Chapter 6
The Derived Category
6.1 The Homotopy Category
S YNOPSIS . Homotopy category; (co)product; triangulation; universal property; unique lifting
properties.
Let U be a category and let ≈ be an equivalence relation on each hom-set in U that
is compatible with composition, i.e. a congruence relation. The quotient category
U/ ≈ has the same objects as U, and for two such objects M, N the hom-set in
U/ ≈ is the set U(M, N)/ ≈ of congruence classes. Evidently, the canonical functor
Q : U → U/ ≈ has the following universal property: If F : U → V is any functor such
that F(α) = F(β) holds for every pair of parallel morphisms in U, then there exists a
unique functor F0 that makes the following diagram commutative,
U
F
}
V.
Q
/ U/ ≈
F0
This section is focused on a special quotient category: the homotopy category K(R).
It is the quotient of C(R) modulo homotopy equivalence. While C(R) is Abelian, the
category K(R) is, in general, not. However, the mapping cone construction in C(R)
facilitates a triangulated structure on K(R).
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).
193
194
6 The Derived Category
6.1.2. By 2.3.12 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 L is also an R-complex, then the composition K(R)(M, N) × K(R)(L, M) → K(R)(L, N) maps ([α], [β]) to [αβ]; it follows
from 2.3.4 that [αβ] does not depend on the choice of representatives for [α] and [β].
6.1.3. The homotopy equivalence class [α] of a morphism in C(R) is an isomorphism in K(R) if and only if α is a homotopy equivalence; see 2.2.25.
6.1.4. The zero complex is evidently a zero object in the homotopy category. It
follows that a zero morphism in K(R) is the homotopy equivalence class [0] of
a zero morphism in C(R). Thus, the class [α] of a morphism in C(R) is the zero
morphism in K(R) if and only if α is null-homotopic; see 2.2.20.
The next result shows that the zero objects in the K(R) are exactly the contractible complexes. Further characterizations of such complexes are given in 4.1.10.
6.1.5 Proposition. An R-complex is isomorphic to 0 in K(R) if and only if it is
contractible.
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).
6.1.6. 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.
6.1.7 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
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.
Recall that a category is said to have (co)products if all set-indexed (co)products
exist in the category.
6.1.8 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 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).
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195
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. The zero complex is a zero object in K(R), and
K(R) has biproducts by 6.1.7. 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
homomorphism τu as a morphism M u\ → 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.
R EMARK . Let U be a category Q
that has products and coproducts.
For a family {M u }u∈U of objects
`
u
u
u
in U, the canonical morphisms u∈U M → M and M → u∈U M u are referred to as projections
and embeddings; MacLane [35] uses the term injections for the latter. Embeddings need not be
monomorphisms and projections need not be epimorphisms. Our use of the term embedding for
inclusions of subobjects is thus a slight abuse of terminology.
6.1.9. Let {[αu ] : M u → N u }u∈U and {[βu ] : N u → M u }u∈U be families of morphisms in K(R). By the universal properties of (co)products there are unique morphisms [α] and [β] that make the following diagrams commutative for every u ∈ U,
Mu
[αu ]
Nu
/ `u∈U M u
[α]
/ `u∈U N u
Q
u∈U
Nu
[β]
/ Qu∈U M u
and
Nu
[βu ]
/ Mu .
The horizontal morphisms in the left-hand diagram are embeddings and the vertical
morphisms in the right-hand diagram are projections. The morphism [α] is called
`
the coproduct in K(R) of {[αu ]}u∈U and denoted u∈U [αu ]. Similarly, [β] is called
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6 The Derived Category
the product in K(R) of {[βu ]}u∈U and denoted u∈U [βu ]. From the construction of
(co)products in K(R) given in 6.1.8, it follows that one has
` u
Q u
` u
Q u
[α ] =
α
and
[β ] =
β ,
`
u∈U
u∈U
u∈U
u∈U
where u∈U αu and u∈U βu are the coproduct and the product of {αu }u∈U and
{βu }u∈U in C(R); see 3.1.5 and 3.1.18.
`
Q
6.1.10 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.
We apply the terminology from Definitions 5.1.6, 5.2.12, and 5.3.12 to quasi'
isomorphisms in the homotopy category. That is, a quasi-isomorphism X −→ M in
K(R), where X is a semi-free/-projective complex is called a semi-free/-projective
'
resolution of M; similarly a quasi-isomorphism X −→ Y , where Y is semi-injective,
is called a semi-injective resolution of M.
The next result is immediate from 4.2.7 and 6.1.9.
6.1.11 Proposition. Let {[αu ] : M u → N u }u∈U be a family of morphisms in K(R).
`
If [αu ] is a quasi-isomorphism for every u ∈ U, then the coproduct u∈U [αu ] and the
Q
product u∈U [αu ] are quasi-isomorphisms.
6.1.12 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 an isomorphism. For a morphism [α]
in K(R) one has ΣK ([α]) = [Σ α].
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 6.1.8, equipped with the k-linear autofunctor Σ = ΣK from 6.1.12. One may now speak of candidate triangles in K(R) in the
sense of A.1.
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197
6.1.13 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
α
M
/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
ϕ=
α=
, ψ = 0 1Σ M
= 0 , and χ = (Σ α) 0 1Σ M = 0 Σ α
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
%= M
and τ = ς1N 0 ,
ς1
where ςsM : M → Σs M is the map introduced in 2.2.3. From the fact that ςsM 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 ΣM 0
0
∂ ας−1
α
M
+
∂
=
0
ς1M
ς1M
0 ∂ΣM
and
∂
ΣN
ς1N
0 +
ς1N
ΣM
∂ N ας−1
0
= 0 Σα .
0 ∂ΣM
6.1.14 Definition. A candidate triangle in K(R) of the form considered in 6.1.13 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.
Triangulated categories are defined in A.3.
6.1.15 Theorem. The homotopy category K(R), equipped with the autofunctor Σ
and the collection of distinguished triangles defined in 6.1.14, 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 4.1.10 and 6.1.5 that application of the
canonical functor Q : C(R) → K(R) to the following diagram in C(R),
M
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1M
/M
M
1
0
/ Cone(1M )
( 0 1Σ M )
/
ΣM ,
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198
6 The Derived Category
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. 6.1.13.
(TR2’): By A.4 it it sufficient to verify that (TR2) holds. Thus, let
γ0
β0
α0
∆ = 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
β0
α0
∆00 = Σ−1 X 0 −−−−→ M 0 −−→ N 0 −−→ X 0
and
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),
N
N
1
0
( 0 1Σ M )
/
/ Cone α
ΣM
−Σα
/ ΣN ,
and
Σ−1 Cone α
( 0 −1M )
/M
α
/N
N
1
0
/ 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 6.1.14. 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 the
vertical morphisms in both diagrams are homotopy equivalences.
N
N
ι= 1
0
/ Cone α
/ ΣM
O
ϑ =( 0 1Σ M 0 )
(?)
N
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( 0 1Σ M )
N
ι= 1
0
/ Cone α
1N
0
0 1Σ M
0 0
!
ϕ=
/ Cone ι
−Σα
0
1Σ M
−Σα
/ ΣN
!
( 0 0 1Σ N )
/ ΣN
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6.1 The Homotopy Category
Σ−1 Cone α
199
π =( 0 −1M )
α
/M
/N
O
ξ =( α 1N 0 )
(‡)
Σ−1 Cone α
π =( 0 −1M )
/M
M
1
0
0
N
1
0
/ 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 ΣM ΣN 
 

∂ ας−1 ς−1
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 = 0 ∂ Σ M 0 = 0 1Σ M
 N ΣM ΣN 
ς−1
∂ αςΣ−1
0  0 ∂ M 0  = ϑ∂ Cone ι .
0
0 ∂ΣN
Next we argue that ϕ is a homotopy equivalence with homotopy inverse ϑ. Evidently one has ϑϕ = 1Σ M , 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. The degree 1 homomorphism σ : Cone ι → Cone ι given by


0 00
σ =  0 0 0
ς1N 0 0
is the desired homotopy. Indeed, one has

 N Σ M Σ N 
 
 N Σ N Σ N   N
∂ ας−1 ς−1
∂ ας−1 ς−1
1 0 0
0 00
0 00
 0 ∂ Σ M 0  0 0 0 +  0 0 0 0 ∂ Σ M 0  =  0 0 0  ;
ς1N 0 0
ς1N 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, given by
 

 N

 N
0
1 0
1 0
β =  0 1Σ M  −  1Σ M  0 1Σ M =  0 0  ,
0 Σα
−Σα
0 0
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6 The Derived Category
is null-homotopic. Let τ : Cone α → Cone ι be the degree one homomorphism


