Chapter 6
The Derived Category
6.1 The Homotopy Category
S YNOPSIS . Homotopy category; (co)product; triangulation; universal property; unique lifting
properties.
O BJECTS AND M ORPHISMS
6.1.1 Definition. The homotopy categeory K(R) has the same objects as C(R), that
is, R-complexes, and the morphisms in K(R) are homotopy equivalence classes of
morphisms in C(R).
6.1.2. By 2.3.11 there is an equality K(R)(M, N) = H0 (HomR (M, N)) of k-modules
for R-complexes M and N. In accordance with 2.2.12 we write [α] for the homotopy
equivalence class of a morphism α in C(R). If K is yet another R-complex then the
composition K(R)(M, N) × K(R)(K, M) → K(R)(K, N) maps ([α], [β]) to [αβ].
6.1.3. Let α be a morphism in C(R). By definition, [α] is the zero morphism in K(R)
if and only if α is null-homotopic; see 2.2.20. Furthermore, [α] is an isomorphism
in K(R) if and only if α is a homotopy equivalence; see 2.2.25.
6.1.4. There is a canonical full functor Q : C(R) → K(R); it is the identity on objects
and it maps a morphism α in C(R) to its homotopy equivalence class Q(α) = [α].
P RODUCTS , C OPRODUCTS , AND k- LINEARITY
The lemma below follows immediately from the definitions.
6.1.5 Lemma. Let U and V be k-prelinear categories that have the same objects,
and let F : U → V be a k-linear functor that is the identity on objects. If M and N
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6 The Derived Category
are objects and if the tuple (M ⊕ N, πM , ιM , πN , ιN ) is a biproduct in U then the tuple
(M ⊕ N, F(πM ), F(ιM ), F(πN ), F(ιN )) is a biproduct in V. In particular, if every pair
of objects has a biproduct in U then every pair of objects has a biproduct in V.
When we say that a category “has (co)products”, we mean that all set-indexed
(co)products exist in the category; such objects are unique up to isomorphism.
6.1.6 Theorem. The homotopy category K(R) and the functor Q : C(R) → K(R)
are k-linear. For every family {M u }u∈U of R-complexes the next assertions hold.
(a) If M with embeddings {ιu : M u M}u∈U is the coproduct of {M u}u∈U in C(R),
then M with the morphisms {[ιu ]}u∈U is the coproduct of {M u }u∈U in K(R).
(b) If M with projections {πu : M u M}u∈U is the product of {M u }u∈U in C(R),
then M with the morphisms {[πu ]}u∈U is the product of {M u }u∈U in K(R).
In particular, the homotopy category K(R) has coproducts and products, and the
canonical functor Q preserves coproducts and products.
P ROOF . It is straightforward to verify that the category K(R) is k-prelinear and that
the canonical functor Q is k-linear. Evidently, the zero complex is a zero object in
K(R), and K(R) has biproducts by 6.1.5. Thus K(R) is a k-linear category.
(a): Let {[αu ] : M u → N}u∈U be morphisms in K(R). The task is to show that
there exixts a unique morphism [α] : M → N in K(R) with [αιu ] = [αu ] for all u ∈ U.
Existence is straightforward; indeed, by the universal property of coproducts in
C(R), there exists a (unique) morphism α : M → N with αιu = αu for all u ∈ U.
Applying Q to these identities one gets [αιu ] = [αu ].
For uniqueness, assume that [αιu ] = [0] holds for all u ∈ U; it must be shown
that [α] is [0]. Since each αιu is null-homotopic there are degree 1 homomorphisms
u
τu : M u → N such that αιu = ∂ N τu + τu ∂ M holds for all u ∈ U. Now consider each
u
u\
homomorphism τ as a morphism M → Σ N \ of graded R-modules. Since M \ together with the embeddings {ιu : M u\ M \ }u∈U is a coproduct of {M u\ }u∈U in
Mgr (R), there is a morphism τ : M \ → Σ N \ with τιu = τu for all u ∈ U. Viewing τ
as a degree 1 homomorphism M → N, it follows that one has
u
u
αιu = ∂ N τu + τu ∂ M = ∂ N τιu + τιu ∂ M = (∂ N τ + τ∂ M )ιu ,
where the second equality is by definition of τ, and the third equality holds as ιu is
a morphism in C(R). As ∂ N τ + τ∂ M is a morphism of R-complexes, it follows from
the universal property of coproducts in C(R) that one has α = ∂ N τ + τ∂ M . Thus α
is null-homotopic, that is, [α] is [0] as desired.
(b): Similar to the proof of part (a).
By construction, the canonical functor Q preserves (co)products.
The following result shows that the zero objects in the homotopy category are
exactly the contractible complexes. Further characterizations of such complexes are
given in 4.1.23.
6.1.7 Proposition. An R-complex is isomorphic to 0 in K(R) if and only if it is
contractible.
