## 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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