Lemma 13.14.7 (05SD)—The Stacks project

Lemma 13.14.7. Assumptions and notation as in Situation 13.14.1. Let $X, Y$ be objects of $\mathcal{D}$.

If $RF$ is defined at $X$ and $Y$, then $RF$ is defined at $X \oplus Y$.
If $\mathcal{D}'$ is Karoubian and $RF$ is defined at $X \oplus Y$, then $RF$ is defined at both $X$ and $Y$.

In either case we have $RF(X \oplus Y) = RF(X) \oplus RF(Y)$. Similarly for $LF$.

Proof. If $RF$ is defined at $X$ and $Y$, then the distinguished triangle $X \to X \oplus Y \to Y \to X[1]$ (Lemma 13.4.11) and Lemma 13.14.6 shows that $RF$ is defined at $X \oplus Y$ and that we have a distinguished triangle $RF(X) \to RF(X \oplus Y) \to RF(Y) \to RF(X)[1]$. Applying Lemma 13.4.11 to this once more we find that $RF(X \oplus Y) = RF(X) \oplus RF(Y)$. This proves (1) and the final assertion.

Conversely, assume that $RF$ is defined at $X \oplus Y$ and that $\mathcal{D}'$ is Karoubian. Since $S$ is a saturated system $S$ is the set of arrows which become invertible under the additive localization functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$, see Categories, Lemma 4.27.21. Thus for any $s : X \to X'$ and $s' : Y \to Y'$ in $S$ the morphism $s \oplus s' : X \oplus Y \to X' \oplus Y'$ is an element of $S$. In this way we obtain a functor

\[ X/S \times Y/S \longrightarrow (X \oplus Y)/S \]

Recall that the categories $X/S, Y/S, (X \oplus Y)/S$ are filtered (Categories, Remark 4.27.7). By Categories, Lemma 4.22.12 $X/S \times Y/S$ is filtered and $F|_{X/S} : X/S \to \mathcal{D}'$ (resp. $G|_{Y/S} : Y/S \to \mathcal{D}'$) is essentially constant if and only if $F|_{X/S} \circ \text{pr}_1 : X/S \times Y/S \to \mathcal{D}'$ (resp. $G|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'$) is essentially constant. Below we will show that the displayed functor is cofinal, hence by Categories, Lemma 4.22.11. we see that $F|_{(X \oplus Y)/S}$ is essentially constant implies that $F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2 : X/S \times Y/S \to \mathcal{D}'$ is essentially constant. By Homology, Lemma 12.30.3 (and this is where we use that $\mathcal{D}'$ is Karoubian) we see that $F|_{X/S} \circ \text{pr}_1 \oplus F|_{Y/S} \circ \text{pr}_2$ being essentially constant implies $F|_{X/S} \circ \text{pr}_1$ and $F|_{Y/S} \circ \text{pr}_2$ are essentially constant proving that $RF$ is defined at $X$ and $Y$.

Proof that the displayed functor is cofinal. To do this pick any $t : X \oplus Y \to Z$ in $S$. Using MS2 we can find morphisms $Z \to X'$, $Z \to Y'$ and $s : X \to X'$, $s' : Y \to Y'$ in $S$ such that

\[ \xymatrix{ X \ar[d]^ s & X \oplus Y \ar[d] \ar[l] \ar[r] & Y \ar[d]_{s'} \\ X' & Z \ar[l] \ar[r] & Y' } \]

commutes. This proves there is a map $Z \to X' \oplus Y'$ in $(X \oplus Y)/S$, i.e., we get part (1) of Categories, Definition 4.17.1. To prove part (2) it suffices to prove that given $t : X \oplus Y \to Z$ and morphisms $s_ i \oplus s'_ i : Z \to X'_ i \oplus Y'_ i$, $i = 1, 2$ in $(X \oplus Y)/S$ we can find morphisms $a : X'_1 \to X'$, $b : X'_2 \to X'$, $c : Y'_1 \to Y'$, $d : Y'_2 \to Y'$ in $S$ such that $a \circ s_1 = b \circ s_2$ and $c \circ s'_1 = d \circ s'_2$. To do this we first choose any $X'$ and $Y'$ and maps $a, b, c, d$ in $S$; this is possible as $X/S$ and $Y/S$ are filtered. Then the two maps $a \circ s_1, b \circ s_2 : Z \to X'$ become equal in $S^{-1}\mathcal{D}$. Hence we can find a morphism $X' \to X''$ in $S$ equalizing them. Similarly we find $Y' \to Y''$ in $S$ equalizing $c \circ s'_1$ and $d \circ s'_2$. Replacing $X'$ by $X''$ and $Y'$ by $Y''$ we get $a \circ s_1 = b \circ s_2$ and $c \circ s'_1 = d \circ s'_2$.