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193
P ROOF . Let M be an R-complex. If M is isomorphic to 0 in K(R) then K(R)(M, M)
consists of a single element. In particular, [1M ] = [0] holds, so 1M is null-homotopic,
i.e. M is contractible. Conversely, if M is contractible then the morphism M → 0 in
C(R) is a homotopy equivalence, whence it represents an isomorphism in K(R).
6.1.8 Proposition. There is a unique endofunctor on K(R) that makes the following
diagram commutative,
C(R)
Q
Σ
C(R)
Q
/ K(R)
/ K(R) .
This functor is denoted ΣK ; it is k-linear and invertible.
P ROOF . It is elementary to verify that the endofunctor on K(R) that maps an object
M to Σ M and a morphism [α] to [Σ α] has the asserted properties.
When there is no risk of ambiguity, we write Σ for the functor ΣK .
T RIANGULATION
Consider the k-linear category K(R), see 6.1.6, equipped with the k-linear and invertible endofunctor Σ = ΣK from 6.1.8. One may now speak of candidate triangles
in K(R) in the sense of A.1.
6.1.9 Lemma. Let α : M → N be a morphism in C(R). The image under the canonical functor Q : C(R) → K(R) of the diagram
M
α
/N
N
1
0
/ Cone α
( 0 1Σ M )
/
ΣM
is a candidate triangle in K(R).
P ROOF . We must prove that the three composites in C(R),
N
1N
α
1
Σ
M
α=
, ψ= 0 1
= 0 , and χ = (Σ α) 0 1Σ M = 0 Σ α
ϕ=
0
0
0
are null-homotopic. Since ψ is even zero in C(R), we are left to consider ϕ and χ.
Define degree 1 homomorphisms % : M → Cone α and τ : Cone α → Σ N by
0
%=
and τ = σN1 0 ,
σM
1
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6 The Derived Category
s
M
where σM
s : M → Σ M is the map introduced in 2.2.3. From the fact that σs is a
degree s chain map and from commutativity of the diagram (2.2.3.1), it follows that
there are equalities ∂ Cone α % + %∂ M = ϕ and ∂ Σ N τ + τ∂ Cone α = χ. Indeed, one has
N ΣN ∂ σ−1 Σ α
0
0
α
M
+
M ∂ = 0
σ
σM
0 ∂ΣM
1
1
and
∂
ΣN
∂ N σΣ N Σ α
N
−1
0 + σ1 0
= 0 Σα .
0 ∂ΣM
σN1
6.1.10 Definition. A candidate triangle in K(R) of the form considered in 6.1.9 is
called a strict triangle. A candidate triangle in K(R) that is isomorphic, in the sense
of A.1, to a strict triangle is called a distinguished triangle.
6.1.11 Theorem. The category K(R), equipped with the invertible endofunctor Σ
and the notion of distinguished triangles from 6.1.10, is triangulated.
P ROOF . We verify the axioms in A.3.
(TR0): Evidently, the collection of distinguished triangles is closed under isomorphisms. Furthermore, it follows from 4.1.23 and 6.1.7 that application of the
canonical functor Q : C(R) → K(R) to the following diagram in C(R),
M
1M
/M
M
1
0
/ Cone(1M )
( 0 1Σ M )
/
ΣM ,
1M
yields, up to isomorphism in K(R), the candidate triangle M −→ M −→ 0 −→ Σ M
which, therefore, is distinguished.
(TR1): By the definition of morphisms in K(R), every morphism in this category
fits into a distinguished (even a strict) triangle; see. 6.1.9.
(TR2’): By A.4 it it sufficient to verify that (TR2) holds. Thus, let
β0
α0
γ0
∆ = M 0 −−→ N 0 −−→ X 0 −−→ Σ M 0
be a distinguished triangle in K(R). We must argue that the candidate triangles
β0
γ0
− Σ α0
∆0 = N 0 −−→ X −−→ Σ M 0 −−−→ Σ N 0
− Σ−1 γ0
α0
β0
and ∆00 = Σ−1 X 0 −−−−→ M 0 −−→ N 0 −−→ X 0
are distinguished. Up to isomorphism, ∆ is given by application of the canonical
functor Q to a diagram in C(R) of the form,
M
α
/N
N
1
0
/ Cone α
( 0 1Σ M )
/
ΣM .
Thus, the candidate triangles ∆0 and ∆00 are, up to isomorphism, given by application
of Q to the following diagrams in C(R),
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195
N
Σ−1 Cone α
( 0 −1M )
α
/M
/N
N
1
0
/ Cone α
N
1
0
( 0 1Σ M )
/
ΣM
−Σα
/ ΣN ,
/ Cone α .
These two diagrams in C(R) are the top rows in (?) and (‡) below. By definition,
the bottom rows in (?) and (‡) give strict triangles in K(R) when the functor Q
is applied; see 6.1.10. Thus, to show that ∆0 and ∆00 are distinguished triangles in
K(R), it suffices to argue that (?) and (‡) are commutative up to homotopy, and that
all vertical morphisms are homotopy equivalences.
N
N
ι= 1
0
/ Cone α
( 0 1Σ M )
ϑ =( 0 1Σ M 0 )
(?)
N
N
ι= 1
0
Σ−1 Cone α
/ Cone α
π =( 0 −1M )
/M
1N 0
0 1Σ M
0 0
α
!
ϕ=
/ Cone ι
Σ−1 Cone α
π =( 0 −1M )
/M
M
1
0
0
0
1Σ M
−Σα
/ ΣN
!
( 0 0 1Σ N )
/ ΣN
N
1
0
/N
O
ξ =( α 1N 0 )
(‡)
−Σα
/ ΣM
O
/ Cone α
0 ψ = 1N
0
/ Cone π
0 1N 0
0 0 1Σ M
/
Cone α
First consider the diagram (?). Note that ϕ and ϑ are morphisms, as one has
 N

