Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading Grothendieck's Tōhoku paper, and I was curious about the reasoning behind the terms "effaceable functor" and "injective effacement". I know that in English, to efface something means to erase it, but I'm not sure if there's another meaning in either conversational or mathematical French.

Given an abelian category $C$ and an additive category $D$, an effaceable functor $F:C\rightarrow D$ is an additive functor such that for all $A\in\text{Ob}(C)$, there exists some monomorphism $u:A\rightarrow M$ such that $F(u)=0$. I'm guessing that the "effacing" going on here is the functor $F$ killing the map $u$.

What I'm less sure about is the reasoning behind "injective effacement". Given an abelian category $C$ and $A\in\text{Ob}(C)$, an injective effacement of $A$ is a monomorphism $f:A\rightarrow M$ such that for any monomorphism $g:B\rightarrow C$, and any map $h:B\rightarrow A$, there is a map $t:C\rightarrow M$ such that the diagram commutes: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lll} B & \ra{g} & C \\ \da{h} & & \da{t} \\ A & \ra{f} & M \\ \end{array} $$

Grothendieck points out that $\text{id}_M:M\rightarrow M$ is an injective effacement of $M$ if and only if $M$ is an injective object of $C$, and that any monomorphism of $A$ to an injective object $M$ is an injective effacement of $A$. I don't really see what we're erasing here; my guess is that an injective effacement $f:A\rightarrow M$ is somehow "erasing" the cohomology of $A$, but this is only a vague sense.

share|improve this question
1  
Zev: I've substantially expanded on my answer, I hope I'm done now. –  t.b. Jun 10 '11 at 13:00
    
@Theo: Your answer is superb! –  Zev Chonoles Jun 10 '11 at 20:13
    
Many thanks, that is very nice to hear! –  t.b. Jun 11 '11 at 15:38
add comment

2 Answers

up vote 14 down vote accepted

First of all: French effacer means "to erase", and it is of course a composite of the prefix e- (latin ex-) [away from] face. I'm not aware of a different meaning than erase (maybe wipe up or annihilate are also viable translations in some circumstances), and what you're describing is probably more or less what Grothendieck had in mind. As far as I know Grothendieck is the inventor of this terminology.

Let me focus on the simplest case:

Let us assume that the abelian category has enough injectives and let $F$ be a (left exact) functor to another abelian category. As you know, the associated right derived functors $R^nF$ can be constructed by choosing an injective resolution of each object, applying $F$ to the resolution and taking homology.

The right derived functors are a universal $\delta$-functor due to the fact that each $A$ admits an injection $A \to I$ such that $R^{n}F(I) = 0$ for $n \geq 1$.

Let us look at a specific instance of the arguments involved: The first step in proving that given any other $\delta$-functor $T^n$ and a natural transformation $\phi : F \Rightarrow T^0$ there is a unique sequence of natural transformations $\phi^n: R^nF \Rightarrow T^n$ extending $\phi$ and compatible with the connecting morphisms of both $R^nF$ and $T^n$.

This sequence of natural transformations is built by induction. For each $A$ choose an injective effacement $0 \to A \to I \to C \to 0$ and apply both $R^nF$ and $T^n$ and the induction hypotheses. We get to the following diagram:

$$\begin{array}{ccccccc} R^nF(I) & \to & R^{n}F(C) & \to & R^{n+1}F(A) & \to & \mathbf{0} \\ \downarrow & & \downarrow & & \downarrow \exists? & & \\ T^n(I) & \to & T^n(C) & \to & T^{n+1}(A) \end{array}$$ The bold zero is due to effaceability of $R^{n+1}F$. The two left hand vertical maps are already constructed and the map marked with a question mark exists and is unique by an easy diagram chase because the bold zero implies that the connecting morphism $R^{n}F(C) \to R^{n+1}F(A)$ is an epimorphism. The map just constructed is the $A$-component $\phi_{A}^{n+1}$ of $\phi^{n+1}$. Such arguments are applied some further times when checking that this construction actually defines a natural transformation $\phi^{n+1} : R^{n+1} F \Rightarrow T^{n+1}$, that the $\phi^n$'s are compatible with the connecting morphisms and that they are unique in that they don't depend on the chosen effacement.

