# The Stacks Project

## Tag 05S7

### 13.15. Derived functors in general

A reference for this section is Deligne's exposé XVII in [SGA4]. A very general notion of right and left derived functors exists where we have an exact functor between triangulated categories, a multiplicative system in the source category and we want to find the ''correct'' extension of the exact functor to the localized category.

Situation 13.15.1. Here $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor of triangulated categories and $S$ is a saturated multiplicative system in $\mathcal{D}$ compatible with the structure of triangulated category on $\mathcal{D}$.

Let $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$. Recall from Categories, Remark 4.26.7 the filtered category $X/S$ of arrows $s : X \to X'$ in $S$ with source $X$. Dually, in Categories, Remark 4.26.15 we defined the cofiltered category $S/X$ of arrows $s : X' \to X$ in $S$ with target $X$.

Definition 13.15.2. Assumptions and notation as in Situation 13.15.1. Let $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$.

1. we say the right derived functor $RF$ is defined at $X$ if the ind-object $$(X/S) \longrightarrow \mathcal{D}', \quad (s : X \to X') \longmapsto F(X')$$ is essentially constant1; in this case the value $Y$ in $\mathcal{D}'$ is called the value of $RF$ at $X$.
2. we say the left derived functor $LF$ is defined at $X$ if the pro-object $$(S/X) \longrightarrow \mathcal{D}', \quad (s: X' \to X) \longmapsto F(X')$$ is essentially constant; in this case the value $Y$ in $\mathcal{D}'$ is called the value of $LF$ at $X$.

By abuse of notation we often denote the values simply $RF(X)$ or $LF(X)$.

It will turn out that the full subcategory of $\mathcal{D}$ consisting of objects where $RF$ is defined is a triangulated subcategory, and $RF$ will define a functor on this subcategory which transforms morphisms of $S$ into isomorphisms.

Lemma 13.15.3. Assumptions and notation as in Situation 13.15.1. Let $f : X \to Y$ be a morphism of $\mathcal{D}$.

1. If $RF$ is defined at $X$ and $Y$ then there exists a unique morphism $RF(f) : RF(X) \to RF(Y)$ between the values such that for any commutative diagram $$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$ with $s, s' \in S$ the diagram $$\xymatrix{ F(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & RF(X) \ar[d] \\ F(Y) \ar[r] & F(Y') \ar[r] & RF(Y) }$$ commutes.
2. If $LF$ is defined at $X$ and $Y$ then there exists a unique morphism $LF(f) : LF(X) \to LF(Y)$ between the values such that for any commutative diagram $$\xymatrix{ X' \ar[d]_{f'} \ar[r]_s & X \ar[d]^f \\ Y' \ar[r]^{s'} & Y }$$ with $s, s'$ in $S$ the diagram $$\xymatrix{ LF(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & F(X) \ar[d] \\ LF(Y) \ar[r] & F(Y') \ar[r] & F(Y) }$$ commutes.

Proof. Part (1) holds if we only assume that the colimits $$RF(X) = \mathop{\rm colim}\nolimits_{s : X \to X'} F(X') \quad\text{and}\quad RF(Y) = \mathop{\rm colim}\nolimits_{s' : Y \to Y'} F(Y')$$ exist. Namely, to give a morphism $RF(X) \to RF(Y)$ between the colimits is the same thing as giving for each $s : X \to X'$ in $\mathop{\rm Ob}\nolimits(X/S)$ a morphism $F(X') \to RF(Y)$ compatible with morphisms in the category $X/S$. To get the morphism we choose a commutative diagram $$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$ with $s, s'$ in $S$ as is possible by MS2 and we set $F(X') \to RF(Y)$ equal to the composition $F(X') \to F(Y') \to RF(Y)$. To see that this is independent of the choice of the diagram above use MS3. Details omitted. The proof of (2) is dual. $\square$

Lemma 13.15.4. Assumptions and notation as in Situation 13.15.1. Let $s : X \to Y$ be an element of $S$.

1. $RF$ is defined at $X$ if and only if it is defined at $Y$. In this case the map $RF(s) : RF(X) \to RF(Y)$ between values is an isomorphism.
2. $LF$ is defined at $X$ if and only if it is defined at $Y$. In this case the map $LF(s) : LF(X) \to LF(Y)$ between values is an isomorphism.

Proof. Omitted. $\square$

Lemma 13.15.5. Assumptions and notation as in Situation 13.15.1. Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.

1. $RF$ is defined at $X$ if and only if it is defined at $X[n]$. In this case there is a canonical isomorphism $RF(X)[n]= RF(X[n])$ between values.
2. $LF$ is defined at $X$ if and only if it is defined at $X[n]$. In this case there is a canonical isomorphism $LF(X)[n] \to LF(X[n])$ between values.

Proof. Omitted. $\square$

Lemma 13.15.6. Assumptions and notation as in Situation 13.15.1. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$. If $RF$ is defined at two out of three of $X, Y, Z$, then it is defined at the third. Moreover, in this case $$(RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h))$$ is a distinguished triangle in $\mathcal{D}'$. Similarly for $LF$.

Proof. Say $RF$ is defined at $X, Y$ with values $A, B$. Let $RF(f) : A \to B$ be the induced morphism, see Lemma 13.15.3. We may choose a distinguished triangle $(A, B, C, RF(f), b, c)$ in $\mathcal{D}'$. We claim that $C$ is a value of $RF$ at $Z$.

To see this pick $s : X \to X'$ in $S$ such that there exists a morphism $\alpha : A \to F(X')$ as in Categories, Definition 4.22.1. We may choose a commutative diagram $$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$ with $s' \in S$ by MS2. Using that $Y/S$ is filtered we can (after replacing $s'$ by some $s'' : Y \to Y''$ in $S$) assume that there exists a morphism $\beta : B \to F(Y')$ as in Categories, Definition 4.22.1. Picture $$\xymatrix{ A \ar[d]_{RF(f)} \ar[r]_-\alpha & F(X') \ar[r] \ar[d]^{F(f')} & A \ar[d]^{RF(f)} \\ B \ar[r]^-\beta & F(Y') \ar[r] & B }$$ It may not be true that the left square commutes, but the outer and right squares commute. The assumption that the ind-object $\{F(Y')\}_{s' : Y' \to Y}$ is essentially constant means that there exists a $s'' : Y \to Y''$ in $S$ and a morphism $h : Y' \to Y''$ such that $s'' = h \circ s'$ and such that $F(h)$ equal to $F(Y') \to B \to F(Y') \to F(Y'')$. Hence after replacing $Y'$ by $Y''$ and $\beta$ by $F(h) \circ \beta$ the diagram will commute (by direct computation with arrows).

