
## 12.11 Cohomological delta-functors

Definition 12.11.1. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. A cohomological $\delta$-functor or simply a $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$ is given by the following data:

1. a collection $F^ n : \mathcal{A} \to \mathcal{B}$, $n \geq 0$ of additive functors, and

2. for every short exact sequence $0 \to A \to B \to C \to 0$ of $\mathcal{A}$ a collection $\delta _{A \to B \to C} : F^ n(C) \to F^{n + 1}(A)$, $n \geq 0$ of morphisms of $\mathcal{B}$.

These data are assumed to satisfy the following axioms

1. for every short exact sequence as above the sequence

$\xymatrix{ 0 \ar[r] & F^0(A) \ar[r] & F^0(B) \ar[r] & F^0(C) \ar[lld]^{\delta _{A \to B \to C}} \\ & F^1(A) \ar[r] & F^1(B) \ar[r] & F^1(C) \ar[lld]^{\delta _{A \to B \to C}} \\ & F^2(A) \ar[r] & F^2(B) \ar[r] & \ldots }$

is exact, and

2. for every morphism $(A \to B \to C) \to (A' \to B' \to C')$ of short exact sequences of $\mathcal{A}$ the diagrams

$\xymatrix{ F^ n(C) \ar[d] \ar[rr]_{\delta _{A \to B \to C}} & & F^{n + 1}(A) \ar[d] \\ F^ n(C') \ar[rr]^{\delta _{A' \to B' \to C'}} & & F^{n + 1}(A') }$

are commutative.

Note that this in particular implies that $F^0$ is left exact.

Definition 12.11.2. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $(F^ n, \delta _ F)$ and $(G^ n, \delta _ G)$ be $\delta$-functors from $\mathcal{A}$ to $\mathcal{B}$. A morphism of $\delta$-functors from $F$ to $G$ is a collection of transformation of functors $t^ n : F^ n \to G^ n$, $n \geq 0$ such that for every short exact sequence $0 \to A \to B \to C \to 0$ of $\mathcal{A}$ the diagrams

$\xymatrix{ F^ n(C) \ar[d]_{t^ n} \ar[rr]_{\delta _{F, A \to B \to C}} & & F^{n + 1}(A) \ar[d]^{t^{n + 1}} \\ G^ n(C) \ar[rr]^{\delta _{G, A \to B \to C}} & & G^{n + 1}(A) }$

are commutative.

Definition 12.11.3. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $F = (F^ n, \delta _ F)$ be a $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$. We say $F$ is a universal $\delta$-functor if an only if for every $\delta$-functor $G = (G^ n, \delta _ G)$ and any morphism of functors $t : F^0 \to G^0$ there exists a unique morphism of $\delta$-functors $\{ t^ n\} _{n \geq 0} : F \to G$ such that $t = t^0$.

Lemma 12.11.4. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $F = (F^ n, \delta _ F)$ be a $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$. Suppose that for every $n > 0$ and any $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ there exists an injective morphism $u : A \to B$ (depending on $A$ and $n$) such that $F^ n(u) : F^ n(A) \to F^ n(B)$ is zero. Then $F$ is a universal $\delta$-functor.

Proof. Let $G = (G^ n, \delta _ G)$ be a $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$ and let $t : F^0 \to G^0$ be a morphism of functors. We have to show there exists a unique morphism of $\delta$-functors $\{ t^ n\} _{n \geq 0} : F \to G$ such that $t = t^0$. We construct $t^ n$ by induction on $n$. For $n = 0$ we set $t^0 = t$. Suppose we have already constructed a unique sequence of transformation of functors $t^ i$ for $i \leq n$ compatible with the maps $\delta$ in degrees $\leq n$.

Let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. By assumption we may choose a embedding $u : A \to B$ such that $F^{n + 1}(u) = 0$. Let $C = B/u(A)$. The long exact cohomology sequence for the short exact sequence $0 \to A \to B \to C \to 0$ and the $\delta$-functor $F$ gives that $F^{n + 1}(A) = \mathop{\mathrm{Coker}}(F^ n(B) \to F^ n(C))$ by our choice of $u$. Since we have already defined $t^ n$ we can set

$t^{n + 1}_ A : F^{n + 1}(A) \to G^{n + 1}(A)$

equal to the unique map such that

$\xymatrix{ \mathop{\mathrm{Coker}}(F^ n(B) \to F^ n(C)) \ar[r]_{t^ n} \ar[d]_{\delta _{F, A \to B \to C}} & \mathop{\mathrm{Coker}}(G^ n(B) \to G^ n(C)) \ar[d]^{\delta _{G, A \to B \to C}} \\ F^{n + 1}(A) \ar[r]^{t^{n + 1}_ A} & G^{n + 1}(A) }$

commutes. This is clearly uniquely determined by the requirements imposed. We omit the verification that this defines a transformation of functors. $\square$

Lemma 12.11.5. Let $\mathcal{A}, \mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor. If there exists a universal $\delta$-functor $(F^ n, \delta _ F)$ from $\mathcal{A}$ to $\mathcal{B}$ with $F^0 = F$, then it is determined up to unique isomorphism of $\delta$-functors.

Proof. Immediate from the definitions. $\square$

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