0 0
τ =  0 0 .
ς1N 0
It is straightforward to verify the equality
 N ΣM ΣN 
 
∂ ας−1 ς−1
0 0
0
 0 ∂ Σ M 0   0 0 +  0
ς1N 0
ς1N
0
0 ∂ΣN

 N

0 N ΣM
1 0
∂ ας−1
0
=0 0 ;
0 ∂ΣM
0
0 Σα
that is, ∂ Cone ι τ + τ∂ Cone α = β holds, and hence β is null-homotopic.
Similar arguemnts show that the second diagram (‡) is commutative up to homotopy, in particular, that ψ is a homotopy equivalence with homotopy inverse ξ.
(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
χ
χ = 21
χ
ψ
M0
/ Cone α
1N
0
0
/ Cone α0
χ12
/ ΣM
χ22
( 0 1Σ M 0 )
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 the
0
equality ψα − α0 ϕ = ∂ N σ + σ∂ M holds. Define χ : Cone α → Cone α0 as follows:
ΣM
ψ σς−1
χ=
.
0 Σϕ
It is straightforward verify that it is a morphism; that is, one has
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6.1 The Homotopy Category
0
∂ Cone α χ =
201
N 0 0 Σ M0 N Σ M ΣM
ΣM
∂ α ς−1
∂ ας−1
ψ σς−1
ψ σς−1
=
= 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 the vertical morphisms
are homotopy equivalences.
M0
⊕
N
α0 ψ
θ = 0 −1N
0 0

0
1N ψ σςΣ M
 0 0 Σ−1
ϕ 
0 0 −1Σ M

!
N0
/ ⊕
Cone α
Cone α0
/ ⊕
ΣM
O
0
0 1Σ M Σ ϕ
0 0 −Σα

0
ΣM
1N
0 σς−1


0
0 
 0
ξ =
0 −1Σ M 
 0

0


0 1Σ M Σ ϕ
0
0 −Σα
Σ M0
/ ⊕
ΣN


0
1N

ψ
0 0


η = 0 0 Σ ϕ 1Σ M 0 0 
0 0 −1Σ M 0 0
M0
⊕
N
θ=
α0 ψ
0 −1N
0 0
!
/
N0
⊕
Cone α
ΣM
σς−1

0
1N 0 0
 0 1N 0 


 0 0 1Σ M 
0 0 0
0 0 0

/ 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
 N 0 0 Σ M0

∂ α ς−1 0
 0 ∂ Σ M0 0 
0
0 ∂ΣM
and
 N0
∂
 0

 0

 0
0

Σ M 0 ψς Σ N
0
0 α0 ς−1
−1
ΣM
ΣN
∂ N ας−1
0 −ς−1

Σ
M
0 ∂
0
0 
.
0
Σ
M
0
0
∂
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 ς1N 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.
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6 The Derived Category
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 difference
morphism γ : N 0 ⊕ Cone α → Cone θ, given by

 N0
 
 N0

ΣM
1
0 σς−1
0 ψ 0
1 0 0
 N0

ΣM
 0 1N 0  0 −1N 0 
 0 0
0 


 

 1 ψ σς−1
Σ
M
ΣM






−
γ =  0 0 −1  0 0 Σ ϕ
 0 0 1  = 0 0 0  ,
0
 0 0 0  0 0 0 
 0 1Σ M Σ ϕ  0 0 −1Σ M
0 0 Σα
0 0 0
0 0 −Σα
is null-homotopic. This follows as % : N 0 ⊕ Cone α → Cone θ, given by


0 0 0
0 0 0



%=
0 0 0 ,
0 0 0
0 ς1N 0
0
is a degree 1 homomorphism with ∂ Cone θ % + %∂ N ⊕Cone α = γ.
T HE U NIVERSAL P ROPERTY
6.1.16 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.
6.1.17 Example. The canonical functor Q : C(R) → K(R) is quasi-triangulated.
The quasi-triangulated functor Q : C(R) → K(R) has a universal property described in the next theorem.
6.1.18 Theorem. Let U be a category and let F : C(R) → U be a functor. If F maps
homotopy equivalences to isomorphisms, then there exists a unique functor F0 that
makes the following diagram commutative,
C(R)
F
{
U;
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here Q is the canonical functor from 6.1.6. For an R-complex M there is an equality
F0 (M) = F(M), and for a morphism [α] in K(R) one has F0 ([α]) = F(α). Furthermore, the following assertions hold.
(a) Assume that U is k-prelinear; then F0 is k-linear if and only if F is k-linear.
(b) Assume that U has (co)products; then F0 preserves (co)products if and only if
F 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 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,
M\
⊕
C\ = M \
⊕
Σ M\
and
 M

ΣM
∂
0 −ς−1
ΣM  .
∂ C =  0 ∂ M ς−1
0 0 ∂ΣM
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 .
ς1M 0 0
One readily verifies the equality ∂ C σ + σ∂ C = ιπ − 1C , that is,
 M

 
  M
 M

ΣM
ΣM
∂
0 −ς−1
∂
0 −ς−1
−1 0 0
0 00
0 00
Σ M  0 0 0+ 0 0 0 0 ∂ M ς Σ M  =  1M 0
 0 ∂ M ς−1
0 .
−1
ς1M 0 0
ς1M 0 0
0 0 −1Σ M
0 0 ∂ΣM
0 0 ∂ΣM
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6 The Derived Category
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
Σ M : C → N is a
β − α = ∂ N % + %∂ M . The degree 0 homomorphism γ = α β %ς−1
morphism, as one has

