Delta functor

Definition

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. A cohomological $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$ consists of additive functors $T^n \colon \mathcal{A} \to \mathcal{B}$ for $n \ge 0$ and of connecting morphisms. For every short exact sequence $0 \to A' \to A \to A'' \to 0$ in $\mathcal{A}$ and every $n \ge 0$, there is a morphism $\delta^n \colon T^n(A'') \to T^{n+1}(A')$. These data must satisfy:

1. for every short exact sequence, the sequence $$0 \to T^0(A') \to T^0(A) \to T^0(A'') \xrightarrow{\delta^0} T^1(A') \to T^1(A) \to T^1(A'') \xrightarrow{\delta^1} T^2(A') \to \cdots$$ is exact;
2. the maps $\delta^n$ are natural. Let a morphism from $0 \to A' \to A \to A'' \to 0$ to $0 \to B' \to B \to B'' \to 0$ have components $f' \colon A' \to B'$ and $f'' \colon A'' \to B''$. Then $\delta^n \circ T^n(f'') = T^{n+1}(f') \circ \delta^n$.

A $\delta$-functor $(T^n)$ is universal if the following holds for every $\delta$-functor $(S^n)$ and every natural transformation $T^0 \to S^0$. There is a unique family of natural transformations $T^n \to S^n$ extending it and commuting with the maps $\delta^n$.