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 (4)

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.

Comment #8344 by on

It is actually enough if for every , and there exists an injective morphism (depending on ) such that maps to zero under . The proof mostly doesn't change.

A reference is Prop. 4.2 in Buchsbaum's paper "On satellites and universal functors" (https://people.brandeis.edu/~buchsbau/miscpapers/027.pdf). This refined effaceability criterion is very useful in cases that there don't exist uniform -killing injections .

A non-example for this situation would be cohomology with respect to Zariski (hyper-)coverings, where there exists a single hypercovering computing the correct cohomology, and an example would be étale cohomology, where we truly need to take the colimit over all hypercoverings.

Comment #8345 by on

@#8344 The version you mention can only be used if is from into the category of abelian groups or modules or something like that.


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'.