 

∂ Σ α 1Σ N
0
0
∂ Cone ι ϕ =  0 ∂ Σ M 0   1Σ M  =  1Σ M  ∂ Σ M = ϕ∂ Σ M
−Σα
−Σα
0 0 ∂ΣN
and
∂ Σ M ϑ = ∂ Σ M 0 1Σ M
 N

∂ Σ α 1Σ N
0 = 0 1Σ M 0  0 ∂ Σ M 0  = ϑ∂ Cone ι .
0 0 ∂ΣN
To make sense of these computations, recall from 4.1.10 that in the matrix ∂ Cone ι ,
the maps Σ α and 1Σ N are viewed as a degree −1 chain maps Σ M → N and Σ N → N.
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6 The Derived Category
We claim that ϕ is a homotopy equivalence with homotopy inverse ϑ. It is clear
that ϑϕ = 1Σ M holds, so it remains to show that the morphism
 N
 

 N

1 0 0
0
1 0 0
1Cone ι − ϕϑ =  0 1Σ M 0  −  1Σ M  0 1Σ M 0 =  0 0 0 
−Σα
0 Σ α 1Σ N
0 0 1Σ N
is null-homotopic. Consider the map σ : Cone ι → Cone ι given by


0 00
σ =  0 0 0 ;
1N 0 0
here 1N is viewed as a degree 1 chain map N → Σ N, and thus σ is a degree 1
homomorphism; cf. 2.2.3. It is elementary to verify that one has
 N