 M
ΣM
∂
0 −ς−1
Σ M = α β %ς Σ M  0 ∂ M ς Σ M  = γ∂ C .
∂ N γ = ∂ N α β %ς−1
−1
−1
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).
It remains to prove the assertions (a), (b), and (c).
(a): If F0 is k-linear then so is F = F0 Q, as a composite 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 6.1.8, 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(
`
M u ) = F(
u∈U
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
F
u∈U (M )
`
u) ,
F(M
u∈U
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 6.1.16. By 6.1.12 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,
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205
that is, to a diagram of the form
Q(α)
M
N
Q 1
0
/N
/ 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)
F0 Q(α)
/
F0 (N)
N
F0 Q 1
0 /
F0 (Cone α)
φ0M ◦ F0 Q( 0 1Σ M )
/ ΣU F0 (M) .
However, this diagram is identical to the one in 6.1.16, which is distinguished triangle in U since F is assumed to be quasi-triangulated.
R EMARK . The crux of 6.1.18 is that the category K(R) is the localization of C(R) with respect
to the collection of homotopy equivalences. In the next section, we treat the further localization of
K(R) with respect to the collection of quasi-isomorphisms; this leads to the derived category.
6.1.19 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 6.1.16. Explicitly, this
op
means that there is a natural isomorphism ψ : Σ−1
V G → GΣ of functors C(R) → V,
such that the candidate triangle
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).
6.1.20 Theorem. Let V be a category and let G : C(R)op → V be a functor. If G maps
homotopy equivalences to isomorphisms, then there exists a unique functor G0 that
makes the following diagram commutative,
C(R)op
G
z
V;
Qop
/ K(R)op
G0
here Q is the canonical functor from 6.1.6. For an R-complex M there is an equality
G0 (M) = G(M), and for a morphism [α] in K(R)op one has G0 ([α]) = G(α). Furthermore, the following assertions hold.
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6 The Derived Category
(a) Assume that V is k-prelinear; then G0 is k-linear if and only if G is k-linear.
(b) Assume that V has (co)products; then G0 preserves (co)products if and only if
G preserves (co)products.
(c) If V is triangulated and G is quasi-triangulated, then G0 is triangulated.
P ROOF . Apply 6.1.18 to the functor Gop : C(R) → Vop .
U NIQUE L IFTING P ROPERTIES
In the balance 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).
Now we rephrase 5.2.16 and 5.3.21 in the language of the homotopy category.
6.1.21. 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
6.1.22 Lemma. Let P be a semi-projective R-complex. For morphisms in K(R),
α
P
β
/
/M
ϕ
'
/N,
where ϕ is a quasi-isomorphism, one has α = β if ϕα = ϕβ holds.
6.1.23. 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
6.1.24 Lemma. Let I be a semi-injective R-complex. For morphisms in K(R),
M
ϕ
'
/N
α
β
/
/I,
where ϕ is a quasi-isomorphism, one has α = β if αϕ = βϕ holds.
We also recast 5.2.17 and 5.3.22 in the language of the homotopy category.
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6.1.25. If β : M → P is a quasi-isomorphism in K(R) and P is semi-projective, then
β has a right inverse in K(R) which is also a quasi-isomorphism.
6.1.26. If β : I → M is a quasi-isomorphism in K(R) and I is semi-injective, then β
has a left inverse in K(R) which is also a quasi-isomorphism.
E XERCISES
E 6.1.1
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.
E 6.1.2 Show that the homotopy category K(Z) is not Abelian.
E 6.1.3 Show that the homotopy category K(R) is Abelian if R is semi-simple.
E 6.1.4 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), 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 projective precover P M. Show that Ω is a well-defined k-linear endofunctor on M(R).
(c) Show that the category M(R) is triangulated if R is quasi-Frobenius.
E 6.1.5 Show that the stable module category M(Q[x]/(x2 )) is not Abelian.
E 6.1.6 Let α : M → N be a morphism in K(R). Show that the complex X in a distinguished
α
triangle M −→ N −→ X −→ Σ M in K(R) is unique up to isomorphism.
E 6.1.7 (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 6.1.8 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 (S, Σ) is a
triangulated category and the embedding functor S → T is full and triangulated.
E 6.1.9 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 6.1.10 Show that the full subcategory of K(R) whose objects are all acyclic R-complexes is
triangulated.
E 6.1.11 Show that the full subcategories of K(R) defined by specifying their objects as follows:
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 6.1.12 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). Show
that the inclusion functor K(Prj R) → K(R) is triangulated.
E 6.1.13 Show that the full subcategory of Kprj (R) defined by specifying its objects,
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6 The Derived Category
Kprj (R) = {P ∈ K(Prj R) | P is semi-projective} ,
is a triangulated subcategory of K(Prj R).
E 6.1.14 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). Show
that the inclusion functor K(Inj R) → K(R) is triangulated.
E 6.1.15 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.
Let U be a category and let S be a collection of morphisms in U. One may seek
a category S−1 U—called the localization of U with respect to S—together with a
functor Q : U → S−1 U that has the following universal property: The functor Q maps
S to isomorphisms in S−1 U, and for any functor F : U → V that maps S to isomorphisms in V there is a unique functor F0 that makes the next diagram commutative,
U
F
|
V.
Q
/ S−1 U
F0
There is a formal way to construct S−1 U. However, this constuction may result
in a “category” in which the hom-sets are not sets but proper classes. Thus, the
localization of U with respect to S might not exist. An early motivation for the
theory of model categories was to avoid such such set theoretic problems.
As asserted in 6.1.18, the homotopy category K(R) is the localization of C(R)
with respect to the collection of homotopy equivalences. Now we proceed with the
further localization of K(R) with respect to the collection of quasi-isomorphisms.
The result is a category D(R)—the existence of semi-projective resolutions can be
harnessed to show that the hom-sets in D(R) are actual sets—called the derived
category over R; it inherits a triangulated structure fom K(R).
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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209
6.2.1. If β : M → V is a morphism and ψ : N → V is a quasi-isomorphism in K(R),
then there exist a morphism α and a quasi-isomorphism ϕ such that the following
diagram in K(R) is commutative,
α
U
/N
ϕ '
(6.2.1.1)
' ψ
M
/V .
β
'
For example, choose by 5.2.13 a semi-projective resolution ϕ : U −→ M and apply
6.1.21 to get the morphism α.
Similarly, from the existence of semi-injective resolutions 5.3.18, it follows that
given a morphism α : U → N and a quasi-isomorphism ϕ : U → M in K(R) there
exist a morphism β and a quasi-isomorphism ψ such that (6.2.1.1) is commutative.
R EMARK . One does not need semi-projective and semi-injective resolutions to prove the assertions in 6.2.1; in fact, they may be proved using only that the homotopy category is triangulated;
see E 6.2.8–E 6.2.10.
6.2.2 Definition. Let M and N be objects in K(R). A left prefraction from M to N
is a pair (α, ϕ) of morphisms in K(R) such that α and ϕ have the same source, the
target of ϕ is M, the target of α is N, and ϕ is a quasi-isomorphism:
ϕ
α
M ←−− U −−→ N .
'
Two left prefractions (α1 , ϕ1 ) and (α2 , ϕ2 ) from M to N are equivalent, in symbols
(α1 , ϕ1 ) ≡ (α2 , ϕ2 ), if there exist a third left prefraction (α, ϕ) from M to N and
morphisms µ1 and µ2 that make the following diagram in K(R) commutative,
UO 1
ϕ1
(6.2.2.1)
~
M `o
µ1
ϕ
α
U
µ2
ϕ2
α1
U2
/N.
>
α2
Note that the morphisms µ1 and µ2 in (6.2.2.1) are quasi-isomorphisms.
6.2.3 Lemma. Let M and N be objects in K(R). The relation ≡ introduced in 6.2.2
is an equivalence relation on the class of left prefractions from M to N.
P ROOF . It is evident that ≡ is reflexive and symmetric. To prove transitivity, assume
that there are relations (α1 , ϕ1 ) ≡ (α2 , ϕ2 ) and (α2 , ϕ2 ) ≡ (α3 , ϕ3 ); that is, there exist
commutative diagrams in K(R),
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6 The Derived Category
UO 1
ϕ1
µ1
~
M`o
ϕ2
ϕ
α
U
µ2
/N
>
µ02

M_o
and
α2
U2
UO 2
ϕ2
α1
ϕ0
ϕ3
U0
µ03
U3
α2
α0
/N,
?
α3
where (α, ϕ) and (α0 , ϕ0 ) are left prefractions. By 6.2.1 there exist quasi-isomorphisms θ and θ0 that make the following diagram in K(R) commutative,
θ0
'
W
/U
' µ2
θ '
U0
/ U2 .
'
µ02
Note that ϕθ0 = ϕ0 θ holds since one has ϕθ0 = ϕ2 µ2 θ0 = ϕ2 µ02 θ = ϕ0 θ; similarly, the
equality αθ0 = α0 θ holds. Thus, there is a commutative diagram,
UO 1
ϕ1
{
M co
α1
µ1 θ 0
'
ϕθ0 =ϕ0 θ
W
αθ0 =α0 θ
µ03 θ
ϕ3
#
/N,
;
α3
U3
which shows that there is a relation (α1 , ϕ1 ) ≡ (α3 , ϕ3 ).
6.2.4 Definition. Let M and N be objects in K(R). For a left prefraction (α, ϕ) from
M to N, denote by α/ϕ the equivalence class containing (α, ϕ). The class α/ϕ is called
a left fraction from M to N, and the collection of all such is denoted D(R)(M, N).
The notation introduced in 6.2.4 is suggestive and, indeed, we shall shortly prove
that there is a category D(R) whose objects are all R-complexes and in which the
hom-set D(R)(M, N) is the collection of all left fractions from M to N.
6.2.5. Consider a diagram in K(R) of the form
Mo
ϕ
'
UO
α
/N
ψ '
V,
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211
where ϕ and ψ are quasi-isomorphisms. The commutative diagram
UO
ϕ
ψ
~
M`o
ϕψ
V
ϕψ
α
αψ
/N
?
αψ
V
shows that the left prefractions (α, ϕ) and (αψ, ϕψ) are equivalent; whence there is
an equality α/ϕ = (αψ)/(ϕψ).
6.2.6 Lemma. Let M and N be objects in K(R) and let α1/ϕ1 and α2/ϕ2 be left
fractions from M to N. There exist morphisms α01 , α02 and a quasi-isomorphism ϕ in
K(R) such that the equalities α1/ϕ1 = α01/ϕ and α2/ϕ2 = α02/ϕ hold.
P ROOF . Since the quasi-isomorphisms ϕ1 : U1 → M and ϕ2 : U2 → M have the same
target, 6.2.1 yields quasi-isomorphisms ψ1 and ψ2 , illustrated below,
Mo
ϕ1
'
UO 1
ψ1 '
α1
Mo
/N
ϕ2
'
and
UO 2
α2
/N
ψ2 '
V,
V
such that ϕ1 ψ1 = ϕ2 ψ2 holds. Note that this composite is a quasi-isomorphism, and
denote it by ϕ. Now 6.2.5 yields αi/ϕi = (αi ψi )/(ϕi ψi ) = (αi ψi )/ϕ for i = 1, 2.
The collection of all left prefractions from M to N is a proper class (as opposed to
a set); however, the collection of equivalence classes of these left prefractions turns
out to be a set.
6.2.7 Lemma. For R-complexes M and N, the collection D(R)(M, N) of left fractions from M to N is a set. Moreover, D(R)(M, N) is k-module with addition and
k-multiplication defined as follows.
• For α1/ϕ1 and α2/ϕ2 in D(R)(M, N) set
α1/ϕ + α2/ϕ = (α01 + α02 )/ϕ
1
2
for any choice of left prefractions (α0i , ϕ) with αi/ϕi = α0i/ϕ for i = 1, 2; cf. 6.2.6.
• For x in k and α/ϕ in D(R)(M, N) set
x (α/ϕ) = (xα)/ϕ .
1M
0
The equivalence class 0/1M containing the left prefraction M ←− M −→ N is the zero
element in the k-module D(R)(M, N).
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6 The Derived Category
'
P ROOF . To show that D(R)(M, N) is a set, let π : P −→ M be a semi-projective resolution of M and consider the map
K(R)(P, N) −→ D(R)(M, N) given by β 7−→ β/π .
(?)
Let (α, ϕ) be a left prefraction from M to N, and denote by U the common source of
α and ϕ. By 6.1.21 there exists a morphism γ : P → U in K(R) with ϕγ = π. Note
that γ is a quasi-isomorphism. It follows from 6.2.1 that α/ϕ = (αγ)/(ϕγ) = (αγ)/π
holds, and hence (?) is surjective. Since K(R)(P, N) is a set, so is D(R)(M, N).
To prove that addition in D(R)(M, N) is well-defined, assume that α01/ϕ = α001/ψ
and α02/ϕ = α002/ψ hold. It must be argued that one has (α01 + α02 )/ϕ = (α001 + α002 )/ψ. By
assumption, there exist commutative diagrams,
UO 0
ϕ
α01
µ1
~
M `o
'
χ1
W1
δ1
/N
>
ν1
ψ
U 00
UO 0
ϕ
and
~
M `o
'
χ2
ψ
α001
α02
µ2
δ2
W2
/N.
>
ν2
U 00
α002
'
Let π : P −→ M be a semi-projective resolution of M. By 6.1.21 there are morphisms
γ1 : P → W1 and γ2 : P → W2 such that χ1 γ1 = π = χ2 γ2 holds. Hence there are commutative diagrams,
UO 0
ϕ
M `o
µ1 γ1
~
π
ψ
P
α01
δ1 γ1
ϕ
/N
>
ν1 γ1
and
µ 2 γ2
~
M `o
π
ψ
α001
U 00
UO 0
P
α02
δ2 γ2
/N.
>
ν2 γ2
U 00
α002
By the uniqueness part in 6.1.21, it follows that µ1 γ1 = µ2 γ2 and ν1 γ1 = ν2 γ2 hold.
Set µ = µi γi and ν = νi γi ; the commutative diagram
ϕ
y
M oe
UO 0
µ
π
P
α01 +α02
δ1 γ1 +δ2 γ2
%/
9N
ν
ψ
U 00
α001 +α002
shows that (α01 + α02 )/ϕ = (α001 + α002 )/ψ holds, as desired.
It is straightforward to see that k-multiplication is well-defined.
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213
Finally, to see that D(R)(M, N) is a k-module with zero element 0/1M notice that
the surjective map (?) preserves addition and k-multiplication.
6.2.8 Lemma. Let L, M, and N be R-complexes. There is a map
D(R)(M, N) × D(R)(L, M) −→ D(R)(L, N) ,
given by
(α/ϕ, β/ψ) 7−→ (α/ϕ)(β/ψ) = (αγ)/(ψχ) ,
where γ/χ is any left fraction that makes the following diagram in K(R) commutative, cf. 6.2.1,
W
γ
χ
V
ψ
L

