bullet $ and $\alpha : E^\bullet \to L^\bullet $ be maps of complexes. Let $n \in \mathbf{Z}$. Assume
$P^\bullet $ is a bounded complex of projectives and $P^ i = 0$ for $i < n$,
$H^ i(\alpha )$ is an isomorphism for $i > n$ and surjective for $i = n$.
Then there exists a map of complexes $\gamma : P^\bullet \to E^\bullet $ such that $\alpha \circ \gamma $ and $\beta $ are homotopic.
Proof. Consider the cone $C^\bullet = C(\alpha )^\bullet $ with map $i : L^\bullet \to C^\bullet $. Note that $i \circ \beta $ is zero by Lemma 13.19.10. Hence we can lift $\beta $ to $E^\bullet $ by Lemma 13.4.2. $\square$
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