Using MS6 choose a morphism of triangles $$(s, s', s'') : (X, Y, Z, f, g, h) \longrightarrow (X', Y', Z', f', g', h')$$ with $s'' \in S$. By TR3 choose a morphism of triangles $$(\alpha, \beta, \gamma) : (A, B, C, RF(f), b, c) \longrightarrow (F(X'), F(Y'), F(Z'), F(f'), F(g'), F(h'))$$

By Lemma 13.15.4 it suffices to prove that $RF(Z')$ is defined and has value $C$. Consider the category $\mathcal{I}$ of Lemma 13.5.8 of triangles $$\mathcal{I} = \{(t, t', t'') : (X', Y', Z', f', g', h') \to (X'', Y'', Z'', f'', g'', h'') \mid (t, t', t'') \in S\}$$ To show that the system $F(Z'')$ is essentially constant over the category $Z'/S$ is equivalent to showing that the system of $F(Z'')$ is essentially constant over $\mathcal{I}$ because $\mathcal{I} \to Z'/S$ is cofinal, see Categories, Lemma 4.22.8 (cofinality is proven in Lemma 13.5.8). For any object $W$ in $\mathcal{D}'$ we consider the diagram $$\xymatrix{ \mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(X'')) & \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, A) \ar[l] \\ \mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(Y'')) \ar[u] & \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, B) \ar[u] \ar[l] \\ \mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(Z'')) \ar[u] & \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, C) \ar[u] \ar[l] \\ \mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(X''[1])) \ar[u] & \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, A[1]) \ar[u] \ar[l] \\ \mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(Y''[1])) \ar[u] & \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, B[1]) \ar[u] \ar[l] }$$ where the horizontal arrows are given by composing with $(\alpha, \beta, \gamma)$. Since filtered colimits are exact (Algebra, Lemma 10.8.8) the left column is an exact sequence. Thus the $5$ lemma (Homology, Lemma 12.5.20) tells us the $$\mathop{\rm colim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, F(Z'')) \longrightarrow \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(W, C)$$ is bijective. Choose an object $(t, t', t'') : (X', Y', Z') \to (X'', Y'', Z'')$ of $\mathcal{I}$. Applying what we just showed to $W = F(Z'')$ and the element $\text{id}_{F(X'')}$ of the colimit we find a unique morphism $c_{(X'', Y'', Z'')} : F(Z'') \to C$ such that for some $(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$ $$F(Z'') \xrightarrow{c_{(X'', Y'', Z'')}} C \xrightarrow{\gamma} F(Z') \to F(Z'') \to F(Z''') \quad\text{equals}\quad F(Z'') \to F(Z''')$$ The family of morphisms $c_{(X'', Y'', Z'')}$ form an element $c$ of $\mathop{\rm lim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(F(Z''), C)$ by uniqueness (computation omitted). Finally, we show that $\mathop{\rm colim}\nolimits_\mathcal{I} F(Z'') = C$ via the morphisms $c_{(X'', Y'', Z'')}$ which will finish the proof by Categories, Lemma 4.22.6. Namely, let $W$ be an object of $\mathcal{D}'$ and let $d_{(X'', Y'', Z'')} : F(Z'') \to W$ be a family of maps corresponding to an element of $\mathop{\rm lim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(F(Z''), W)$. If $d_{(X', Y', Z')} \circ \gamma = 0$, then for every object $(X'', Y'', Z'')$ of $\mathcal{I}$ the morphism $d_{(X'', Y'', Z'')}$ is zero by the existence of $c_{(X'', Y'', Z'')}$ and the morphism $(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$ satisfying the displayed equality above. Hence the map $$\mathop{\rm lim}\nolimits_\mathcal{I} \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(F(Z''), W) \longrightarrow \mathop{\rm Mor}\nolimits_{\mathcal{D}'}(C, W)$$ (coming from precomposing by $\gamma$) is injective. However, it is also surjective because the element $c$ gives a left inverse. We conclude that $C$ is the colimit by Categories, Remark 4.14.4. $\square$

Lemma 13.15.7. Assumptions and notation as in Situation 13.15.1. Let $X, Y$ be objects of $\mathcal{D}$.

1. If $RF$ is defined at $X$ and $Y$, then $RF$ is defined at $X \oplus Y$.
2. 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 13.4.10) and Lemma 13.15.6 shows that $RF$ is defined at $X \oplus Y$ and that we have have a distinguished triangle $RF(X) \to RF(X \oplus Y) \to RF(Y) \to RF(X)[1]$. Applying Lemma 13.4.10 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 4.26.21. 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 4.26.7). By Categories, Lemma 4.22.9 $X/S \times Y/S$ 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 4.22.8. 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 12.27.3 (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. $\square$

Proposition 13.15.8. Assumptions and notation as in Situation 13.15.1.

1. The full subcategory $\mathcal{E}$ of $\mathcal{D}$ consisting of objects at which $RF$ is defined is a strictly full triangulated subcategory of $\mathcal{D}$.
2. We obtain an exact functor $RF : \mathcal{E} \longrightarrow \mathcal{D}'$ of triangulated categories.
3. Elements of $S$ with either source or target in $\mathcal{E}$ are morphisms of $\mathcal{E}$.
4. The functor $S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}$ is a fully faithful exact functor of triangulated categories.
5. Any element of $S_\mathcal{E} = \text{Arrows}(\mathcal{E}) \cap S$ is mapped to an isomorphism by $RF$.
6. We obtain an exact functor $$RF : S_\mathcal{E}^{-1}\mathcal{E} \longrightarrow \mathcal{D}'.$$
7. If $\mathcal{D}'$ is Karoubian, then $\mathcal{E}$ is a saturated triangulated subcategory of $\mathcal{D}$.