~
'
β
U
ϕ
'
M
~
α
'
N.
P ROOF . To prove that the map is well-defined, it must be verified that given a commutative diagram
W0
γ0
χ0
VO 0
ψ0
L _o
'
ψ
'
'

ψ00
'
}
β0
ν0
/Mo
= a
α0
'
ϕ
'
'
}
ν00
ϕ00
β00
V 00 a
UO 0
ϕ0
!
β
V
!
µ0
α
U
/N
>
µ00
00
U
=
α00
'
χ00
γ00
W 00
in K(R) one can construct a commutative diagram of the form
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6 The Derived Category
WO 0
α0 γ0
ψ0 χ0
~
L o`
(?)
ξ0
'
ω
'
'
ψ00 χ00
δ
P
/N.
>
ξ00
α00 γ00
W 00
'
Choose a semi-projective resolution π : P −→ V and let β̃ be the morphism, see
6.1.21, that makes the following diagram commutative,
P
β̃
' ϕ
π '
V
/U
β
/M.
Set ω = ψπ and δ = αβ̃. Let ξ0 and ξ00 be the morphisms that make the diagrams
P
ξ0
W0
~
ν0 π
χ0
'
/ V0
P
ξ00
and
W0
}
χ00
'
ν00 π
/ V 00
commutative. The two left-hand triangles in (?) are commutative, as one has
ψ0 χ0 ξ0 = ψ0 ν0 π = ψπ = ω and ψ00 χ00 ξ00 = ψ00 ν00 π = ψπ = ω. Next we argue that the
upper right-hand triangle in (?) is commutative, i.e. that α0 γ0 ξ0 = δ holds. As one
has δ = αβ̃ = α0 µ0 β̃, it suffices to verify the identity γ0 ξ0 = µ0 β̃. By 6.1.22 it is sufficient to show that one has ϕ0 γ0 ξ0 = ϕ0 µ0 β̃, and that is a straightforward computation:
ϕ0 γ0 ξ0 = β0 χ0 ξ0 = β0 ν0 π = βπ = ϕβ̃ = ϕ0 µ0 β̃. A similar argument shows that the lower
right-hand triangle in (?) is commutative.
P RODUCTS , C OPRODUCTS , AND k- LINEARITY
The following definition is justified by the subsequent lemma.
6.2.9 Definition. The derived category D(R) has the same objects as C(R) and
K(R), that is, R-complexes. For R-complexes M and N, the hom-set D(R)(M, N) is
the set of all left fractions from M to N; cf. 6.2.4. Composition in D(R) is given by
the map in 6.2.8. Isomorphisms in D(R) are marked by the symbol ’'’.
6.2.10. Notice the following special cases of composition in D(R).
(a) For composable morphisms α/1, β/ψ in D(R) one has (α/1)(β/ψ) = (αβ)/ψ.
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(b) For composable morphisms 1/ϕ, β/ψ in D(R) one has (β/ψ)(1/ϕ) = β/(ϕψ).
6.2.11 Lemma. The derived category D(R) is a category. The identity morphism in
D(R) for an R-complex M is 1M/1M .
P ROOF . It follows from 6.2.10 that 1M/1M is an identity for M. Let α1/ϕ1 , α2/ϕ2 ,
and α3/ϕ3 be composable morphisms in D(R). Apply 6.2.1 to get morphisms β1/ψ1 ,
β2/ψ2 , and γ/χ in D(R) that make the following diagram in K(R) commutative,
W
χ
V1
ψ1
U1
ϕ1
M1
}
'
}
}
!
'
!
!
β1
ϕ2
α1
M2
γ
'
}
ψ2
U2
}
β2
!
'
ϕ3
α2
!
'
V2
M3
}
'
U3
α3
!
M4 .
One has ((α3/ϕ3 )(α2/ϕ2 ))(α1/ϕ1 ) = (α3 β2 γ)/(ϕ1 ψ1 χ) = (α3/ϕ3 )((α2/ϕ2 )(α1/ϕ1 )) by
6.2.8, and hence composition in D(R) is associative.
6.2.12. There is a canonical functor V : K(R) → D(R); it is the identity on objects
and it maps a morphism α : M → N in K(R) to the left fraction α/1M . Indeed, for
composable morphisms α : M → N and β : L → M in K(R), it follows from 6.2.10
that V(αβ) = (αβ)/1L = (α/1M )(β/1L ) = V(α)V(β) holds. Furtheremore, for every
R-complex M, the morphism V(1M ) = 1M/1M in D(R) is the identity for M.
6.2.13 Lemma. A left fraction α/ϕ in D(R)(M, N) equals 0/1M if and only if there
exists a quasi-isomorphism µ in K(R) such that αµ = 0 in K(R).
P ROOF . “If”: Assume that there exists a quasi-isomorphism µ with αµ = 0. Then
6.2.5 yields α/ϕ = (αµ)/(ϕµ) = 0/(ϕµ) = (0ϕµ)/(1M ϕµ) = 0/1M .
“Only if”: Assume that there is an equality α/ϕ = 0/1M , that is, there exists a
commutative diagram in K(R) of the following form,
UO
ϕ
M`o
~
1M
µ
'
α
/N.
>
X
M
0
It follows that µ is a quasi-isomorphism with αµ = 0.
6.2.14 Theorem. The derived category D(R) and the functor V : K(R) → D(R) are
k-linear. For every family {M u }u∈U of R-complexes the following assertions hold.
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6 The Derived Category
(a) If M with embeddings {ιu : M u → M}u∈U is the coproduct of {M u}u∈U in K(R),
then M with the morphisms { ιu/1Mu }u∈U is the coproduct of {M u }u∈U in D(R).
(b) If M with projections {πu : M u → M}u∈U is the product of {M u }u∈U in K(R),
then M with the morphisms { πu/1Mu }u∈U is the product of {M u }u∈U in D(R).
In particular, the derived category D(R) has coproducts and products, and the canonical functor V preserves coproducts and products.
P ROOF . By 6.2.7 the hom-sets in D(R) are k-modules, and it is straightforward to
verify that composition of morphisms in D(R) is k-bilinear. Thus the category D(R)
is k-prelinear. The functor V is k-linear; indeed, for x in k and a morphism α in K(R)
one has V(xα) = (xα)/1 = x(α/1 ) = xV(α). To show that D(R) is k-linear, it must
be argued that it has biproducts and a zero object. Since K(R) has biproducts, see
6.1.8, it follows from 6.1.7 applied to the canonical functor V : K(R) → D(R) that
D(R) has biproducts as well. The zero complex 0 is a zero object in D(R), that is, 0
is both an initial and a terminal object in D(R). Indeed, the commutative diagrams
M`o
1U
~
1M
UO
UO
ϕ
ϕ
/0
?
U
ϕ

