13.12 The canonical delta-functor
The derived category should be the receptacle for the universal cohomology functor. In order to state the result we use the notion of a \delta -functor from an abelian category into a triangulated category, see Definition 13.3.6.
Consider the functor \text{Comp}(\mathcal{A}) \to K(\mathcal{A}). This functor is not a \delta -functor in general. The easiest way to see this is to consider a nonsplit short exact sequence 0 \to A \to B \to C \to 0 of objects of \mathcal{A}. Since \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(C[0], A[1]) = 0 we see that any distinguished triangle arising from this short exact sequence would look like (A[0], B[0], C[0], a, b, 0). But the existence of such a distinguished triangle in K(\mathcal{A}) implies that the extension is split. A contradiction.
It turns out that the functor \text{Comp}(\mathcal{A}) \to D(\mathcal{A}) is a \delta -functor. In order to see this we have to define the morphisms \delta associated to a short exact sequence
0 \to A^\bullet \xrightarrow {a} B^\bullet \xrightarrow {b} C^\bullet \to 0
of complexes in the abelian category \mathcal{A}. Consider the cone C(a)^\bullet of the morphism a. We have C(a)^ n = B^ n \oplus A^{n + 1} and we define q^ n : C(a)^ n \to C^ n via the projection to B^ n followed by b^ n. Hence a morphism of complexes
q : C(a)^\bullet \longrightarrow C^\bullet .
It is clear that q \circ i = b where i is as in Definition 13.9.1. Note that, as a^\bullet is injective in each degree, the kernel of q is identified with the cone of \text{id}_{A^\bullet } which is acyclic. Hence we see that q is a quasi-isomorphism. According to Lemma 13.9.14 the triangle
(A, B, C(a), a, i, -p)
is a distinguished triangle in K(\mathcal{A}). As the localization functor K(\mathcal{A}) \to D(\mathcal{A}) is exact we see that (A, B, C(a), a, i, -p) is a distinguished triangle in D(\mathcal{A}). Since q is a quasi-isomorphism we see that q is an isomorphism in D(\mathcal{A}). Hence we deduce that
(A, B, C, a, b, -p \circ q^{-1})
is a distinguished triangle of D(\mathcal{A}). This suggests the following lemma.
Lemma 13.12.1. Let \mathcal{A} be an abelian category. The functor \text{Comp}(\mathcal{A}) \to D(\mathcal{A}) defined has the natural structure of a \delta -functor, with
\delta _{A^\bullet \to B^\bullet \to C^\bullet } = - p \circ q^{-1}
with p and q as explained above. The same construction turns the functors \text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A}), \text{Comp}^{-}(\mathcal{A}) \to D^{-}(\mathcal{A}), and \text{Comp}^ b(\mathcal{A}) \to D^ b(\mathcal{A}) into \delta -functors.
Proof.
We have already seen that this choice leads to a distinguished triangle whenever given a short exact sequence of complexes. We have to show that given a commutative diagram
\xymatrix{ 0 \ar[r] & A^\bullet \ar[r]_ a \ar[d]_ f & B^\bullet \ar[r]_ b \ar[d]_ g & C^\bullet \ar[r] \ar[d]_ h & 0 \\ 0 \ar[r] & (A')^\bullet \ar[r]^{a'} & (B')^\bullet \ar[r]^{b'} & (C')^\bullet \ar[r] & 0 }
we get the desired commutative diagram of Definition 13.3.6 (2). By Lemma 13.9.2 the pair (f, g) induces a canonical morphism c : C(a)^\bullet \to C(a')^\bullet . It is a simple computation to show that q' \circ c = h \circ q and f[1] \circ p = p' \circ c. From this the result follows directly.
\square
Lemma 13.12.2. Let \mathcal{A} be an abelian category. Let
\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & D^\bullet \ar[r] & E^\bullet \ar[r] & F^\bullet \ar[r] & 0 }
be a commutative diagram of morphisms of complexes such that the rows are short exact sequences of complexes, and the vertical arrows are quasi-isomorphisms. The \delta -functor of Lemma 13.12.1 above maps the short exact sequences 0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0 and 0 \to D^\bullet \to E^\bullet \to F^\bullet \to 0 to isomorphic distinguished triangles.
Proof.
Trivial from the fact that K(\mathcal{A}) \to D(\mathcal{A}) transforms quasi-isomorphisms into isomorphisms and that the associated distinguished triangles are functorial.
\square
Lemma 13.12.3. Let \mathcal{A} be an abelian category. Let
\xymatrix{ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 }
be a short exact sequences of complexes. Assume this short exact sequence is termwise split. Let (A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta ) be the distinguished triangle of K(\mathcal{A}) associated to the sequence. The \delta -functor of Lemma 13.12.1 above maps the short exact sequences 0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0 to a triangle isomorphic to the distinguished triangle
(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta ).
Proof.
Follows from Lemma 13.9.14.
\square
Lemma 13.12.5. Let \mathcal{A} be an abelian category. Let
K_0^\bullet \to K_1^\bullet \to \ldots \to K_ n^\bullet
be maps of complexes such that
H^ i(K_0^\bullet ) = 0 for i > 0,
H^{-j}(K_ j^\bullet ) \to H^{-j}(K_{j + 1}^\bullet ) is zero.
Then the composition K_0^\bullet \to K_ n^\bullet factors through \tau _{\leq -n}K_ n^\bullet \to K_ n^\bullet in D(\mathcal{A}). Dually, given maps of complexes
K_ n^\bullet \to K_{n - 1}^\bullet \to \ldots \to K_0^\bullet
such that
H^ i(K_0^\bullet ) = 0 for i < 0,
H^ j(K_{j + 1}^\bullet ) \to H^ j(K_ j^\bullet ) is zero,
then the composition K_ n^\bullet \to K_0^\bullet factors through K_ n^\bullet \to \tau _{\geq n}K_ n^\bullet in D(\mathcal{A}).
Proof.
The case n = 1. Since \tau _{\leq 0}K_0^\bullet = K_0^\bullet in D(\mathcal{A}) we can replace K_0^\bullet by \tau _{\leq 0}K_0^\bullet and K_1^\bullet by \tau _{\leq 0}K_1^\bullet . Consider the distinguished triangle
\tau _{\leq -1}K_1^\bullet \to K_1^\bullet \to H^0(K_1^\bullet )[0] \to (\tau _{\leq -1}K_1^\bullet )[1]
(Remark 13.12.4). The composition K_0^\bullet \to K_1^\bullet \to H^0(K_1^\bullet )[0] is zero as it is equal to K_0^\bullet \to H^0(K_0^\bullet )[0] \to H^0(K_1^\bullet )[0] which is zero by assumption. The fact that \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(K_0^\bullet , -) is a homological functor (Lemma 13.4.2), allows us to find the desired factorization. For n = 2 we get a factorization K_0^\bullet \to \tau _{\leq -1}K_1^\bullet by the case n = 1 and we can apply the case n = 1 to the map of complexes \tau _{\leq -1}K_1^\bullet \to \tau _{\leq -1}K_2^\bullet to get a factorization \tau _{\leq -1}K_1^\bullet \to \tau _{\leq -2}K_2^\bullet . The general case is proved in exactly the same manner.
\square
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