A similar result holds for $LF$.

Proof. Since $S$ is saturated it contains all isomorphisms (see remark following Categories, Definition 4.26.20). Hence (1) follows from Lemmas 13.15.4, 13.15.6, and 13.15.5. We get (2) from Lemmas 13.15.3, 13.15.5, and 13.15.6. We get (3) from Lemma 13.15.4. The fully faithfulness in (4) follows from (3) and the definitions. The fact that $S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}$ is exact follows from the fact that a triangle in $S_\mathcal{E}^{-1}\mathcal{E}$ is distinguished if and only if it is isomorphic to the image of a distinguished triangle in $\mathcal{E}$, see proof of Proposition 13.5.5. Part (5) follows from Lemma 13.15.4. The factorization of $RF : \mathcal{E} \to \mathcal{D}'$ through an exact functor $S_\mathcal{E}^{-1}\mathcal{E} \to \mathcal{D}'$ follows from Lemma 13.5.6. Part (7) follows from Lemma 13.15.7. $\square$

Proposition 13.15.8 tells us that $RF$ lives on a maximal strictly full triangulated subcategory of $S^{-1}\mathcal{D}$ and is an exact functor on this triangulated category. Picture: $$\xymatrix{ \mathcal{D} \ar[d]_Q \ar[rrr]_F & & & \mathcal{D}' \\ S^{-1}\mathcal{D} & & S_\mathcal{E}^{-1}\mathcal{E} \ar[ll]_{\text{fully faithful}}^{\text{exact}} \ar[ur]_{RF} }$$

Definition 13.15.9. In Situation 13.15.1. We say $F$ is right derivable, or that $RF$ everywhere defined if $RF$ is defined at every object of $\mathcal{D}$. We say $F$ is left derivable, or that $LF$ everywhere defined if $LF$ is defined at every object of $\mathcal{D}$.

In this case we obtain a right (resp. left) derived functor $$\tag{13.15.9.1} RF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}', \quad\text{(resp. } LF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}'),$$ see Proposition 13.15.8. In most interesting situations it is not the case that $RF \circ Q$ is equal to $F$. In fact, it might happen that the canonical map $F(X) \to RF(X)$ is never an isomorphism. In practice this does not happen, because in practice we only know how to prove $F$ is right derivable by showing that $RF$ can be computed by evaluating $F$ at judiciously chosen objects of the triangulated category $\mathcal{D}$. This warrants a definition.

Definition 13.15.10. In Situation 13.15.1.

1. An object $X$ of $\mathcal{D}$ computes $RF$ if $RF$ is defined at $X$ and the canonical map $F(X) \to RF(X)$ is an isomorphism.
2. An object $X$ of $\mathcal{D}$ computes $LF$ if $LF$ is defined at $X$ and the canonical map $LF(X) \to F(X)$ is an isomorphism.

Lemma 13.15.11. Assumptions and notation as in Situation 13.15.1. Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.

1. $X$ computes $RF$ if and only if $X[n]$ computes $RF$.
2. $X$ computes $LF$ if and only if $X[n]$ computes $LF$.

Proof. Omitted. $\square$

Lemma 13.15.12. Assumptions and notation as in Situation 13.15.1. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$. If $X, Y$ compute $RF$ then so does $Z$. Similar for $LF$.

Proof. By Lemma 13.15.6 we know that $RF$ is defined at $Z$ and that $RF$ applied to the triangle produces a distinguished triangle. Consider the morphism of distinguished triangles $$\xymatrix{ (F(X), F(Y), F(Z), F(f), F(g), F(h)) \ar[d] \\ (RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h)) }$$ Two out of three maps are isomorphisms, hence so is the third. $\square$

Lemma 13.15.13. Assumptions and notation as in Situation 13.15.1. Let $X, Y$ be objects of $\mathcal{D}$. If $X \oplus Y$ computes $RF$, then $X$ and $Y$ compute $RF$. Similarly for $LF$.

Proof. If $X \oplus Y$ computes $RF$, then $RF(X \oplus Y) = F(X) \oplus F(Y)$. In the proof of Lemma 13.15.7 we have seen that the functor $X/S \times Y/S \to (X \oplus Y)/S$, $(s, s') \mapsto s \oplus s'$ is cofinal. We will use this without further mention. Let $s : X \to X'$ be an element of $S$. Then $F(X) \to F(X')$ has a section, namely, $$F(X') \to F(X' \oplus Y) \to RF(X' \oplus Y) = RF(X \oplus Y) = F(X) \oplus F(Y) \to F(X).$$ where we have used Lemma 13.15.4. Hence $F(X') = F(X) \oplus E$ for some object $E$ of $\mathcal{D}'$ such that $E \to F(X' \oplus Y) \to RF(X'\oplus Y) = RF(X \oplus Y)$ is zero (Lemma 13.4.11). Because $RF$ is defined at $X' \oplus Y$ with value $F(X) \oplus F(Y)$ we can find a morphism $t : X' \oplus Y \to Z$ of $S$ such that $F(t)$ annihilates $E$. We may assume $Z = X'' \oplus Y''$ and $t = t' \oplus t''$ with $t', t'' \in S$. Then $F(t')$ annihilates $E$. It follows that $F$ is essentially constant on $X/S$ with value $F(X)$ as desired. $\square$

Lemma 13.15.14. Assumptions and notation as in Situation 13.15.1.

1. If for every object $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$ there exists an arrow $s : X \to X'$ in $S$ such that $X'$ computes $RF$, then $RF$ is everywhere defined.
2. If for every object $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$ there exists an arrow $s : X' \to X$ in $S$ such that $X'$ computes $LF$, then $LF$ is everywhere defined.