0 o_
and
0
α
/N
?
0
M
in K(R) show that one has D(R)(M, 0) = { 0/1M } and D(R)(0, N) = { 0/10 }.
(a): Let { αu/ϕu : M u → N}u∈U be morphisms in D(R). It must be shown that there
is a unique morphism α/ϕ : M → N in D(R) with (α/ϕ)(ιu/1Mu ) = αu/ϕu for all u ∈ U.
For existence, let V u be the common source of αu and ϕu , and denote by V the
coproduct of {V u }u∈U in K(R) with embeddings εu : V u → V . Let α : V → N be the
unique morphism in K(R) with αεu = αu for all u ∈ U, and let ϕ : V → M be the
coproduct in K(R) of the family of quasi-isomorphisms {ϕu : V u → M u }u∈U . Then
ϕ is a quasi-isomorphism by 6.1.11, and there is a commutative diagram
Vu
ϕu
1M
Mu
}
u
Mu
}
εu
'
V
ϕ
ιu
!
M
~
'
α
N,
from which it follows that (α/ϕ)(ιu/1Mu ) = αu/ϕu holds.
For uniqueness, assume that (α/ϕ)(ιu/1Mu ) is zero in D(R) for all u ∈ U; it must
be shown that α/ϕ is zero in D(R). By definition, each composite (α/ϕ)(ιu/1Mu ) is
equal to (αβu )/ψu for any choice of morphism βu and quasi-isomorphism ψu that
make the following diagram in K(R) commutative,
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217
Wu
ψu
u
1M
Mu
Mu
|
|
βu
!
'
ϕ
ιu
"
M
}
V
'
α
N.
Let W denote the coproduct of {W u }u∈U in K(R) with embeddings ωu : W u → W .
Let β : W → V be the unique morphism in K(R) with βωu = βu , and let ψ : W → M
'
be the coproduct in K(R) of the family {ψu : W u −→ M u }u∈U . Then ψ is a quasiisomorphism by 6.1.11, and it is the unique morphism W → M with ψωu = ιu ψu for
all u ∈ U; see 6.1.9. Since one also has (ϕβ)ωu = ϕβu = ιu ψu , we conclude ϕβ = ψ.
Since ϕ and ψ are quasi-isomorphisms, so is β. Hence 6.2.5 yields α/ϕ = (αβ)/(ϕβ). In
view of this, 6.2.13 implies that α/ϕ is zero in D(R) if and only if αβµ is zero in K(R)
for some quasi-isomorphism µ. By assumption, (αβu )/ψu is zero, and thus for every
u ∈ U there is a quasi-isomorphism µu : X u → W u with αβu µu = 0. The coproduct
µ : X → W in K(R) of the quasi-isomorphisms {µu }u∈U is a quasi-isomorphism;
since both αβµ and 0 make the diagrams
Xu
/X
αβu µu =0
~
N
commutative, the universal property of coproducts implies that αβµ = 0 holds.
(b): Let { αu/ϕu : N → M u }u∈U be morphisms in D(R). It must be shown that there
is a unique morphism α/ϕ : N → M in D(R) with (πu/1M )(α/ϕ) = αu/ϕu for all u ∈ U.
'
For every u ∈ U let X u be the common source of αu and ϕu . Let λu : M u −→ I u be
u
u
a semi-injective resolution, and let β : N → I be the unique morphism, see 6.1.23,
that makes the following diagram in K(R) commutative,
Xu
αu
ϕu '
N
/ Mu
' λu
βu
/ Iu .
Let I be the product of {I u }u∈U in K(R) with projections ρu : I → I u . Denote by
λ : M → I the product in K(R) of the quasi-isomorphisms {λu : M u → I u }u∈U . Then
λ is a quasi-isomorphism by 6.1.11, and it is the unique morphism in K(R) with
ρu λ = λu πu for all u ∈ U; cf. 6.1.9. Furtermore, let β : N → I be the unique morphism
in K(R) that satisfies ρu β = βu for all u ∈ U. Choose a semi-projective resolution
'
ϕ : P −→ N and let α : P → M be the unique morphism that makes the following
diagram in K(R) commutative, see 6.1.21,
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6 The Derived Category
α
P
ϕ '
/M
' λ
N
/I.
β
We claim that (πu/1M )(α/ϕ) = αu/ϕu holds for all u ∈ U. Let τu : P → X u be the
unique morphism in K(R) that makes the next diagram commutative, see 6.1.21,
P
τu
~
Xu
ϕ
'
ϕu
/N.
Note that τu is a quasi-isomorphism, as ϕ and ϕu are quasi-isomorphisms. The morphism τu satisfies αu τu = πu α; indeed, by 6.1.22 the equality αu τu = πu α holds if
one has λu αu τu = λu πu α; and that is a simple computation: λu αu τu = βu ϕu τu =
βu ϕ = ρu βϕ = ρu λα = λu πu α. Thus, it follows from 6.2.10 and 6.2.5 that one has
u
u
u u
(π /1M ) (α/ϕ) = (π α)/ϕ = (α τ )/(ϕu τu ) = αu/ϕu .
This proves the existence of the desired morphism α/ϕ.
For uniqueness, assume that (πu/1M )(α/ϕ) = (πu α)/ϕ is zero in D(R) for all u ∈ U;
it must be shown that α/ϕ is zero in D(R). Denote by X the common source of α
and ϕ. As (πu α)/ϕ is zero there is by 6.2.13 a quasi-isomorphism µu : Y u → X with
'
πu αµu = 0. Let ρ : Q −→ X be a semi-projective resolution, and let νu : Q → Y u be
the unique morphism in K(R) that makes the following diagram commutative,
Q
νu
Yu