 
 N
  N

∂ Σ α 1Σ N
∂ Σ α 1Σ N
1 0 0
0 00
0 00
 0 ∂ Σ M 0   0 0 0 +  0 0 0  0 ∂ Σ M 0  =  0 0 0  ;
1N 0 0
1N 0 0
0 Σ α 1Σ N
0 0 ∂ΣN
0 0 ∂ΣN
that is, ∂ Cone ι σ + σ∂ Cone ι = 1Cone ι − ϕϑ holds. Thus ϑ is a homotopy inverse of ϕ.
Now we turn to the issue of commutativity of (?). The left- and right-hand squares
in (?) are even commutative in C(R). For the commutativity, up to homotopy, of the
middle square, it must be proved that the morphism β : Cone α → Cone ι, defined by
 N
 

 N

0
1 0
1 0
β =  0 1Σ M  −  1Σ M  0 1Σ M =  0 0  ,
0 Σα
−Σα
0 0
is null-homotopic. Let τ : Cone α → Cone ι be the map


0 0
τ =  0 0 ,
1N 0
where 1N is viewed as a degree 1 chain map N → Σ N. Thus τ is a degree 1 homomorphism. It is straightforward to verify the equality
 N

 

 N

∂ Σ α 1Σ N
0 0
0 0 N
1 0
 0 ∂ Σ M 0   0 0 +  0 0 ∂ ΣΣαM =  0 0  ;
0 ∂
1N 0
1N 0
0 Σα
0 0 ∂ΣN
that is, ∂ Cone ι τ + τ∂ Cone α = β holds, and hence β is null-homotopic.
For the second diagram (‡), arguments similar to the ones given above show that
ψ is a homotopy equivalence with homotopy inverse ξ, and that (‡) is commutative
up to homotopy.
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197
(TR4’): Consider the following diagram in C(R), where the rows and the morphisms ϕ and ψ are given, and the left-hand square is commutative up to homotopy,
α
M
ϕ
()
/N
N
1
0
α0
/ N0
( 0 1Σ M )
11 12 χ χ
χ = 21 22
χ χ
ψ
M0
/ Cone α
1N
0
0
/ Cone α0
( 0 1Σ M 0 )
/ ΣM
Σϕ
/ Σ M0 .
To verify that (TR4’) holds we are, in view of the definition of distinguished triangles in K(R), required to prove that there exists a morphism χ : Cone α → Cone α0
with the following two properties: First of all, χ must make () commutative up to
homotopy. In this case, (Q(ϕ), Q(ψ), Q(χ)) is a morphism of distinguished triangles in K(R), and its mapping cone candidate triangle is given by application of the
functor Q to the following diagram in C(R),
(§)
M0
⊕
N
α0 ψ
0 −1N
0 0

0
1N χ11 χ12
 0 χ21 χ22 
0 0 −1Σ M /

!
N0
/ ⊕
Cone α
Cone α0
⊕
ΣM
0
0 1Σ M Σ ϕ
0 0 −Σα
Σ M0
/ ⊕ .
ΣN
Secondly, Q applied to (§) must yield a distinguished triangle in K(R).
We start by constructing a morphism χ that makes () commutative up to homotopy. By assumption, there is a degree 1 homomorphism σ : M → N 0 such that
0
the equality ψα − α0 ϕ = ∂ N σ + σ∂ M holds. Define χ : Cone α → Cone α0 to be the
degree 0 homomorphism
ψ Σσ
χ=
,
0 Σϕ
where Σ σ is viewed as a degree 0 homomorphism Σ M → N 0 . It is elementary to
verify that χ is a morphism; that is,
N0
N
0
∂ Σα
∂ Σ α0
ψ Σσ
ψ Σσ
∂ Cone α χ =
=
= χ∂ Cone α .
0
0 Σϕ
0 Σϕ
0 ∂ΣM
0 ∂ΣM
Notice that χ makes the middle and right-hand squares in () commutative in C(R).
Finally, to see that application of the functor Q to (§) yields a distinguished triangle in K(R), note that (§) is the top row in following diagram, and that the bottom
row yields a strict triangle in K(R) when Q is applied. Thus, it suffices to argue that
the diagram below is commutative up to homotopy, and that all vertical morphisms
are homotopy equivalences.
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198
M0
⊕
N
6 The Derived Category
α0 ψ
θ = 0 −1N
0 0

