Situation 13.14.1. 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 $\mathcal{D}$.
13.14 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.
Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Recall from Categories, Remark 4.27.7 the filtered category $X/S$ of arrows $s : X \to X'$ in $S$ with source $X$. Dually, in Categories, Remark 4.27.15 we defined the cofiltered category $S/X$ of arrows $s : X' \to X$ in $S$ with target $X$.
Definition 13.14.2. Assumptions and notation as in Situation 13.14.1. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$.
we say the right derived functor $RF$ is defined at $X$ if the ind-object
is essentially constant1; in this case the value $Y$ in $\mathcal{D}'$ is called the value of $RF$ at $X$.
we say the left derived functor $LF$ is defined at $X$ if the pro-object
is essentially constant; in this case the value $Y$ in $\mathcal{D}'$ is called the value of $LF$ at $X$.
\[ (X/S) \longrightarrow \mathcal{D}', \quad (s : X \to X') \longmapsto F(X') \]
\[ (S/X) \longrightarrow \mathcal{D}', \quad (s: X' \to X) \longmapsto F(X') \]
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 13.14.3. Assumptions and notation as in Situation 13.14.1. Let $f : X \to Y$ be a morphism of $\mathcal{D}$.
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
with $s, s' \in S$ the diagram
commutes.
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
with $s, s'$ in $S$ the diagram
commutes.
\[ \xymatrix{ X \ar[d]_ f \ar[r]_ s & X' \ar[d]^{f'} \\ Y \ar[r]^{s'} & Y' } \]
\[ \xymatrix{ F(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & RF(X) \ar[d] \\ F(Y) \ar[r] & F(Y') \ar[r] & RF(Y) } \]
\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_ s & X \ar[d]^ f \\ Y' \ar[r]^{s'} & Y } \]
\[ \xymatrix{ LF(X) \ar[d] \ar[r] & F(X') \ar[d] \ar[r] & F(X) \ar[d] \\ LF(Y) \ar[r] & F(Y') \ar[r] & F(Y) } \]
Proof. Part (1) holds if we only assume that the colimits
\[ RF(X) = \mathop{\mathrm{colim}}\nolimits _{s : X \to X'} F(X') \quad \text{and}\quad RF(Y) = \mathop{\mathrm{colim}}\nolimits _{s' : Y \to Y'} F(Y') \]
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 $\mathop{\mathrm{Ob}}\nolimits (X/S)$ 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 13.14.4. Assumptions and notation as in Situation 13.14.1. Let $s : X \to Y$ be an element of $S$.
$RF$ 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.
$LF$ 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 13.14.5. Assumptions and notation as in Situation 13.14.1. Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.
$RF$ 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.
$LF$ 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 13.14.6. Assumptions and notation as in Situation 13.14.1. 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 is a distinguished triangle in $\mathcal{D}'$. Similarly for $LF$.
\[ (RF(X), RF(Y), RF(Z), RF(f), RF(g), RF(h)) \]
Proof. Say $RF$ is defined at $X, Y$ with values $A, B$. Let $RF(f) : A \to B$ be the induced morphism, see Lemma 13.14.3. 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 13.14.4 it suffices to prove that $RF(Z')$ is defined and that the arrow $\gamma : C \to F(Z')$ induces an isomorphism $C \to RF(Z')$. Namely, then we will get an isomorphism
\[ (A, B, C, RF(f), b, c) \longrightarrow (RF(X'), RF(Y'), RF(Z'), RF(f'), RF(g'), RF(h')) \]
of triangles and by TR1 we conclude that the target is a distinguished triangle. Consider the category $\mathcal{I}$ of Lemma 13.5.10 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 4.22.11 (cofinality is proven in Lemma 13.5.10). For any object $W$ in $\mathcal{D}'$ we consider the diagram
\[ \xymatrix{ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(X'')) & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, A) \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Y'')) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, B) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Z'')) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, C) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(X''[1])) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, A[1]) \ar[u] \ar[l] \\ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Y''[1])) \ar[u] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, B[1]) \ar[u] \ar[l] } \]
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 map
\[ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, F(Z'')) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(W, C) \]
is bijective. We conclude that $F(Z'')$ is essentially constant over $\mathcal{I}$ with value $C$ by part (4) of Categories, Lemma 4.22.9. $\square$
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. $\square$
Proposition 13.14.8. Assumptions and notation as in Situation 13.14.1.
The full subcategory $\mathcal{E}$ of $\mathcal{D}$ consisting of objects at which $RF$ is defined is a strictly full triangulated subcategory of $\mathcal{D}$.
We obtain an exact functor $RF : \mathcal{E} \longrightarrow \mathcal{D}'$ of triangulated categories.