ρ
'
µu
/X.
Note that πu αρ = πu αµu νu = 0νu = 0 holds for all u ∈ U, and hence the universal property of products in K(R) implies that one has αρ = 0. Since ρ is a quasiisomorphism, another application of 6.2.13 gives that α/ϕ is zero in D(R).
By construction, the canonical functor V preserves (co)products.
R EMARK . Let M and N be R-complexes. A right prefraction from M to N is a diagram in K(R),
(∗)
β
ψ
M −−→ V ←−− N ,
'
where ψ is a quasi-isomorphism. Dually to 6.2.2 one can define an equivalence relation on the
collection of right prefractions from M to N; the equivalence class containing the right prefraction
(∗) is denoted ψ\β and called a right fraction. Like D(R)(M, N), the collection D0 (R)(M, N) of
all right fractions from M to N is a set. The collection of all such sets provide the hom-sets for
a category D0 (R) whose objects are all R-complexes. There is a functor D0 (R) → D(R); it is
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219
the identity on objects and it maps a right fraction ψ\β to the left fraction α/ϕ for any choice of
morphism α and quasi-isomorphism ϕ such that the diagram (6.2.1.1) in K(R) is commutative.
The functor D0 (R) → D(R) is an equivalence and, consequently, the derived category may just as
well be constructed using right fractions. We shall soon prove that D(R) is a triangulated category;
similarly so is D0 (R). The equivalence D0 (R) → D(R) is actually a triangulated functor, and hence
D0 (R) and D(R) are equivalent even as triangulated categories.
We shall not pursue the right fraction point of view beyond this remark, even though it does
have certain advantages. For example, as the proof of 6.2.14 reveals, the argument for existence
of coproducts in D(R) is more straightforward than the one proving existence of products. This is
because left fractions mesh better with coproducts than with products. Dually, it is straightforward
to show existence of products in D0 (R), but slightly more involved to establish the existence of
coproducts. Had we proved the equivalence between D0 (R) and D(R), existence of products in
D(R) would follow immediately from the existence of products in D0 (R).
6.2.15 Lemma. Let α, β, and γ be morphisms in K(R). If αβ and βγ are quasiisomorphisms then α, β, and γ are quasi-isomorphisms.
P ROOF . By 6.1.10, we may assume that α, β, and γ are morphisms in C(R). Since
H(α) H(β) = H(αβ) is an isomorphism, H(β) has a left-inverse; and as H(β) H(γ) =
H(βγ) is an isomorphism, H(β) has a right-inverse. Consequently, H(β) is an isomorphism. It follows that H(α) and H(γ) are isomorphisms as well.
6.2.16. It is straightforward to verify that if (α, ϕ) and (α0 , ϕ0 ) are left prefractions
such that α/ϕ = α0/ϕ0 holds in D(R), then α is a quasi-isomorphism if and only if α0
is a quasi-isomorphism; cf. (6.2.2.1).
The next result describes the isomorphisms in the derived category. It also explains why isomorphisms in D(R) are marked by the same symbol ‘'’ as quasiisomorphisms in C(R) and K(R) and not by the usual ‘∼
=’, which is used for isomorphisms in abstract categories—triangulated categories included; see Appn. A.
6.2.17 Proposition. A morphism α/ϕ in D(R) is an isomorphism if and only if α is
a quasi-isomorphism, in which case one has (α/ϕ)−1 = ϕ/α.
P ROOF . Let α/ϕ be a morphism from M to N.
If α is a quasi-isomorphism, then ϕ/α is a morphism in D(R) from N to M. One
has (ϕ/α)(α/ϕ) = ϕ/ϕ = (1M ϕ)/(1M ϕ) = 1M/1M , where the last equality follows from
6.2.5. Similarly, one has (α/ϕ)(ϕ/α) = α/α = (1N α)/(1N α) = 1N/1N , whence α/ϕ is an
isomorphism with inverse ϕ/α.
Conversely, assume that α/ϕ is an isomorphism. Denote by U the common source
of α and ϕ. By the arguments above, 1U/ϕ is an isomorphism, and since one has
α/ϕ = (α/1U )(1U/ϕ) by 6.2.10, it follows that α/1U is an isomorphism; denote by β/ψ
its inverse. From 6.2.16 and the equalities 1N/1N = (α/1U )(β/ψ) = (αβ)/ψ it follows
that αβ is a quasi-isomorphism. Furthermore, one has 1U/1U = (β/ψ)(α/1U ) = (βγ)/χ
for some morphism γ and quasi-isomorphism χ. Another application of 6.2.16 gives
that βγ is a quasi-isomorphism, and hence α is a quasi-isomorphism by 6.2.15.
Quasi-isomorphisms of complexes yield isomorphisms in the derived category.
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6 The Derived Category
α
6.2.18 Corollary. For every quasi-isomorphism M −→ N of R-complexes the fraction V(α) = α/1M is an isomorphism in D(R) with inverse 1M/α.
For complexes with certain lifting properties there is a conceptual converse to the
corollary. That is, isomorphisms in D(R) yield quasi-isomorphisms of complexes.
6.2.19 Proposition. Let P and M be R-complexes. If P is semi-projective and M
'
and P are isomorphic in D(R), then there is a quasi-isomorphism P −→ M in C(R).
P ROOF . If P and M are isomorphic in D(R), then by 6.2.17 there are quasi-isomor'
'
'
phisms P ←− U −→ M. By 6.1.25 there exists a quasi-isomorphism P −→ U which
'
composes with U −→ M to give the desired quasi-isomorphism.
6.2.20 Proposition. Let I and M be R-complexes. If I is semi-injective and M and
'
I are isomorphic in D(R), then there is a quasi-isomorphism M −→ I in C(R).
P ROOF . If M and I are isomorphic in D(R), then there exist quasi-isomorphisms
'
'
'
'
M ←− U −→ I by 6.2.17. Now 6.2.1 yields quasi-isomorphisms M −→ V ←− I. It
'
follows from 6.1.26 that there is a quasi-isomorphism V −→ I which composes with
'
M −→ V to give the desired quasi-isomorphism.
'
R EMARK . Existence of an isomorphism M −→ N in D(R) does not imply that there is even a
morphism M → N in C(R); see E 6.2.2.
It follows from 6.2.17 that complexes that are isomorphic in the derived category
have isomorphic homology. As the next example shows, the converse is not true.
6.2.21 Example. Over the ring Z/4Z consider the complexes
P=0
/ Z/4Z
2
/ Z/4Z
/ 0 and
M =0
/ Z/2Z
0
/ Z/2Z
/0
concentrated in degrees 0 and 1. Evidently, one has H(P) ∼
=M∼
= H(M). The complex P is semi-projective by 5.2.7, so if P and M were isomorphic in D(Z/4Z), then
by 6.2.19 there would exist a quasi-isomorphism P → M. However, it is straightforward to verify that every morphism α : P → M in C(Z/4Z) has H1 (α) = 0.
The complexes that are isomorphic to 0 in the homotopy category are precisely
the contractible complexes. The next result explains what is means for a complex to
be isomorphic to 0 in the derived category.
6.2.22 Proposition. An R-complex is isomorphic to 0 in D(R) if and only if it is
acyclic.
P ROOF . If M is acyclic, then the morphism 0 → M in C(R), and hence also 0 → M
=
'
in K(R), is a quasi-isomorphism. By 6.2.17 the left prefraction 0 ←− 0 −→ M in
K(R) gives an isomorphism from 0 to M in D(R).
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221
If M is isomorphic to 0 in D(R), then D(R)(M, M) consists of a single element. In
particular, 1M/1M = 0/1M holds, and 6.2.16 implies that the zero morphism M → M
in K(R), and hence in C(R), is a quasi-isomorphism. Thus M is acyclic.
6.2.23 Lemma. Assume that R is a left principal ideal domain. For every R-complex
'
L of free modules there is a quasi-isomorphism L −→ H(L).
P ROOF . For every v ∈ Z the exact sequence 0 → Zv (L) → Lv → Bv−1 (L) → 0 splits.
Indeed, Bv−1 (L) is a submodule of the free module Lv−1 , so by 1.3.10 it is itself free
and, in particular, projective. Thus, by 1.3.17 there are isomorphisms
Lv −→ Zv (L) ⊕ Bv−1 (L) .
(?)
For every v let K v be the complex 0 → Bv (L) → Zv (L) → 0 concentrated in degrees v + 1 and v; notice that one has H(K v ) = Hv (L). The isomorphisms (?) yield
`
an isomorphism of complexes L → v∈Z K v = K, and there is an obvious quasi'
isomorphism K −→ H(K) ∼
= H(L).
6.2.24 Proposition. Assume that R is a left principal ideal domain. For every Rcomplex M there is an isomorphism M ' H(M) in D(R).
'
P ROOF . Pick by 5.1.7 a semi-free resolution π : L −→ M. It follows from 6.2.23 that
there there is a quasi-isomorphism α : L → H(L); now H(π)α/π is an isomorphism in
D(R) from M to H(M).
6.2.25 Proposition. There is a unique endofunctor on D(R) that makes the following diagram commutative,
V /
D(R)
K(R)
Σ
K(R)
V
/ D(R) .
This functor is denoted ΣD ; it is k-linear and an isomorphism. For a morphism α/ϕ
in D(R) one has ΣD (α/ϕ) = (Σ α)/(Σ ϕ).
P ROOF . It is elementary to verify that the endofunctor on D(R) that maps an object
M to Σ M and a morphism α/ϕ in D(R) to (Σ α)/(Σ ϕ) has the asserted properties.
When there is no risk of ambiguity, we write Σ for the functor ΣD .
T RIANGULATION
Consider the k-linear category D(R), see 6.2.14, equipped with the k-linear autofunctor Σ = ΣD from 6.2.25. One may now speak of candidate triangles in D(R) in
the sense of A.1.
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6 The Derived Category
6.2.26 Definition. A candidate triangle in D(R) is called a distinguished triangle if
it is isomorphic to the image of a distinguished triangle in K(R) under the canonical
functor V : K(R) → D(R).
The canoncal functor V : K(R) → D(R) maps a morphism α : M → N to α/1M . It
is often convenient to omit the superscript M and use the abridged notation α/1 .
'
'
6.2.27 Lemma. Let π : P −→ M and ρ : F −→ N be semi-projective resolutions. For
every morphism α/ϕ : M → N in D(R) there exists a unique morphism α̃ : P → F in
K(R) such that the following diagram in D(R) is commutative,
π/1
P
α̃/1
/M
α/ϕ
F
/N.
ρ/1
P ROOF . For uniqueness, it must be shown that the only morphism β : P → F in
K(R) with (ρ/1 )(β/1 ) = 0/1 is β = 0. The composite (ρ/1 )(β/1 ) is equal to (ρβ)/1 ,
and thus 6.2.13 yields a quasi-isomorphism µ with ρβµ = 0. Since µ is a quasiisomorphism with the semi-projective complex P as target, µ has a right inverse by
by 6.1.25. Thus the equality ρβµ = 0 implies ρβ = 0. As β has the semi-projective
complex P as source, it follows from ρβ = 0 and 6.1.22 that one has β = 0.
For existence, denote by U the domain of α and ϕ, and apply 6.1.21 to get morphisms γ and α̃ that make the following diagrams in K(R) commutative,
P
γ
U
~
P
π
and
/M
'
ϕ
α̃
F

αγ
/N.
'
ρ
Now one has (α/ϕ)(π/1 ) = (αγ)/1 , since there is a commutative diagram,
P
P
P

1
'