Proof. This is clear from the definitions. $\square$

Lemma 13.15.15. Assumptions and notation as in Situation 13.15.1. If there exists a subset $\mathcal{I} \subset \mathop{\rm Ob}\nolimits(\mathcal{D})$ such that

1. for all $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$ there exists $s : X \to X'$ in $S$ with $X' \in \mathcal{I}$, and
2. for every arrow $s : X \to X'$ in $S$ with $X, X' \in \mathcal{I}$ the map $F(s) : F(X) \to F(X')$ is an isomorphism,

then $RF$ is everywhere defined and every $X \in \mathcal{I}$ computes $RF$. Dually, if there exists a subset $\mathcal{P} \subset \mathop{\rm Ob}\nolimits(\mathcal{D})$ such that

1. for all $X \in \mathop{\rm Ob}\nolimits(\mathcal{D})$ there exists $s : X' \to X$ in $S$ with $X' \in \mathcal{P}$, and
2. for every arrow $s : X \to X'$ in $S$ with $X, X' \in \mathcal{P}$ the map $F(s) : F(X) \to F(X')$ is an isomorphism,

then $LF$ is everywhere defined and every $X \in \mathcal{P}$ computes $LF$.

Proof. Let $X$ be an object of $\mathcal{D}$. Assumption (1) implies that the arrows $s : X \to X'$ in $S$ with $X' \in \mathcal{I}$ are cofinal in the category $X/S$. Assumption (2) implies that $F$ is constant on this cofinal subcategory. Clearly this implies that $F : (X/S) \to \mathcal{D}'$ is essentially constant with value $F(X')$ for any $s : X \to X'$ in $S$ with $X' \in \mathcal{I}$. $\square$

Lemma 13.15.16. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be triangulated categories. Let $S$, resp. $S'$ be a saturated multiplicative system in $\mathcal{A}$, resp. $\mathcal{B}$ compatible with the triangulated structure. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be exact functors. Denote $F' : \mathcal{A} \to (S')^{-1}\mathcal{B}$ the composition of $F$ with the localization functor.

1. If $RF'$, $RG$, $R(G \circ F)$ are everywhere defined, then there is a canonical transformation of functors $t : R(G \circ F) \longrightarrow RG \circ RF'$.
2. If $LF'$, $LG$, $L(G \circ F)$ are everywhere defined, then there is a canonical transformation of functors $t : LG \circ LF' \to L(G \circ F)$.

Proof. In this proof we try to be careful. Hence let us think of the derived functors as the functors $$RF' : S^{-1}\mathcal{A} \to (S')^{-1}\mathcal{B}, \quad R(G \circ F) : S^{-1}\mathcal{A} \to \mathcal{C}, \quad RG : (S')^{-1}\mathcal{B} \to \mathcal{C}.$$ Let us denote $Q_A : \mathcal{A} \to S^{-1}\mathcal{A}$ and $Q_B : \mathcal{B} \to (S')^{-1}\mathcal{B}$ the localization functors. Then $F' = Q_B \circ F$. Note that for every object $Y$ of $\mathcal{B}$ there is a canonical map $$G(Y) \longrightarrow RG(Q_B(Y))$$ in other words, there is a transformation of functors $t' : G \to RG \circ Q_B$. Let $X$ be an object of $\mathcal{A}$. We have \begin{align*} R(G \circ F)(Q_A(X)) & = \mathop{\rm colim}\nolimits_{s : X \to X' \in S} G(F(X')) \\ & \xrightarrow{t'} \mathop{\rm colim}\nolimits_{s : X \to X' \in S} RG(Q_B(F(X'))) \\ & = \mathop{\rm colim}\nolimits_{s : X \to X' \in S} RG(F'(X')) \\ & = RG(\mathop{\rm colim}\nolimits_{s : X \to X' \in S} F'(X')) \\ & = RG(RF'(X)). \end{align*} The system $F'(X')$ is essentially constant in the category $(S')^{-1}\mathcal{B}$. Hence we may pull the colimit inside the functor $RG$ in the third equality of the diagram above, see Categories, Lemma 4.22.5 and its proof. We omit the proof this this defines a transformation of functors. The case of left derived functors is similar. $\square$

1. For a discussion of when an ind-object or pro-object of a category is essentially constant we refer to Categories, Section 4.22.

The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 4352–5045 (see updates for more information).

\section{Derived functors in general}
\label{section-derived-functors}

\noindent
A reference for this section is Deligne's expos\'e XVII in \cite{SGA4}.
A very general notion of right and left derived functors exists where
we have an exact functor between triangulated categories, a multiplicative
system in the source category and we want to find the correct'' extension
of the exact functor to the localized category.

\begin{situation}
\label{situation-derived-functor}
Here $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor of triangulated
categories and $S$ is a saturated multiplicative
system in $\mathcal{D}$ compatible with the structure
of triangulated category on $\mathcal{D}$.
\end{situation}

\noindent
Let $X \in \Ob(\mathcal{D})$. Recall from
Categories, Remark \ref{categories-remark-left-localization-morphisms-colimit}
the filtered category $X/S$ of arrows $s : X \to X'$ in $S$ with source $X$.
Dually, in
Categories, Remark \ref{categories-remark-right-localization-morphisms-colimit}
we defined the cofiltered category $S/X$ of arrows $s : X' \to X$ in $S$
with target $X$.

\begin{definition}
\label{definition-right-derived-functor-defined}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $X \in \Ob(\mathcal{D})$.
\begin{enumerate}
\item we say the {\it right derived functor $RF$ is defined at}
$X$ if the ind-object
$$(X/S) \longrightarrow \mathcal{D}', \quad (s : X \to X') \longmapsto F(X')$$
is essentially constant\footnote{For a discussion of when an ind-object
or pro-object of a category is essentially constant we refer to
Categories, Section \ref{categories-section-essentially-constant}.};
in this case the value
$Y$ in $\mathcal{D}'$ is called the {\it value of $RF$ at $X$}.
\item we say the {\it left derived functor $LF$ is defined at} $X$
if the pro-object
$$(S/X) \longrightarrow \mathcal{D}', \quad (s: X' \to X) \longmapsto F(X')$$
is essentially constant; in this case the value $Y$ in $\mathcal{D}'$
is called the {\it value of $LF$ at $X$}.
\end{enumerate}
By abuse of notation we often denote the values simply
$RF(X)$ or $LF(X)$.
\end{definition}

\noindent
It will turn out that the full subcategory of $\mathcal{D}$ consisting
of objects where $RF$ is defined is a triangulated subcategory, and
$RF$ will define a functor on this subcategory which transforms morphisms
of $S$ into isomorphisms.