Elements of $S$ with either source or target in $\mathcal{E}$ are morphisms of $\mathcal{E}$.
Any element of $S_\mathcal {E} = \text{Arrows}(\mathcal{E}) \cap S$ is mapped to an isomorphism by $RF$.
The set $S_\mathcal {E}$ is a saturated multiplicative system in $\mathcal{E}$ compatible with the triangulated structure.
The functor $S_\mathcal {E}^{-1}\mathcal{E} \to S^{-1}\mathcal{D}$ is a fully faithful exact functor of triangulated categories.
We obtain an exact functor
If $\mathcal{D}'$ is Karoubian, then $\mathcal{E}$ is a saturated triangulated subcategory of $\mathcal{D}$.
\[ RF : S_\mathcal {E}^{-1}\mathcal{E} \longrightarrow \mathcal{D}'. \]
A similar result holds for $LF$.
Proof. Since $S$ is saturated it contains all isomorphisms (see remark following Categories, Definition 4.27.20). Hence (1) follows from Lemmas 13.14.4, 13.14.6, and 13.14.5. We get (2) from Lemmas 13.14.3, 13.14.5, and 13.14.6. We get (3) from Lemma 13.14.4. Part (4) follows from Lemma 13.14.4. Part (5) follows from the definitions and part (3). The fully faithfulness in (6) 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 13.5.6. 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 13.5.7. Finally, part (8) follows from Lemma 13.14.7. $\square$
Proposition 13.14.8 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 13.14.9. In Situation 13.14.1. 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
13.14.9.1
\begin{equation} \label{derived-equation-everywhere} RF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}', \quad \text{(resp. } LF : S^{-1}\mathcal{D} \longrightarrow \mathcal{D}'), \end{equation}
see Proposition 13.14.8. 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 13.14.10. In Situation 13.14.1.
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.
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 13.14.11. Assumptions and notation as in Situation 13.14.1. Let $X$ be an object of $\mathcal{D}$ and $n \in \mathbf{Z}$.
$X$ computes $RF$ if and only if $X[n]$ computes $RF$.
$X$ computes $LF$ if and only if $X[n]$ computes $LF$.
Proof. Omitted. $\square$
Lemma 13.14.12. Assumptions and notation as in Situation 13.14.1. 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 13.14.6 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 13.14.13. Assumptions and notation as in Situation 13.14.1. 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 13.14.7 we have seen that the functor $X/S \times Y/S \to (X \oplus Y)/S$, $(s, s') \mapsto s \oplus s'$ is cofinal. Thus by Categories, Lemma 4.22.11 and by characterization (4) of Categories, Lemma 4.22.9 we know that for any object $W$ in $\mathcal{D}'$ the map
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X \oplus Y), W) \longrightarrow \mathop{\mathrm{colim}}\nolimits _{s : X \to X', s' : Y \to Y'} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X' \oplus Y'), W) \]
is bijective. Since this arrow is clearly compatible with direct sum decompositions on both sides, we conclude that the map
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X), W) \longrightarrow \mathop{\mathrm{colim}}\nolimits _{s : X \to X'} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X'), W) \]
is bijective (minor detail omitted). Thus by Categories, Lemma 4.22.9 we conclude $RF$ is defined at $X$ with value $F(X)$. Similarly for $Y$. $\square$
Lemma 13.14.14. Assumptions and notation as in Situation 13.14.1.
If for every object $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ there exists an arrow $s : X \to X'$ in $S$ such that $X'$ computes $RF$, then $RF$ is everywhere defined.
If for every object $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ 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 13.14.15. Assumptions and notation as in Situation 13.14.1. If there exists a subset $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that
for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ there exists $s : X \to X'$ in $S$ with $X' \in \mathcal{I}$, and
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 $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that
for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ there exists $s : X' \to X$ in $S$ with $X' \in \mathcal{P}$, and
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 13.14.16. 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.
If $RF'$, $RG$, $R(G \circ F)$ are everywhere defined, then there is a canonical transformation of functors $t : R(G \circ F) \longrightarrow RG \circ RF'$.
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{\mathrm{colim}}\nolimits _{s : X \to X' \in S} G(F(X')) \\ & \xrightarrow {t'} \mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} RG(Q_ B(F(X'))) \\ & = \mathop{\mathrm{colim}}\nolimits _{s : X \to X' \in S} RG(F'(X')) \\ & = RG(\mathop{\mathrm{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 4.22.8 and its proof. We omit the proof this defines a transformation of functors. The case of left derived functors is similar. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (4)
Comment #454 by Keenan Kidwell on
Comment #514 by Keenan Kidwell on
Comment #516 by Keenan Kidwell on
Comment #518 by Johan on