In summary and conclusion:

Universality follows from effaceability due to the fact that the connecting morphisms $R^{n}F(C) \to R^{n+1}F(A)$ associated to the sequence $0 \to A \to I \to C \to 0$ are forced to be epimorphisms, as $R^{n+1}F(I) = 0$. In other words, because $R^{n+1}F$ gets erased at $I$.

I hope that this is clear enough, if not, I strongly recommend to go through the proof of universality of derived functors once again, e.g. in Theorem 2.4.7 of Weibel. (Don't miss the exercises at the end of the section!)

The effaceability condition can be used to show more generally: If $S^n$ is a $\delta$-functor and $S^{n}$ is effaceable for $n \geq 1$, then the $\delta$-functor $S^n$ is universal.

This is used when one applies dimension shifting: Suppose we have an effacement $0 \to A \to I \to C \to 0$ and $S^{n}(I) = 0 = S^{n+1}(I)$. Then the sequence $0 \to S^n(C) \to S^{n+1}(A) \to 0$ shows that the connecting morphism $S^n(C) \to S^{n+1}(A)$ is an isomorphism because of this effacement.

For instance, the definition of flatness implies that $Tor_{n}(A,F)$ is zero for a flat module $F$ and $n \geq 1$. A variant of the previous argument can be used to show that $Tor_{n+1}(A,B)$ can be constructed from $Tor_{n}(A,-)$ using a flat (co-)effacement of $B$, and the resulting $\delta$-functor is universal due to the above.

Sometimes a slightly more general variant of the effaceability is called Buchsbaum's criterion (or weak effaceability) due to Proposition 4.2 in Buchsbaum's Annals paper Satellites and Universal Functors.


Added:

Here is, to the best of my knowledge, the most useful general criterion for $F$ to admit a (classical) right derived functor, which I learned from Bernhard Keller. It covers all the "oodles of applications" explained by Akhil in his answer:

Suppose $\mathcal{C}$ is an abelian (or more generally an exact) category and assume that $F:\mathcal{C} \to \mathcal{D}$ is a functor to an abelian category $\mathcal{D}$.

Theorem. If there exists a full subcategory $\mathcal{A}$ of $\mathcal{C}$ such that

  1. $\mathcal{A}$ is closed under extensions: if $0 \to A' \to C \to A'' \to 0$ is exact in $\mathcal{C}$ then $C$ is an object of $\mathcal{A}$.
  2. The restriction of $F$ to $\mathcal{A}$ is exact.
  3. For every object of $\mathcal{C}$ there exists a monomorphism $C \to A$ with $A$ in $\mathcal{A}$.
  4. For every short exact sequence $0 \to A' \to C \to C''$ with $A'$ in $\mathcal{A}$ there exists a commutative diagram $$\begin{array}{ccccccccc} 0 & \to & A' & \to & C & \to & C'' & \to & 0 \\ & & \parallel & & \downarrow & & \downarrow \\ 0 & \to & A' & \to & A & \to & A'' & \to & 0 \end{array}$$ where the second row is exact and $A, A''$ are in $\mathcal{A}$.

Then $F$ admits a right derived functor.

Notes:

  1. The technical condition 4. is satisfied in presence of 3. if $\mathcal{A}$ is supposed to be closed under quotients in $\mathcal{C}$. That is: Whenever $0 \to A' \to A \to C \to 0$ is short exact then $C$ is an object of $\mathcal{A}$.
  2. No left exactness of $F$ on $\mathcal{C}$ is imposed: this is only needed to guarantee that $R^0F = F$. In general, $R^{0}F$ is the best left exact approximation to $F$. In highbrow terms: $R^{0}$ can be seen as reflection of the inclusion of the left exact functors into all additive functors.
  3. The right derived functors of $F$ can be computed by choosing for each $C$ an $\mathcal{A}$-resolution — which exists due to the effaceability conditions 2 and 3.
  4. The objects of $\mathcal{A}$ are called $F$-acyclic or adapted to $F$ if $\mathcal{A}$ satisfies the hypotheses of the theorem.