γ
1
'
π
U
ϕ
M
~
α
'
N.
Thus there are equalities (ρ/1 )(α̃/1 ) = (ρα̃)/1 = (αγ)/1 = (α/ϕ)(π/1 ).
α0
β0
γ0
6.2.28 Lemma. Consider a distinguished triangle M 0 −→ N 0 −→ X −→ Σ M 0 in K(R),
and a commutative diagram in D(R),
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M0
α0/1
'
(6.2.28.1)
M
β0/1
/ N0
γ0/1
/ X0
/ ΣD M 0
'
/N
α/ϕ
/X
β/ψ
'
/ ΣD M ,
γ/χ
where the bottom row is a distinguished triangle, M 0 → M and N 0 → N are isomorphisms, and the isomorphism ΣD M 0 → ΣD M is induced by M 0 → M. There exists
an isomorphism X 0 → X in D(R) that makes (6.2.28.1) commutative.
P ROOF . By the definition of distinguished triangles in D(R), see 6.2.26, there exist
a distinguished triangle in K(R), say,
β00
α00
γ00
M 00 −−→ N 00 −−→ X 00 −−→ Σ M 00 ,
and an isomorphism of distinguished triangles in D(R),
α/ϕ
M
(?)
'
M 00
β/ψ
/N
α00/1
/ ΣD M
'
'
/ N 00
γ/χ
/X
β00/1
/ X 00
γ00/1
'
/ ΣD M 00 .
Denote the composite isomorphisms M 0 → M → M 00 and N 0 → N → N 00 by µ/ζ and
ν/η, respectively. Then there is a commutative diagram in D(R),
M0
(‡)
α0/1
' µ/ζ
M 00
/ N0
β0/1
' ν/η
α00/1
/ N 00
β00/1
/ X0
/ X 00
γ0/1
/ ΣD M 0
' ΣD (µ/ζ )
γ00/1
/ ΣD M 00 .
If one can construct an isomorphism X 0 → X 00 in D(R) that makes (‡) commutative, then the composite of this isomorphism with the inverse of the isomorphism
X → X 00 from (?) evidently gives an isomorphism X 0 → X that makes (6.2.28.1)
commutative, as desired. To find X 0 → X 00 , apply [40, Lemma 2.1.38].
6.2.29 Theorem. The derived category D(R), equipped with the autofunctor Σ and
the collection of distinguished triangles defined in 6.2.26, is triangulated. Furthermore, the canonical functor V : K(R) → D(R) is triangulated.
P ROOF . By 6.2.25 one has VΣ = ΣDV. Thus, once it has been established that the
category D(R) is triangulated, the functor V is evidently triangulated by 6.2.26.
(TR0): Follows immediately from the definition of distinguished triangles in
D(R) and the fact that (K(R), Σ) satisfies (TR0).
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6 The Derived Category
(TR1): Let α/ϕ : M → N be a morphism in D(R) and denote by U the common
source of α and ϕ. As (K(R), Σ) satisfies (TR1), the morphism α fits in a distinguished triangle in K(R), say,
β
α
γ
U −−→ N −−→ X −−→ ΣU .
Applying V to this distinguished triangle in K(R), one gets by 6.2.26 a distinguished
triangle in D(R), namely the top row in the commutative diagram
V(α)
U
V(β)
/N
V(γ)
/X
/ ΣDU
V(ϕ)
M
/N
α/ϕ
/X
V(β)
ΣDV(ϕ) ◦ V(γ)
ΣDV(ϕ)
/ ΣD M .
Since V(ϕ) = ϕ/1U is an isomorphism by 6.2.17, the bottom candidate triangle is
also distinguished.
(TR2’): Follows immediately since (K(R), Σ) satisfies (TR2’).
(TR4’): Consider a commutative diagram in D(R),
M1
α1/ϕ1
/ N1
µ/
ζ
(?)
M2
β1/ψ1
/ X1
γ1/χ1
/ ΣD M 1
ν/
η
α2/ϕ2
/ N2
ΣD (µ/ζ )
β2/ψ2
/ X2
γ2/χ2
/ ΣD M 2 ,
where the rows are distinguished triangles. A morphism λ/θ : X 1 → X 2 must be constructed that makes (?) commutative, and such that the mapping cone candidate
triangle of (µ/ζ , ν/η, λ/θ) in D(R) is distinguished. Choose in K(R) semi-projective
'
'
resolutions πi : Pi −→ M i and ρi : F i −→ N i for i = 1, 2, and let
α̃i : Pi −→ F i ,
µ̃ : P1 −→ P2 ,
and ν̃ : F 1 −→ F 2
be the unique morphisms in K(R) that make the following diagrams commutative,
Pi
πi/1
α̃i/1
Fi
/ Mi
αi/ϕi
ρi/
1
/ Ni
P1
,
π1/1
µ/ζ
µ̃/1
P2
/ M1
π2/
1
/ M2
F1
,
and
ρ1/1
ν/η
ν̃/1
F2
/ N1
ρ2/
1
;
/ N2
see 6.2.27. Since the homotopy category is triangulated, the morphisms α̃1 and α̃2
fit in distinguished triangles in K(R), which are the rows in the following diagram,
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225
P1
(‡)
α̃1
/ F1
α̃2
/ F2
µ̃
β̃1
ν̃
P2
γ̃1
/ Y1
β̃2
/ Σ P1
Σ µ̃
λ̃
γ̃2
/ Y2
/ Σ P2 .
The left-hand square in (‡) is commutative. Indeed, as (α2/ϕ2 )(µ/ζ ) = (ν/η)(α1/ϕ1 )
holds, both morphisms α̃2 µ̃ and ν̃α̃1 make the diagram
/ F2
P1
π1/1
ρ2/1
M1
(α2/ϕ2 )(µ/ζ ) = (ν/η)(α1/ϕ1 )
/ N2
in D(R) commutative, and it follows from 6.2.27 that one has α̃2 µ̃ = ν̃α̃1 . As K(R)
is triangulated, there exists a morphism λ̃ : Y 1 → Y 2 in K(R) that makes (‡) commutative, and such that the mapping cone candidate triangle ∆˜ in K(R) of the morphism
(µ̃, ν̃, λ̃) is distinguished.
In the diagram () below, the “front” is the image of (‡) under the canonical
functor V : K(R) → D(R), and the “back” is the diagram (?),
π1/
1
α1/ϕ1
1
M
?
ρ1/
1
'
α̃1/
1
P1
µ̃/1
P2
2
M
?
'
π2/1
α̃2/1
γ1/χ1
/ X1
?
'
ΣD
'
β̃1/
/ F1
γ̃1/
/ Y1
1
µ/ζ
()
β1/ψ1
/ N1
?
1
(π1/
1)
/ N2
?
ν̃/1
/ F2
'
β2/ψ2
/ X2
?
λ̃/1
ρ2/1
'
/ Y2
β̃2/1
γ̃2/1
'
/ ΣD P1
ν/η
α2/ϕ2
/ ΣD M 1
?
γ2/χ2
ΣD (µ̃/1 )
'
ΣD (µ/ζ )
/ ΣD M 2 .
?
ΣD (π2/1 )
/ ΣD P2
The mapping cone candidate triangle ∆0 in D(R) of the morphism (µ̃/1 , ν̃/1 , λ̃/1 ) is
isomorphic to the image of the mapping cone distinguished triangle ∆˜ of (µ̃, ν̃, λ̃)
under the canonical functor V : K(R) → D(R). Hence, ∆0 is distinguished.
The morphisms πi/1 : Pi → M i and ρi/1 : F i → N i for i = 1, 2 are isomorphisms in
D(R) by 6.2.17, and thus 6.2.28 yields (dotted) isomorphisms Y1 → X1 and Y2 → X2
that make the “top” and “bottom” in () commutative. Now, let λ/θ be the unique
morphism in D(R) that makes the “interior wall”
Y1
λ̃/1
Y2
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'
/ X2
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6 The Derived Category
in () commutative. A straightforward diagram chase reveals that this morphism λ/θ
makes the “back” in () commutative. It follows from A.8 that the mapping cone
candidate triangle ∆ of (µ/ζ , ν/η, λ/θ) is isomorphic to the mapping cone candidate
triangle ∆0 of (µ̃/1 , ν̃/1 , λ̃/1 ), and since ∆0 is distinguished, so is ∆.
R EMARK . Let Kprj (R) denote the full subcategory of K(R) whose objects are the semi-projective
R-complexes. The proof of 6.1.15 shows that (Kprj (R), Σ) is a triangulated category, albeit not a
triangulated subcategory of K(R); see E 6.1.13. The composite functor Kprj (R) → K(R) → D(R)
is a triangulated equivalence; see E 6.2.4.
T HE U NIVERSAL P ROPERTY
The triangulated functor V : K(R) → D(R) has a universal property described in the
next theorem.
6.2.30 Theorem. Let U be a category and let F : K(R) → U be a functor. If F
maps quasi-isomorphisms to isomorphisms, then there exists a unique functor F0
that makes the following diagram commutative,
K(R)
F
{
U;
V
/ D(R)
F0
here V is the canonical functor from 6.2.12. For an R-complex M there is an equality
F0 (M) = F(M), and for a morphism α/ϕ in D(R) one has F0 (α/ϕ) = F(α)F(ϕ)−1 .
Furthermore, the following assertions hold.
