The Stacks project

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. $\square$


Comments (2)

Comment #373 by Fan on

The last commutative diagram doesn't seem right: the left column should be mapping to Coker, and the same for the right column

Comment #384 by on

@#373: Huh? This seems correct to me.


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