0
1N ψ Σ σ
 0 0 Σϕ 
0 0 −1Σ M

!
N0
/ ⊕
Cone α

0
1N ψ Σ σ
0
0



η = 0 0 Σ ϕ 1Σ M 0 0 
Σ
M
0 0 −1
0 0
M0
⊕
N
θ=
α0 ψ
0 −1N
0 0
!
/
N0
⊕
Cone α

0
1N 0 0
N
 0 1
0 


 0 0 1Σ M 
0 0 0
0 0 0

Cone α0
/ ⊕
ΣM
O
0
0 1Σ M Σ ϕ
0 0 −Σα

0
1N
0
Σσ
 0
0
0 


ξ = 0
0 −1Σ M 


0
0 1Σ M Σ ϕ
0
0 −Σα
Σ M0
/ ⊕
ΣN

/ Cone θ
0
0 0 0 1Σ M 0
0 0 0 0 1Σ N
Σ M0
/ ⊕
ΣN
The differentials on the complexes Cone α0 ⊕ Σ M and Cone θ are given by
 N0

∂
0 0 Σ α0 Σ ψ
 N0

 0 ∂ N Σ α 0 −1Σ N 
∂ Σ α0 0


0
Cone α0 ⊕Σ M
Cone θ
ΣM
Σ
M


0
0 
∂
=
and ∂
=
0 ∂
0
 0 0 ∂
.
 0 0 0 ∂ Σ M0 0 
0
0 ∂ΣM
0 0 0
0 ∂ΣN
It is straightforward to verify that ξ and η are morphisms. Evidently there is an equal0
ity ηξ = 1Cone α ⊕Σ M . Furthermore, the morphism ξη − 1Cone θ is null-homotopic, as
the degree 1 homomorphism τ : Cone θ → Cone θ given by


0 0 000
0 0 0 0 0



τ=
0 0 0 0 0
0 0 0 0 0
0 1N 0 0 0
satisfies the identity ∂ Cone θ τ + τ∂ Cone θ = ξη − 1Cone θ . Hence ξ is a homotopy equivalence with homotopy inverse η.
The left-hand and right-hand squares in the diagram are commutative in C(R).
The diagram’s middle square is commutative up to homotopy, indeed, the morphism
γ : N 0 ⊕ Cone α → Cone θ, defined by
 N0

 N0
 

1
0
Σσ
0 ψ 0
1 0 0
 N0

 0 0
 0 1N 0  0 −1N 0 
0 

 1 ψ Σσ

 

Σ
M




0 0 0
γ =  0 0 −1  0 0 Σ ϕ
−
=
0 0 1Σ M 



,
 0 1Σ M0 Σ ϕ  0 0 −1Σ M
 0 0 0  0 0 0 
0 0 Σα
0 0 0
0 0 −Σα
is null-homotopic. This follows as % : N 0 ⊕ Cone α → Cone θ, given by
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199

0
0

%=
0
0
0
0
0
0
0
1N

0
0

0
,
0
0
0
is a degree 1 homomorphism with ∂ Cone θ % + %∂ N ⊕Cone α = γ.
T HE U NIVERSAL P ROPERTY
6.1.12 Definition. Let (U, ΣU ) be a triangulated category. A functor F : C(R) → U is
called quasi-triangulated if there is a natural isomorphism ϕ : FΣ → ΣU F such that
F(M)
F(α)
/ F(N)
N
F 1
0 /
F(Cone α)
ϕM ◦ F( 0 1Σ M )
/ ΣU F(M)
is a distinguished triangle in U for every morphism of R-complexes α : M → N.
The canonical functor Q : C(R) → K(R) has the following universal property.
6.1.13 Theorem. If F : C(R) → U is a functor that maps homotopy equivalences
to isomorphisms, then there exists a unique functor F0 : K(R) → U that makes the
following diagram commutative,
C(R)
F
{
U;
Q
/ K(R)
F0
here Q is the canonical functor from 6.1.4. Furthermore, the next assertions hold.
(a) Assume that U is k-prelinear; then F is k-linear if and only if F0 is k-linear.
(b) Assume that U has (co)products; then F preserves (co)products if and only if
F0 preserves (co)products.
(c) If U is triangulated and F is quasi-triangulated, then F0 is triangulated.
P ROOF . Uniqueness of F0 follows as Q is the identity on objects and it is full.
For existence of F0 , set F0 (M) = F(M) for every R-complex M and F0 ([α]) = F(α)
for every morphism α of R-complexes. To see that this makes sense—in which case
the the identity F0 Q = F evidently holds—let α, β : M → N be homotopic morphisms
of R-complexes. It must be shown that one has F(α) = F(β). To this end, define an
R-complex C as follows,
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6 The Derived Category
M\
⊕
C\ = M \
⊕
Σ M\
and
 M