\begin{lemma}
\label{lemma-derived-functor}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $f : X \to Y$ be a morphism of $\mathcal{D}$.
\begin{enumerate}
\item If $RF$ is defined at $X$ and $Y$ then there exists a unique
morphism $RF(f) : RF(X) \to RF(Y)$ between the values such that
for any commutative diagram
$$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$
with $s, s' \in S$ the diagram
$$\xymatrix{ F(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & RF(X) \ar[d] \\ F(Y) \ar[r] & F(Y') \ar[r] & RF(Y) }$$
commutes.
\item If $LF$ is defined at $X$ and $Y$ then there exists a unique
morphism $LF(f) : LF(X) \to LF(Y)$ between the values such that
for any commutative diagram
$$\xymatrix{ X' \ar[d]_{f'} \ar[r]_s & X \ar[d]^f \\ Y' \ar[r]^{s'} & Y }$$
with $s, s'$ in $S$ the diagram
$$\xymatrix{ LF(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & F(X) \ar[d] \\ LF(Y) \ar[r] & F(Y') \ar[r] & F(Y) }$$
commutes.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) holds if we only assume that the colimits
$$RF(X) = \colim_{s : X \to X'} F(X') \quad\text{and}\quad RF(Y) = \colim_{s' : Y \to Y'} F(Y')$$
exist. Namely, to give a morphism $RF(X) \to RF(Y)$ between the colimits
is the same thing as giving for each $s : X \to X'$ in $\Ob(X/S)$
a morphism $F(X') \to RF(Y)$ compatible with morphisms in the category
$X/S$. To get the morphism we choose a commutative diagram
$$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$
with $s, s'$ in $S$ as is possible by MS2 and we set
$F(X') \to RF(Y)$ equal to the composition $F(X') \to F(Y') \to RF(Y)$.
To see that this is independent of the choice of the diagram above use
MS3. Details omitted. The proof of (2) is dual.
\end{proof}

\begin{lemma}
\label{lemma-derived-inverts}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $s : X \to Y$ be an element of $S$.
\begin{enumerate}
\item $RF$ is defined at $X$ if and only if it is defined at $Y$.
In this case the map $RF(s) : RF(X) \to RF(Y)$ between values
is an isomorphism.
\item $LF$ is defined at $X$ if and only if it is defined at $Y$.
In this case the map $LF(s) : LF(X) \to LF(Y)$ between values
is an isomorphism.
\end{enumerate}
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-derived-shift}
\begin{slogan}
Derived functors are compatible with shifts
\end{slogan}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.
\begin{enumerate}
\item $RF$ is defined at $X$ if and only if it is defined at $X[n]$.
In this case there is a canonical isomorphism
$RF(X)[n]= RF(X[n])$ between values.
\item $LF$ is defined at $X$ if and only if it is defined at $X[n]$.
In this case there is a canonical isomorphism
$LF(X)[n] \to LF(X[n])$ between values.
\end{enumerate}
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-2-out-of-3-defined}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$.
If $RF$ is defined at two out of three of $X, Y, Z$, then it is defined
at the third. Moreover, in this case
$$(RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h))$$
is a distinguished triangle in $\mathcal{D}'$. Similarly for $LF$.
\end{lemma}

\begin{proof}
Say $RF$ is defined at $X, Y$ with values $A, B$.
Let $RF(f) : A \to B$ be the induced morphism, see
Lemma \ref{lemma-derived-functor}.
We may choose a distinguished triangle
$(A, B, C, RF(f), b, c)$
in $\mathcal{D}'$. We claim that $C$ is a value of $RF$ at $Z$.

\medskip\noindent
To see this pick $s : X \to X'$ in $S$ such that there exists a morphism
$\alpha : A \to F(X')$ as in
Categories,
Definition \ref{categories-definition-essentially-constant-diagram}.
We may choose a commutative diagram
$$\xymatrix{ X \ar[d]_f \ar[r]_s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' }$$
with $s' \in S$ by MS2. Using that $Y/S$ is filtered we can (after replacing
$s'$ by some $s'' : Y \to Y''$ in $S$) assume that there exists
a morphism $\beta : B \to F(Y')$ as in
Categories,
Definition \ref{categories-definition-essentially-constant-diagram}.
Picture
$$\xymatrix{ A \ar[d]_{RF(f)} \ar[r]_-\alpha & F(X') \ar[r] \ar[d]^{F(f')} & A \ar[d]^{RF(f)} \\ B \ar[r]^-\beta & F(Y') \ar[r] & B }$$
It may not be true that the left square commutes, but the outer and
right squares commute.
The assumption that the ind-object $\{F(Y')\}_{s' : Y' \to Y}$
is essentially constant means that there exists a $s'' : Y \to Y''$
in $S$ and a morphism $h : Y' \to Y''$ such that $s'' = h \circ s'$ and
such that $F(h)$ equal to $F(Y') \to B \to F(Y') \to F(Y'')$. Hence
after replacing $Y'$ by $Y''$ and $\beta$ by $F(h) \circ \beta$ the
diagram will commute (by direct computation with arrows).

\medskip\noindent
Using MS6 choose a morphism of triangles
$$(s, s', s'') : (X, Y, Z, f, g, h) \longrightarrow (X', Y', Z', f', g', h')$$
with $s'' \in S$. By TR3 choose a morphism of triangles
$$(\alpha, \beta, \gamma) : (A, B, C, RF(f), b, c) \longrightarrow (F(X'), F(Y'), F(Z'), F(f'), F(g'), F(h'))$$

