## 13.19 Projective resolutions

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:

1. We have $P^ n = 0$ for $n > 0$.

2. Each $P^ n$ is an projective object of $\mathcal{A}$.

3. The map $P^0 \to A$ induces an isomorphism $\mathop{\mathrm{Coker}}(d^{-1}) \to A$.

4. 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

$\ldots \to P^{-1} \to P^0 \to A \to 0 \to \ldots$

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

1. We have $P^ n = 0$ for $n \gg 0$, i.e., $P^\bullet$ is bounded above.

2. Each $P^ n$ is an projective object of $\mathcal{A}$.

3. 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 $\mathcal{A}$.

1. If $K^\bullet$ has a projective resolution then $H^ n(K^\bullet ) = 0$ for $n \gg 0$.

2. 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.

1. Any object of $\mathcal{A}$ has a projective resolution.

2. If $H^ n(K^\bullet ) = 0$ for all $n \gg 0$ then $K^\bullet$ has a projective resolution.

3. 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

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet )$

is an isomorphism. This is dual to Remark 13.18.5.

Lemma 13.19.6. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

$\xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_\beta }$

where $P^\bullet$ is bounded above and consists of projective objects, and $\alpha$ is a quasi-isomorphism.

1. There exists a map of complexes $\beta$ making the diagram commute up to homotopy.

2. If $\alpha$ is surjective in every degree then we can find a $\beta$ which makes the diagram commute.

Proof. Dual to Lemma 13.18.6. $\square$

Lemma 13.19.7. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

$\xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_{\beta _ i} }$

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.

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

$\mathop{Mor}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet ) = \mathop{Mor}\nolimits _{D(\mathcal{A})}(P^\bullet , L^\bullet ).$

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 $\text{Comp}^{+}(\mathcal{A})$

$\xymatrix{ 0 \ar[r] & P_1^\bullet \ar[r] \ar[d] & P_2^\bullet \ar[r] \ar[d] & P_3^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 }$

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 $P_3^\bullet = P^\bullet$.

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

1. $P^\bullet$ is a bounded complex consisting of projective objects,

2. $P^ i = 0$ for $i < n$, and

3. $H^ i(K^\bullet ) = 0$ 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

$(\tau _{\leq n - 1}K^\bullet , K^\bullet , \tau _{\geq n}K^\bullet , f, g, h)$

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

1. $P^\bullet$ is a bounded complex of projectives and $P^ i = 0$ for $i < n$,

2. $H^ i(\alpha )$ 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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