∂
0 −1Σ M
∂ C =  0 ∂ M 1Σ M  ,
0 0 ∂ΣM
where the maps ±1Σ M in the right-hand column of the matrix ∂ C are viewed as
degree −1 chain maps Σ M → M. The following three maps are morphisms in C(R),
M
1
ε= 0
0
M
0 ι = 1M
0
/
/C
π =( 1M 1M 0 )
/M.
Moreover, ι is a homotopy equivalence with homotopy inverse π, indeed, the equality πι = 1M evidently holds. To prove that the morphism
 
 M
  M

1 0 0
−1 0 0
0
ιπ − 1C = 1M  1M 1M 0 −  0 1M 0  =  1M 0 0 
0
0 0 1Σ M
0 0 −1Σ M
is null-homotopic, consider the degree 1 homomorphism σ : C → C given by


0 00
σ =  0 0 0 ,
1M 0 0
where 1M is viewed as degree 1 chain map M → Σ M. One readily verifies the equality ∂ C σ + σ∂ C = ιπ − 1C , that is,
 M

 
 M
  M

∂
0 −1Σ M
∂
0 −1Σ M
−1 0 0
0 00
0 00
 0 ∂ M 1Σ M  0 0 0+ 0 0 0 0 ∂ M 1Σ M  =  1M 0 0  .
1M 0 0
1M 0 0
0 0 −1Σ M
0 0 ∂ΣM
0 0 ∂ΣM
These arguments show that ι is a homotopy equivalence with homotopy inverse π. It
follows that F(ι) is an isomorphism in U with inverse F(π). As πε = 1M , and hence
F(π)F(ε) = 1F(M) , holds it follows that one has F(ε) = F(ι). Since the morphisms
α, β : M → N are homotopic, there exists a degree 1 homomorphism % : M → N with
β − α = ∂ N % + %∂ M . Viewing Σ % as a degree 0 homomorphism Σ M → N, the degree
0 homomorphism γ = α β Σ % : C → N is a morphism, as one has