\medskip\noindent
By
Lemma \ref{lemma-derived-inverts}
it suffices to prove that $RF(Z')$ is defined and has value $C$.
Consider the category $\mathcal{I}$ of
Lemma \ref{lemma-limit-triangles}
of triangles
$$\mathcal{I} = \{(t, t', t'') : (X', Y', Z', f', g', h') \to (X'', Y'', Z'', f'', g'', h'') \mid (t, t', t'') \in S\}$$
To show that the system $F(Z'')$ is essentially constant over the category
$Z'/S$ is equivalent to showing that the system of $F(Z'')$ is essentially
constant over $\mathcal{I}$ because $\mathcal{I} \to Z'/S$ is cofinal, see
Categories, Lemma \ref{categories-lemma-cofinal-essentially-constant}
(cofinality is proven in Lemma \ref{lemma-limit-triangles}).
For any object $W$ in $\mathcal{D}'$ we
consider the diagram
$$\xymatrix{ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(X'')) & \Mor_{\mathcal{D}'}(W, A) \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Y'')) \ar[u] & \Mor_{\mathcal{D}'}(W, B) \ar[u] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Z'')) \ar[u] & \Mor_{\mathcal{D}'}(W, C) \ar[u] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(X''[1])) \ar[u] & \Mor_{\mathcal{D}'}(W, A[1]) \ar[u] \ar[l] \\ \colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Y''[1])) \ar[u] & \Mor_{\mathcal{D}'}(W, B[1]) \ar[u] \ar[l] }$$
where the horizontal arrows are given by composing with
$(\alpha, \beta, \gamma)$. Since filtered colimits are exact
(Algebra, Lemma \ref{algebra-lemma-directed-colimit-exact}) the left column
is an exact sequence. Thus the $5$ lemma
(Homology, Lemma \ref{homology-lemma-five-lemma}) tells us the
$$\colim_\mathcal{I} \Mor_{\mathcal{D}'}(W, F(Z'')) \longrightarrow \Mor_{\mathcal{D}'}(W, C)$$
is bijective. Choose an object
$(t, t', t'') : (X', Y', Z') \to (X'', Y'', Z'')$ of $\mathcal{I}$.
Applying what we just showed to $W = F(Z'')$ and the element
$\text{id}_{F(X'')}$ of the colimit we find a unique morphism
$c_{(X'', Y'', Z'')} : F(Z'') \to C$ such that for some
$(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$
$$F(Z'') \xrightarrow{c_{(X'', Y'', Z'')}} C \xrightarrow{\gamma} F(Z') \to F(Z'') \to F(Z''') \quad\text{equals}\quad F(Z'') \to F(Z''')$$
The family of morphisms $c_{(X'', Y'', Z'')}$ form an element $c$ of
$\lim_\mathcal{I} \Mor_{\mathcal{D}'}(F(Z''), C)$ by uniqueness
(computation omitted). Finally, we show that
$\colim_\mathcal{I} F(Z'') = C$ via the morphisms $c_{(X'', Y'', Z'')}$
which will finish the proof by
Categories, Lemma \ref{categories-lemma-characterize-essentially-constant-ind}.
Namely, let $W$ be an object of $\mathcal{D}'$ and let
$d_{(X'', Y'', Z'')} : F(Z'') \to W$ be a family of maps corresponding
to an element of $\lim_\mathcal{I} \Mor_{\mathcal{D}'}(F(Z''), W)$.
If $d_{(X', Y', Z')} \circ \gamma = 0$, then for every object
$(X'', Y'', Z'')$ of $\mathcal{I}$ the morphism $d_{(X'', Y'', Z'')}$
is zero by the existence of $c_{(X'', Y'', Z'')}$ and the
morphism $(X'', Y'', Z'') \to (X''', Y''', Z'')$ in $\mathcal{I}$
satisfying the displayed equality above. Hence the map
$$\lim_\mathcal{I} \Mor_{\mathcal{D}'}(F(Z''), W) \longrightarrow \Mor_{\mathcal{D}'}(C, W)$$
(coming from precomposing by $\gamma$) is injective. However, it is
also surjective because the element $c$ gives a left inverse. We conclude
that $C$ is the colimit by
Categories, Remark \ref{categories-remark-limit-colim}.
\end{proof}

\begin{lemma}
\label{lemma-direct-sum-defined}
Assumptions and notation as in Situation \ref{situation-derived-functor}.
Let $X, Y$ be objects of $\mathcal{D}$.
\begin{enumerate}
\item If $RF$ is defined at $X$ and $Y$, then $RF$ is defined at $X \oplus Y$.
\item If $\mathcal{D}'$ is Karoubian and $RF$ is defined at $X \oplus Y$,
then $RF$ is defined at both $X$ and $Y$.
\end{enumerate}
In either case we have $RF(X \oplus Y) = RF(X) \oplus RF(Y)$.
Similarly for $LF$.
\end{lemma}

\begin{proof}
If $RF$ is defined at $X$ and $Y$, then the distinguished triangle
$X \to X \oplus Y \to Y \to X[1]$ (Lemma \ref{lemma-split}) and
Lemma \ref{lemma-2-out-of-3-defined}
shows that $RF$ is defined at $X \oplus Y$ and that we have
have a distinguished triangle
$RF(X) \to RF(X \oplus Y) \to RF(Y) \to RF(X)[1]$.
Applying Lemma \ref{lemma-split} to this once more we find
that $RF(X \oplus Y) = RF(X) \oplus RF(Y)$.
This proves (1) and the final assertion.

\medskip\noindent
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 \ref{categories-lemma-what-gets-inverted}.
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
\ref{categories-remark-left-localization-morphisms-colimit}).
By Categories, Lemma \ref{categories-lemma-essentially-constant-over-product}
$X/S \times Y/S$ 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 \ref{categories-lemma-cofinal-essentially-constant}.
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
\ref{homology-lemma-direct-sum-from-product-essentially-constant}
(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 $Y$.

\medskip\noindent
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
\ref{categories-definition-cofinal}. 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
$c \circ s'_1 = d \circ s'_2$.

