The Stacks project

Lemma 12.12.4. 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.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 010T. Beware of the difference between the letter 'O' and the digit '0'.