 M
∂
0 −1Σ M
∂ N γ = ∂ N α β Σ % = α β Σ %  0 ∂ M 1Σ M  = γ∂ C .
0 0 ∂ΣM
From the equalities α = γε and γι = β one gets F(α) = F(γ)F(ε) = F(γ)F(ι) = F(β).
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201
It remains to prove the assertions (a), (b), and (c).
(a): If F0 is k-linear then so is F = F0 Q, as a composition of two k-linear functors.
Conversely, if F is k-linear then so is F0 , since the equalities
F0 (x[α] + [β]) = F0 ([xα + β]) = F(xα + β) = xF(α) + F(β) = xF0 ([α]) + F0 ([β])
hold for every pair α, β of parallel morphisms in C(R) and every element x in k.
(b): Let {M u }u∈U be a family of R-complexes. Since the functor Q preserves
coproducts; see 6.1.6, the canonical morphism
Mu =
`
u∈U
`
ψ
Q(M u ) −−→ Q(
u∈U
Mu)
`
u∈U
in K(R) is an isomorphism; cf. 3.1.9. Application of F0 yields an isomorphism
F0 (
`
F0 (ψ)
M u ) −−−→ F0 Q(
u∈U
`
M u ) = F(
u∈U
`
Mu)
u∈U
in U such that there is a commutative diagram
F0 (
`
u∈U
Mu)
F0 (ψ)
∼
=
/ F(`u∈U M u )
ϕ
ϕ0
`
0
u
u∈U F (M )
`
u
u∈U F(M ) ,
where ϕ and ϕ0 are the canonical morphisms. It follows that ϕ is and isomorphism if
and only ϕ0 is an isomorphism; and hence the functor F preserves coproducts if and
only if F0 preserves coproducts.
The assertion about products is proved similarly.
(c): Let ϕ : FΣC → ΣU F be a natural isomorphism as in 6.1.12. By 6.1.8 there
are equalities FΣC = F0 QΣC = F0 ΣK Q, and one has ΣU F = ΣU F0 Q. Since Q is the
identity on objects, ϕ can be viewed as a natural isomorphism ϕ0 : F0 ΣK → ΣU F0 .
We verify that the functor F0 : K(R) → U with the isomorphism ϕ0 , is triangulated.
By definition, every distinguished triangle in K(R) is isomorphic to a strict triangle,
that is, to a diagram of the form
M
Q(α)
/N
N
Q 1
0
/ Cone α
Q( 0 1Σ M )
/
ΣK M ,
where α : M → N is a morphism in C(R). Thus, it must be shown that the following
candidate triangle in U is distinguished,
F0 (M)
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F0 Q(α)
/ F0 (N)
N
F0 Q 1
0 /
F0 (Cone α)
ϕ0M ◦ F0 Q( 0 1Σ M )
/ ΣU F0 (M) .
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6 The Derived Category
However, this diagram is identical to the one in 6.1.12, which is distinguished triangle in U since F is assumed to be quasi-triangulated.
R EMARK . The main assertion in 6.1.13 is that the homotopy category K(R) is the localization of
C(R) with respect to the collection of homotopy equivalences. In the next section, we focus on yet
another important example of a localization, namely the localization of K(R) with respect to the
collection of quasi-isomorphisms; this gives rise to the derived category.
6.1.14 Definition. Let (V, ΣV ) be a triangulated category. A functor G : C(R)op → V
is called quasi-triangulated if the functor Gop from C(R) to the triangulated category (Vop , Σ−1
V ), see A.5, is quasi-triangulated in the sense of 6.1.12. Explicitly, this means that there exists a natural isomorphism ψ : Σ−1
V G → GΣ of functors
C(R)op → V, such that
Σ−1
V G(M)
G( 0 1Σ M ) ◦ ψM
/ G(Cone α)
N
G 1
0 /
G(N)
G(α)
/ G(M)
is a distinguished triangle in V for every morphism of R-complexes α : M → N.
A morphism in C(R)op is called a homotopy equivalence if the corresponding
morphism in C(R) is a homotopy equivalence in the sense of 2.2.25. To parse and
prove the next result, recall further that if F : U → V is a functor between categories
with products (coproducts), then Fop : Uop → Vop is a functor between categories
with coproducts (products), and F preserves products (coproducts) if and only if Fop
preserves coproducts (products).
6.1.15 Theorem. If G : C(R)op → V is a functor that maps homotopy equivalences
to isomorphisms, then there exists a unique functor G0 : K(R)op → V that makes the
following diagram commutative,
C(R)op
G
z
V;
Qop
/ K(R)op
G0
here Q is the canonical functor from 6.1.4. Furthermore, the next assertions hold.
(a) Assume that V is k-prelinear; then G is k-linear if and only if G0 is k-linear.
(b) Assume that V has (co)products; then G preserves (co)products if and only if
G0 preserves (co)products.
(c) If V is triangulated and G is quasi-triangulated, then G0 is triangulated.
P ROOF . Apply 6.1.13 to the functor Gop : C(R) → Vop .
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203
U NIQUE L IFTING P ROPERTIES
Throughout the rest of this chapter, we use Greek letters for morphisms in K(R);