\medskip\noindent
The proof of the corresponding statements for $LF$ are dual.
\end{proof}

\begin{proposition}
\label{proposition-derived-functor}
Assumptions and notation as in Situation \ref{situation-derived-functor}.
\begin{enumerate}
\item The full subcategory $\mathcal{E}$ of $\mathcal{D}$ consisting of
objects at which $RF$ is defined is a strictly full triangulated
subcategory of $\mathcal{D}$.
\item We obtain an exact functor
$RF : \mathcal{E} \longrightarrow \mathcal{D}'$
of triangulated categories.
\item Elements of $S$ with either source or target
in $\mathcal{E}$ are morphisms of $\mathcal{E}$.
\item The functor $S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}$
is a fully faithful exact functor of triangulated categories.
\item Any element of $S_\mathcal{E} = \text{Arrows}(\mathcal{E}) \cap S$
is mapped to an isomorphism by $RF$.
\item We obtain an exact functor
$$RF : S_\mathcal{E}^{-1}\mathcal{E} \longrightarrow \mathcal{D}'.$$
\item If $\mathcal{D}'$ is Karoubian, then $\mathcal{E}$ is a saturated
triangulated subcategory of $\mathcal{D}$.
\end{enumerate}
A similar result holds for $LF$.
\end{proposition}

\begin{proof}
Since $S$ is saturated it contains all isomorphisms (see remark
following Categories, Definition
\ref{categories-definition-saturated-multiplicative-system}). Hence
(1) follows from Lemmas \ref{lemma-derived-inverts},
\ref{lemma-2-out-of-3-defined}, and
\ref{lemma-derived-shift}. We get (2) from
Lemmas \ref{lemma-derived-functor}, \ref{lemma-derived-shift}, and
\ref{lemma-2-out-of-3-defined}. We get (3) from
Lemma \ref{lemma-derived-inverts}. The fully faithfulness in (4) follows
from (3) and the definitions. The fact that
$S_\mathcal{E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}$ is exact
follows from the fact that a triangle in $S_\mathcal{E}^{-1}\mathcal{E}$
is distinguished if and only if it is isomorphic to the image of a
distinguished triangle in $\mathcal{E}$, see proof of
Proposition \ref{proposition-construct-localization}.
Part (5) follows from Lemma \ref{lemma-derived-inverts}.
The factorization of $RF : \mathcal{E} \to \mathcal{D}'$
through an exact functor $S_\mathcal{E}^{-1}\mathcal{E} \to \mathcal{D}'$
follows from Lemma \ref{lemma-universal-property-localization}.
Part (7) follows from Lemma \ref{lemma-direct-sum-defined}.
\end{proof}

\noindent
Proposition \ref{proposition-derived-functor}
tells us that $RF$ lives on a maximal strictly full
triangulated subcategory of $S^{-1}\mathcal{D}$ and is an exact functor
on this triangulated category. Picture:
$$\xymatrix{ \mathcal{D} \ar[d]_Q \ar[rrr]_F & & & \mathcal{D}' \\ S^{-1}\mathcal{D} & & S_\mathcal{E}^{-1}\mathcal{E} \ar[ll]_{\text{fully faithful}}^{\text{exact}} \ar[ur]_{RF} }$$

\begin{definition}
\label{definition-everywhere-defined}
In
Situation \ref{situation-derived-functor}.
We say $F$ is {\it right derivable}, or that $RF$ {\it everywhere defined}
if $RF$ is defined at every object of $\mathcal{D}$.
We say $F$ is {\it left derivable}, or that $LF$ {\it everywhere defined}
if $LF$ is defined at every object of $\mathcal{D}$.
\end{definition}

\noindent
In this case we obtain a right (resp.\ left) derived functor

\label{equation-everywhere}
RF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}',
LF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}'),

see
Proposition \ref{proposition-derived-functor}.
In most interesting situations it is not the case that $RF \circ Q$ is
equal to $F$. In fact, it might happen that the canonical map
$F(X) \to RF(X)$ is never an isomorphism. In practice this does not happen,
because in practice we only know how to prove $F$ is right derivable by
showing that $RF$ can be computed by evaluating $F$ at judiciously chosen
objects of the triangulated category $\mathcal{D}$. This warrants
a definition.

\begin{definition}
\label{definition-computes}
In
Situation \ref{situation-derived-functor}.
\begin{enumerate}
\item An object $X$ of $\mathcal{D}$ {\it computes} $RF$ if $RF$ is defined
at $X$ and the canonical map $F(X) \to RF(X)$ is an isomorphism.
\item An object $X$ of $\mathcal{D}$ {\it computes} $LF$ if $LF$ is defined
at $X$ and the canonical map $LF(X) \to F(X)$ is an isomorphism.
\end{enumerate}
\end{definition}

\begin{lemma}
\label{lemma-computes-shift}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.
\begin{enumerate}
\item $X$ computes $RF$ if and only if $X[n]$ computes $RF$.
\item $X$ computes $LF$ if and only if $X[n]$ computes $LF$.
\end{enumerate}
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-2-out-of-3-computes}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
Let $(X, Y, Z, f, g, h)$ be a distinguished triangle of $\mathcal{D}$.
If $X, Y$ compute $RF$ then so does $Z$. Similar for $LF$.
\end{lemma}

\begin{proof}
By
Lemma \ref{lemma-2-out-of-3-defined}
we know that $RF$ is defined at $Z$ and that $RF$ applied to the
triangle produces a distinguished triangle.
Consider the morphism of distinguished triangles
$$\xymatrix{ (F(X), F(Y), F(Z), F(f), F(g), F(h)) \ar[d] \\ (RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h)) }$$
Two out of three maps are isomorphisms, hence so is the third.
\end{proof}

\begin{lemma}
\label{lemma-direct-sum-computes}
Assumptions and notation as in Situation \ref{situation-derived-functor}.
Let $X, Y$ be objects of $\mathcal{D}$. If $X \oplus Y$ computes $RF$, then
$X$ and $Y$ compute $RF$. Similarly for $LF$.
\end{lemma}

\begin{proof}
If $X \oplus Y$ computes $RF$, then $RF(X \oplus Y) = F(X) \oplus F(Y)$.
In the proof of Lemma \ref{lemma-direct-sum-defined} we have seen that
the functor $X/S \times Y/S \to (X \oplus Y)/S$, $(s, s') \mapsto s \oplus s'$
is cofinal. We will use this without further mention. Let $s : X \to X'$ be
an element of $S$. Then $F(X) \to F(X')$ has a section, namely,
$$F(X') \to F(X' \oplus Y) \to RF(X' \oplus Y) = RF(X \oplus Y) = F(X) \oplus F(Y) \to F(X).$$
where we have used Lemma \ref{lemma-derived-inverts}.
Hence $F(X') = F(X) \oplus E$ for some object $E$ of $\mathcal{D}'$
such that $E \to F(X' \oplus Y) \to RF(X'\oplus Y) = RF(X \oplus Y)$
is zero (Lemma \ref{lemma-when-split}).
Because $RF$ is defined at $X' \oplus Y$ with value
$F(X) \oplus F(Y)$ we can find a morphism $t : X' \oplus Y \to Z$
of $S$ such that $F(t)$ annihilates $E$. We may assume
$Z = X'' \oplus Y''$ and $t = t' \oplus t''$ with $t', t'' \in S$.
Then $F(t')$ annihilates $E$. It follows that $F$ is essentially constant
on $X/S$ with value $F(X)$ as desired.
\end{proof}