(a) Assume that U is k-prelinear; then F0 is k-linear if and only if F is k-linear.
(b) Assume that U has (co)products; then F0 preserves (co)products if and only if
F preserves (co)products.
(c) Assume that U is triangulated; then F0 is triangulated if and only if F is triangulated.
P ROOF . For uniqueness of the functor F0 , assume that F0 V = F holds. Since V is the
identity on objects, one has F0 (M) = F(M) for every R-complex M. A morphism α/ϕ
in D(R) can be written α/ϕ = (α/1 )(1/ϕ) = (α/1 )(ϕ/1 )−1 = V(α)V(ϕ)−1 by 6.2.10
and 6.2.17, and thus there are equalities
F0 (α/ϕ) = F0 (V(α)V(ϕ)−1 ) = (F0 V(α))(F0 V(ϕ))−1 = F(α)F(ϕ)−1 .
Consequently, the functor F0 is uniquely determined by F.
For existence, notice that if α1/ϕ1 = α2/ϕ2 holds in D(R), then there is an equality
F(α1 )F(ϕ1 )−1 = F(α2 )F(ϕ2 )−1 in U. Indeed, if one has α1/ϕ1 = α2/ϕ2 , then by 6.2.2
there exist a morphism α and quasi-isomorphisms ϕ, µ1 , and µ2 in K(R) such that
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227
α1 µ1 = α = α2 µ2 and ϕ1 µ1 = ϕ = ϕ2 µ2 hold. It follows that F(α1 )F(µ1 ) = F(α) and
F(ϕ1 )F(µ1 ) = F(ϕ) hold and, consequently, there are equalities
F(α1 )F(ϕ1 )−1 = F(α)F(µ1 )−1 F(µ1 )F(ϕ)−1 = F(α)F(ϕ)−1 .
Similarly one finds F(α2 )F(ϕ2 )−1 = F(α)F(ϕ)−1 . Thus, one can set F0 (M) = F(M)
for R-complexes M and F0 (α/ϕ) = F(α)F(ϕ)−1 for morphisms α/ϕ in D(R). With
this definition, one evidently has F0 V = F.
In order for F0 to be a functor, it must preserve identity morphisms and respect
0
composition. By definition, F0 (1M/1M ) = F(1M )F(1M )−1 = 1F(M) = 1F (M) holds for
every R-complex M. Let α/ϕ and β/ψ be composable morphisms in D(R). By 6.2.8
the composition (α/ϕ)(β/ψ) is (αγ)/(ψχ) for any choice of morphism γ and quasiisomorphism χ in K(R) with βχ = ϕγ. Thus there are equalities,
F0 ((α/ϕ)(β/ψ)) = F0 ((αγ)/(ψχ))
= F(αγ)F(ψχ)−1
= F(α)F(γ)F(χ)−1 F(ψ)−1
= F(α)F(ϕ)−1 F(β)F(ψ)−1
= F0 (α/ϕ)F0 (β/ψ) .
(a): If F0 is k-linear, then so is F = F0 V as a composite of two k-linear functors.
Conversely, assume that F is k-linear, and let α1/ϕ1 and α2/ϕ2 be parallel morphisms
and let x be an element in k. Write α1/ϕ1 = α01/ϕ and α2/ϕ2 = α02/ϕ for morphisms
α01 , α02 and a quasi-isomorphism ϕ, see 6.2.6. Now 6.2.7 yields
F0 (x(α1/ϕ1 ) + α2/ϕ2 ) = F0 ((xα01 + α02 )/ϕ)
= F(xα01 + α02 )F(ϕ)−1
= (xF(α01 ) + F(α02 ))F(ϕ)−1
= xF(α01 )F(ϕ)−1 + F(α02 )F(ϕ)−1
= xF0 (α01/ϕ) + F0 (α02/ϕ)
= xF0 (α1/ϕ1 ) + F0 (α2/ϕ2 ) ,
and hence F0 is k-linear.
(b): The proof of 6.1.18(b) applies to prove part (b) in this theorem; only one has
to replace the functor Q by V and the reference to 6.1.8 by one to 6.2.14.
(c): If F0 is triangulated, then so is F = F0 V as a composite of two triangulated
functors; cf. 6.2.29. Conversely, assume that F is triangulated. By definition, there
exists a natural isomorphism φ : FΣ → ΣU F such that
(?)
F(α)
F(β)
φM ◦ F(γ)
F(M) −−→ F(N) −−→ F(X) −−−−−→ ΣU F(M)
is a distinguished triangle in U for every distinguished triangle
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6 The Derived Category
β
α
γ
M −−→ N −−→ X −−→ Σ M
(‡)
in K(R). As one has FΣ = F0 VΣ = F0 ΣDV and ΣU F = ΣU F0 V, and since V is the
identity on objects, φ can be viewed as a natural isomorphism φ0 : F0 ΣD → ΣU F0 .
We verify that the functor F0 with the natural isomorphism φ0 , is triangulated. By
definition, a distinguished triangle in D(R) has, up to isomorphism, the form
V(α)
V(β)
V(γ)
M −−−→ N −−→ X −−→ ΣD M
for some distinguished triangle (‡) in K(R). Hence, it must be argued that
()
F0 V(α)
F0 V(β)
φ0M ◦ F0 V(γ)
F0 (M) −−−−→ F0 (N) −−−−→ F0 (X) −−−−−−−→ ΣU F0 (M)
is a distinguished triangle; however, () is exactly the distinguished triangle (?).
A morphism in K(R)op is called a quasi-isomorphism if the corresponding morphism in K(R) is a quasi-isomorphism in the sense of 6.1.10.
6.2.31 Theorem. Let V be a category and let G : K(R)op → V be a functor. If G
maps quasi-isomorphisms to isomorphisms, then there exists a unique functor G0
that makes the following diagram commutative,
K(R)op
G
y
V;
Vop
/ D(R)op
G0
here V is the canonical functor from 6.2.12. For an R-complex M there is an equality
G0 (M) = G(M), and for a morphism α/ϕ in D(R)op one has G0 (α/ϕ) = G(ϕ)−1 G(α).
Furthermore, the following assertions hold.
(a) Assume that V is k-prelinear; then G0 is k-linear if and only if G is k-linear.
(b) Assume that V has (co)products; then G0 preserves (co)products if and only if
G preserves (co)products.
(c) Assume that V is triangulated; then G0 is triangulated if and only if G is
triangulated.
P ROOF . Apply 6.2.30 to the functor Gop : K(R) → Vop .
E XERCISES
E 6.2.1 Let α be a morphism in K(R). Show that there exists a quasi-isomorphism µ with αµ = 0
if and only if there exists a quasi-isomorphism ν with να = 0.
E 6.2.2 Show that the complexes in 4.2.4 are isomorphic in D(R). Hint: E 5.1.8.
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'
E 6.2.3 Let M and N be R-complexes and let P −→ M be a semi-projective resolution. Show that
there is an isomorphisms of k-modules, K(R)(P, N) ∼
= D(R)(M, N).
E 6.2.4 Show that Kprj (R) and D(R) are equivalent as triangulated categories; cf. E 6.1.13.
'
E 6.2.5 Let M and N be R-complexes and let N −→ I be a semi-injective resolution. Show that
there is an isomorphisms of k-modules, K(R)(M, I) ∼
= D(R)(M, N).
E 6.2.6 Show that Kinj (R) and D(R) are equivalent as triangulated categories; cf. E 6.1.15.
E 6.2.7 Let R be left hereditary. Show that there is an isomorphism M ' H(M) in D(R) for every
R-complex M. Hint: E 5.2.3.
E 6.2.8 Let S be a triangulated subcategory of a triangulated category (T, Σ). A morphism
α : M → N in T is called S-trivial if in some, equivalently in every, distinguished triangle,
α
M −−→ N −→ X −→ Σ M ,
the object X belongs to S. Describe the S-trivial morphisms in the category K(R) if S
consists of all acyclic R-complexes; cf. E 6.1.10.
E 6.2.9 Let (T, Σ) be a triangulated category. A commutative square in T,
U
ϕ
α
M
β
/N
/V
ψ
is called homotopy cartesian if there exists a distinguished triangle of the form
U
ϕ
−α
M
/⊕
(β ψ)
/V
γ
/ ΣU .
N
The pair (ϕ, α) is called a homotopy pullback of (β, ψ), and (β, ψ) is called a homotopy
pushout of (ϕ, α). Show that homotopy pushouts and homotopy pullbacks always exist.
E 6.2.10 Let S be a triangulated subcategory of a triangulated category (T, Σ), and consider the
homotopy cartesian square in T from E 6.2.9. Show that the morphism α is S-trivial if
and only if β is S-trivial in the sense of E 6.2.8. Hint: See [40, lem. 1.5.8].
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