that is, α, β, γ, . . . denote homotopy equivalence classes of morphisms in C(R).
6.1.16 Definition. A morphism α in K(R) is called a quasi-isomorphism if some,
equivalently every, morphism in C(R) that represents the homotopy equivalence
class α is a quasi-isomorphism.
Next we reformulate 5.2.16 and 5.3.21 in the language of the homotopy category.
6.1.17. Let P be a semi-projective R-complex. If α : P → N is a morphism and
β : M → N is a quasi-isomorphism in K(R), then there exists a unique morphism γ
that makes the following diagram in K(R) commutative,
P
γ
M
~
α
/N.
'
β
6.1.18. Let I be a semi-injective R-complex. If α : M → I is a morphism and
β : M → N is a quasi-isomorphism in K(R), then there exists a unique morphism γ
that makes the following diagram in K(R) commutative,
M
α
~
I.
β
'
/N
γ
E XERCISES
E 6.1.1
E 6.1.2
E 6.1.3
E 6.1.4
E 6.1.5
Show that a morphism in a triangulated category is a monomorphism (epimorphism)
if and only if it has a left (right) inverse. Conclude that every object in a triangulated
category is both injective and projective.
Show that the homotopy category K(Z) is not Abelian.
Show that the homotopy category K(R) is Abelian if R is semi-simple.
Two homomorphisms of R-modules α, β : M → N are called stably equivalent if α − β
factors through a projective R-module. The stable module category M(R) has as objects
all R-modules. The hom-set M(R)(M, N) in this category, often written as HomR (M, N),
is the set of classes of stably equivalent homomorphisms M → N. (a) Show that M(R)
is a k-linear category with coproducts. (b) For an R-module M, let Ω(M) be the kernel of
any surjective homomorphism α : P → M where P is projective. Show that Ω is a welldefined endofunctor on M(R). (c) Show that the category (M(R), Ω) is triangulated if
R is quasi-Frobenius.
Show that the stable module category M(Z) is not Abelian.
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6 The Derived Category
(Cf. A.5) Let (T, Σ) be a triangulated category. Show that (T op , Σ−1 ) is triangulated in
the canonical way: A candidate triangle M → N → X → Σ−1 M in T op is distinguished
if and only if the candidate triangle Σ−1 M → X → N → M is distinguished in T.
E 6.1.7 Let (T, Σ) be a triangulated category and let S be a subcategory of T that is closed under
isomorphisms. Show that S is a triangulated subcategory if and only if it is a triangulated
category and the embedding functor S → T is full and triangulated.
E 6.1.8 Let S be a triangulated subcategory of (T, Σ) and let M → N → X → Σ M be a distinguished triangle in T. Show that if two of the objects M, N, and X are in S, then the
third object is also in S.
E 6.1.9 Show that the full subcategory of K(R) whose objects are all acyclic R-complexes is
triangulated.
E 6.1.10 Show that the full subcategories of K(R) defined by specifying their objects as follows:
E 6.1.6
K@ (R) = {M ∈ K(R) | there is a bounded above complex M 0 with M ∼
= M 0 in K(R)} ,
0
0
∼
K@
A (R) = {M ∈ K(R) | there is a bounded complex M with M = M in K(R)} , and
KA (R) = {M ∈ K(R) | there is a bounded below complex M 0 with M ∼
= M 0 in K(R)}
are triangulated subcategories of K(R).
E 6.1.11 Show that the full subcategory of K(R) defined by specifying its objects,
K(Prj R) = {P ∈ K(R) | P is a complex of projective modules} ,
is a triangulated category but, in general, not a triangulated subcategory of K(R).
E 6.1.12 Show that the full subcategory of Kprj (R) defined by specifying its objects,
Kprj (R) = {P ∈ K(Prj R) | P is semi-projective} ,
is a triangulated subcategory of K(Prj R).
E 6.1.13 Show that the full subcategory of K(R) defined by specifying its objects,
K(Inj R) = {I ∈ K(R) | I is a complex of injective modules} ,
is a triangulated category but, in general, not a triangulated subcategory of K(R).
E 6.1.14 Show that the full subcategory of K(R) defined by specifying its objects,
Kinj (R) = {I ∈ K(Inj R) | is semi-injective} ,
is a triangulated subcategory of K(Inj R).
6.2 The Derived Category
S YNOPSIS . Derived category; (co)product; triangulation; universal property.
O BJECTS AND M ORPHISMS
Recall that we use Greek letters for morphisms in K(R), that is, α, β, γ, . . . denote
homotopy equivalence classes of morphisms in C(R).
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