CGP DERIVED SEMINAR
GABRIEL C. DRUMMOND-COLE
1. March 28: Morimishi Kawasaki: Motivation
So I’ll talk about bad behavior of the localization. We want to claim that the
derived category is not an Abelian category. C(R), the category of chain complexes
of R-modules is an Abelian category, but D(Z) is not an Abelian category. We
should say that D(Z) has some good structure, it’s an additive category.
Definition 1.1. C is an additive category if it satisfies the same conditions as an
Abelian category except the existence of kernel and cokernels. It satisfies additivity,
so that morphisms are Abelian groups and composition is bilinear, it has a zero
object, and existence of finite direct sums (and products).
Definition 1.2. Suppose that for any X and Y in the objects of C, Hom
C
(X, Y ) is
an Abelian group. Then there is a zero map and we call Z a zero object if 0
ZZ
= id
Z
.
Proposition 1.1. D(R) is an additive category.
Definition 1.3. Let S ⊂ Hom
C
, then S is a multiplicative system if for all X in
Ob C,
(1) id
X
is in S and
(2) S is closed under composition,
(3) and for any f in Hom
C
(X, Y ) and any morphism in S from Y
′
to Y , then
there exists a morphism g in Hom
C
(X
′
, Y
′
) and a t in Hom
C
(X
′
, X) such
that the diagram commutes
X
′
g
//
t
Y
′
X
f
//
Y
and similarly for any g and t there exists f and s
(4) and for any f and g in Hom
C
(X, Y )¡ there exists s ∈ S such that s ○f = s ○g
if and only if ther exists t in S such that f ○ t = g ○ t.
Proposition 1.2. the set of quasi-isomorphisms is a multiplicative system.
1
I won’t prove this statement.
Lemma 1.1. Let C be an additive category and S a multiplicative system. Then
the localisation is also an additive category.
1
[Added note by Damien Lejay] Stated like that, the proposition is not true, instead what
is true is the following statement: define K(R) to be the category whose objects are the chain
complexes but whose morphisms are the homotopy equivalence classes of morphisms of chain
complexes. Then, in this category, the set of quasi-isomorphisms is a multiplicative system.
1
2 GABRIEL C. DRUMMOND-COLE
Then Proposition 1.2 and Lemma 1.1 directly prove Proposition 1.2.
I’ll sketch a proof of the lemma, only explaining what is the Abelian structure
of Hom
s
−1
C
(X, Y ). By definition of the localization, you can write a morphism as
f ○ s
−1
and another as g ○ t
−1
. Fix
ˆ
f and ˆg in Hom
S
−1
C
(X, Y ), then represent these
by [(f, s)] and [(g, t)] and s and t are in S and f and g in Hom
C
. Then define
ˆ
f + ˆg
as [(f ○ h + g ○ h
′
, u)] where u will be in S and h and h
′
are in Hom
C
. By definition
of the multiplicative system, write
̃
X
h
′
//
h
X
1
X
2
//
X
So now I want to argue that the derived category is not Abelian. First, recall that
[Z/2, Z/2[1]] ≅ Ext
1
(Z/2, Z/2) but these are objects in C(Z). Here Z/2[1] has Z/2
in the −1 position. Then [X, Y ] is Hom
D(R )
(X, Y ). Take P , which is the complex
Z → Z, a projective resolution of Z/2. Then Ext
1
(Z/2, Z/2) = {e, 0} where e(n) =
[n] (mod ()2). So we see that [Z/2, Z/2[1]] ≠ 0, where this is Z/2 ← P → Z/2[1].
This is an explict representation of ˆe.
Now we’re ready for the proof. Assume D(Z) is Abelian. Then, we’ll show that
[Z/2, Z/2[1]] = 0. Then this shows that D(Z) is not Abelian. So let us show that
if D(Z) is Abelian then [Z/2, Z/2[1]] is 0.
Take f ∈ [Z/2, Z/2[1]] and assume D(Z) is Abelian. There exists the kernel of
f = (X, ι) with ι ∶ X → Z/2. Since D(Z) is Abelian, 0 → X → Z/2 → Z/2[1] is an
exact sequence in D(Z).
So to prove that X → Z/2 is a quasi-isomorphism, you use the fact that H
n
(C)
is [Z, C[n]], and now you have 0 → X → Z/2 → Z/2[1], and you can shift to get
X[n] → Z/2[n] → Z/2[n + 1], and then we’ll have
[Z, 0] → [Z, X[n]] → [Z, Z/2[n]] → [Z, Z/2[n + 1]]
which is
0 → H
n
(X) → H
n
(Z/2) → H
n+1
(Z/2)
and the last map here is 0 and so H
n
(X) → H
n
(Z/2) is an isomorphism, so X → Z/2
is a quasi-isomorphism, so f is a zero map, a contradiction.
