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:
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
\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
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
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.
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.
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{\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})
\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
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
(\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
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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