\begin{lemma}
\label{lemma-existence-computes}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
\begin{enumerate}
\item If for every object $X \in \Ob(\mathcal{D})$
there exists an arrow $s : X \to X'$ in $S$ such that $X'$ computes
$RF$, then $RF$ is everywhere defined.
\item If for every object $X \in \Ob(\mathcal{D})$
there exists an arrow $s : X' \to X$ in $S$ such that $X'$ computes
$LF$, then $LF$ is everywhere defined.
\end{enumerate}
\end{lemma}

\begin{proof}
This is clear from the definitions.
\end{proof}

\begin{lemma}
\label{lemma-find-existence-computes}
Assumptions and notation as in
Situation \ref{situation-derived-functor}.
If there exists a subset $\mathcal{I} \subset \Ob(\mathcal{D})$
such that
\begin{enumerate}
\item for all $X \in \Ob(\mathcal{D})$
there exists $s : X \to X'$ in $S$ with $X' \in \mathcal{I}$,
and
\item for every arrow $s : X \to X'$ in $S$ with $X, X' \in \mathcal{I}$
the map $F(s) : F(X) \to F(X')$ is an isomorphism,
\end{enumerate}
then $RF$ is everywhere defined and every $X \in \mathcal{I}$
computes $RF$. Dually, if there exists a subset
$\mathcal{P} \subset \Ob(\mathcal{D})$
such that
\begin{enumerate}
\item for all $X \in \Ob(\mathcal{D})$
there exists $s : X' \to X$ in $S$ with $X' \in \mathcal{P}$,
and
\item for every arrow $s : X \to X'$ in $S$ with $X, X' \in \mathcal{P}$
the map $F(s) : F(X) \to F(X')$ is an isomorphism,
\end{enumerate}
then $LF$ is everywhere defined and every $X \in \mathcal{P}$
computes $LF$.
\end{lemma}

\begin{proof}
Let $X$ be an object of $\mathcal{D}$.
Assumption (1) implies that the arrows $s : X \to X'$ in $S$ with
$X' \in \mathcal{I}$ are cofinal in the category $X/S$. Assumption
(2) implies that $F$ is constant on this cofinal subcategory.
Clearly this implies that $F : (X/S) \to \mathcal{D}'$ is essentially
constant with value $F(X')$ for any $s : X \to X'$ in $S$
with $X' \in \mathcal{I}$.
\end{proof}

\begin{lemma}
\label{lemma-compose-derived-functors-general}
Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be triangulated categories.
Let $S$, resp.\ $S'$ be a saturated multiplicative system in
$\mathcal{A}$, resp.\ $\mathcal{B}$ compatible with the triangulated structure.
Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$
be exact functors. Denote $F' : \mathcal{A} \to (S')^{-1}\mathcal{B}$ the
composition of $F$ with the localization functor.
\begin{enumerate}
\item If $RF'$, $RG$, $R(G \circ F)$ are everywhere defined, then there
is a canonical transformation of functors
$t : R(G \circ F) \longrightarrow RG \circ RF'$.
\item If $LF'$, $LG$, $L(G \circ F)$ are everywhere defined, then there
is a canonical transformation of functors
$t : LG \circ LF' \to L(G \circ F)$.
\end{enumerate}
\end{lemma}

\begin{proof}
In this proof we try to be careful. Hence let us think of
the derived functors as the functors
$$RF' : S^{-1}\mathcal{A} \to (S')^{-1}\mathcal{B}, \quad R(G \circ F) : S^{-1}\mathcal{A} \to \mathcal{C}, \quad RG : (S')^{-1}\mathcal{B} \to \mathcal{C}.$$
Let us denote
$Q_A : \mathcal{A} \to S^{-1}\mathcal{A}$ and
$Q_B : \mathcal{B} \to (S')^{-1}\mathcal{B}$
the localization functors. Then $F' = Q_B \circ F$. Note that for
every object $Y$ of $\mathcal{B}$ there is a canonical map
$$G(Y) \longrightarrow RG(Q_B(Y))$$
in other words, there is a transformation of functors
$t' : G \to RG \circ Q_B$. Let $X$ be an object of $\mathcal{A}$.
We have
\begin{align*}
R(G \circ F)(Q_A(X))
& = \colim_{s : X \to X' \in S} G(F(X')) \\
& \xrightarrow{t'} \colim_{s : X \to X' \in S} RG(Q_B(F(X'))) \\
& = \colim_{s : X \to X' \in S} RG(F'(X')) \\
& = RG(\colim_{s : X \to X' \in S} F'(X')) \\
& = RG(RF'(X)).
\end{align*}
The system $F'(X')$ is essentially constant in the category
$(S')^{-1}\mathcal{B}$. Hence we may pull the colimit inside the
functor $RG$ in the third equality of the diagram above, see
Categories, Lemma \ref{categories-lemma-image-essentially-constant}
and its proof. We omit the proof this this defines a transformation
of functors. The case of left derived functors is similar.
\end{proof}

Comment #454 by Keenan Kidwell on March 1, 2014 a 3:58 am UTC

In the last paragraph of 05S9, "will define functor" should be "will define a functor."

Comment #514 by Keenan Kidwell on April 3, 2014 a 12:58 pm UTC

In 05S9, the $s$ in the last sentence should be $S$.

Comment #516 by Keenan Kidwell on April 3, 2014 a 6:09 pm UTC

In the proof of 05SA, at the beginning, the colimits should be over $s:X\to X^\prime$ and $s:Y\to Y^\prime$ instead of the other way around.

Comment #518 by Johan (site) on April 3, 2014 a 6:26 pm UTC

@#514 and #516. Thanks and fixed here.

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