Final remarks (coming back to effaceability):

  1. The above theorem admits a partial converse: if $F$ admits a right derived functor, the category $\mathcal{A}(F)$ of $F$-acyclic objects (a notion I don't want to define in full generality here), satisfies conditions 1., 2. and 4. above. The effaceability of the higher right derived functors is thus seen as the part missing from a complete characterization of derived functors.

  2. The usefulness of injective effacements is of course that injective objects are $F$-acyclic for every additive functor. This is because short exact sequences of injectives split, by definition of injectivity. In particular, every functor on an abelian category with enough injectives can be derived (not only left exact ones).

  3. I'm not aware of a complete proof of the theorem as stated in the literature without making a detour via derived categories, which certainly looks like overkill given its rather elementary nature. See sections 12ff in Keller's article or sections 10.5, 10.6 and 12 in my notes on exact categories.

share|improve this answer
2  
I'd also recommend reading Ch. 2 of Weibel before going through the paper. It's a short chapter written very clearly, and the derived functors approach is essentially based on Grothendieck's paper. –  Jason Polak Jun 9 '11 at 19:32
1  
Dear Theo, I'm assuming the simplest proof of the result you describe is to define the derived functor on the derived category itself (bounded-below?) by replacing a complex by a complex consisting of elements of $\mathcal{A}$? (It seems that the conditions you list are precisely what one needs to ensure that a bounded-below complex admits a quasi-isomorphism into an $\mathcal{A}$-complex and that $F$ preserves quasi-isomorphisms of such complexes.) –  Akhil Mathew Jun 10 '11 at 14:09
    
Dear @Akhil: Yes that's it exactly. The conditions ensure (1.) that $\mathcal{A}$ is an exact category (thus $D(\mathcal{A})$ is defined following Neeman) (2.) the restriction of $F$ trivially descends to $D(\mathcal{A})$ and that the natural functor $D^{+}(\mathcal{A}) \to D^{+}(\mathcal{C})$ is fully faithful (4.) and essentially surjective (3.) and thus an equivalence. The total derived functor $R^{+}F$ is then defined by inverting this equivalence and composing with the restriction. Keller's article I linked to in the last few lines contains a very thorough and clear discussion on this. –  t.b. Jun 10 '11 at 14:18
    
Ah, I see. Thanks for the explanation! –  Akhil Mathew Jun 10 '11 at 14:31
add comment

Theo's answer is excellent. Let me add a bit on how this formalism is used in practice.

Say you want to show that $\mathrm{Tor}$ commutes with flat base change of rings. That is, if $M, N$ are $A$-modules, $B$ a flat $A$-algebra, then $\mathrm{Tor}_B(M_B, N_B) \simeq \mathrm{Tor}_A(M,N) \otimes_A B$ functorially -- it's not even obvious a priori how to get the map. The point is that both are universal $\delta$-functors in either variable, though, and since there is a clear isomorphism in dimension zero, it follows that the two functors are not only isomorphic, but isomorphic as $\delta$-functors. (To see that they are co-effaceable, one uses the fact that base change preserves projectivity, for instance.)

In Hartshorne, for instance, there are oodles of applications of this formalism, e.g. in showing that sheaf cohomology is the same derived functor whether you work in the category of all sheaves or just sheaves of modules over some sheaf of rings (the point is that either way, the derived functors are effaceable by flasque sheaves, which mean the same in either category). This last fact can also be seen if you accept that derived functors can be computed using acyclic (rather than injective) resolutions.

Let me add one more example, which is perhaps more interesting. Suppose $A$ is a noetherian ring and $I \subset A$ an ideal. Then we can consider the functor on the category of $A$-modules that sends an $M$ to the submodule consisting of all $m \in M$ killed by a high power of $I$. This is a left-exact functor, and its derived functors are the local cohomology modules $H^i_I(M)$. It turns out that $H^i_I(M) \simeq \varinjlim_n \mathrm{Ext}^i(R/I^n, M)$ functorially; a quick way of seeing this is that it is clearly true in dimension zero (by inspection), and for higher dimensions we see that both are effaceable $\delta$-functors in $M$ (since filtered colimits are exact for modules).

share|improve this answer
2  
Thanks for the nice words and further explanations. I sincerely appreciate it. –  t.b. Jun 10 '11 at 9:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.