The proof of the corresponding statements for $LF$ are dual. $\square$

Comments (9)

Comment #524 by Nuno on April 04, 2014 at 21:12

I'm probably missing something elementary, but it's not clear to me how can we finish the proof of this lemma. Also, are we considering the arrows $s \oplus s'$ as a full subcategory? I'm asking because, in order to justify the inverse statement, which is needed to show that the category of the next proposition is additive, I had to consider only the morphisms of the form $a \oplus b$ .

Comment #526 by Johan on April 05, 2014 at 01:29

OK, this proof was really incomplete. Thanks for pointing this out. Here is a commit which gives a fair amount of detail, although of course you can always add more. Note that in the usual approach using injective resolutions both this lemma and the even trickier Lemma 37.28 are proved by explicitly producing a short exact sequence of injective resolutions. However, the advantage of Deligne's "pointwise" approach is that it works in complete generality, i.e., you always get $RF$ defined on some saturated triangulated subcategory --- possibly consisiting just of the zero object.

Comment #527 by Nuno on April 05, 2014 at 14:51

It is quite funny that the omitted part of the new proof is exactly what was bothering me in the first place. I was able to justify that the functor $X/S \times Y/S \to (X \oplus Y)/S$ is cofinal. So to achieve our goals it was enough to consider the functor $X/S \times Y/S \to \mathcal{D'}$ . If RF is defined at $X$ and $Y$ , it is clear that the colimit of this functor exists and that it is essentially constant with value $RF(X) \oplus RF(Y)$ . The problem began when I tried to justify the converse. Assuming only that $RF$ is defined at $X \oplus Y$ , it is not clear to me how can we even guarantee the existence of the colimit $colim_{s : X \to X'} F(X')$ . Unless I am misremembering something, this is not true for arbitrary bifunctors.

When I first read parts of this section, I wondered why did the stacks project followed such an abstract approach, but now I agree with you that it has some advantages and is quite elegant.

Comment #528 by Johan on April 05, 2014 at 15:44

OK, so the lemma we should add is the following: Suppose given two filtered categories $\mathcal{C}$ and $\mathcal{D}$ and an additive category $\mathcal{A}$ . Suppose given functors $F : \mathcal{C} \to \mathcal{A}$ and $G : \mathcal{D} \to \mathcal{A}$ . Then (1) $\mathcal{C} \times \mathcal{D}$ is a filtered category and (2) $F \oplus G$ is essentially constant if and only if $F$ and $G$ are essentially constant. Do you agree?

Comment #529 by Nuno on April 05, 2014 at 16:03

Yes, that is what we need, specially the implication (->)in (2). Thanks for the attention.

Comment #530 by Johan on April 05, 2014 at 17:14

Thanks for keeping at it. I now finally understand what the problem is and I think there may indeed be a problem. Namely, it seems easy to prove this when the target additive category $\mathcal{A}$ is Karoubian, but I don't see how to do it when $\mathcal{A}$ isn't Karoubian!!! Let me think some more.

Comment #532 by Johan on April 05, 2014 at 22:53

OK, you win a Stacks project mug as this was actually an error. (I still have a few left; email me your postal address privately.) It doesn't affect almost any results as the lemma is correct when the target triangulated category is Karoubian. See commits first and second and third.

Comment #543 by Nuno on April 08, 2014 at 03:09

There are two "confinal" in the text and here "(resp. $G|_{Y/S} : Y/S \to \mathcal{D}'$ )" we should replace $G|_{Y/S}$ by $F|_{Y/S}$ . I believe everything is fine now. Thanks again for the attention and let me say that I am impressed how fast you did all this.

Comment #553 by Johan on April 09, 2014 at 01:01

Thanks. Fixed here.

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4 comment(s) on Section 13.14: Derived functors in general

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