CGP DERIVED SEMINAR
GABRIEL C. DRUMMOND-COLE
1. June 12: Kyoung-Seog Lee
Thank you for giving me this opportunity to give a talk. What Damien asked
me to do was solve some exercises. Before that, let me briefly tell you what we’re
going to do. We want to give a model category structure on dg categories. We’ll do
it by giving two of the three classes. The weak equivalences are quasi-equivalences.
We’ll define the fibrations, and thes will be f ∶ T → T
′
satisfying two conditions.
(1) f
x,y
∶ T (x, y) → T
′
(f(x), f (y)) is surjective in chain complexes, and
(2) for any u
′
∶ x
′
→ y
′
in [T
′
], there exists y ∈ [T ] such that f(y) = y
′
, there
exists an isomorphism u ∶ x → y in [T ] such that [f](u) = u
′
Theorem 1.1. (Tabuada) these classes (W, F ) determine a model category struc-
ture on dg categories.
I want to prove exercise 14, my mission given by Damien, this is fun. This is four
parts.
(1) Let 1 have a single object and the morphisms just k in degree 0. Show that
1 is cofibrant. Recall that A is cofibrant if and only if ∅ → A is cofibration.
This is true if and only if for every map X → Y that is a trivial fibration,
there is a lifting
∅
//
X
A
>>
//
Y
Let us prove this. So we have a commutative diagram, so our object in A
hits something y in Y , since this is a trivial fibration, the map from [X]
to [Y ] is essentially surjective. Then there is an object quasi-isomorphic to
y, and x
′
that hits it. By the second property, then there is some x which
hits y. I then just define a dg functor which sends my object to x. You
can just directly check that this works. You have to check that this is a
chain morphism. I can define a map because the differential of the identity
is always zero.
(2) Let me introduce another category and show it has another lifting property.
This category ∆
1
k
is the k-linear category with two objects 0 and 1 and
a morphism. The claim is that this is again cofibrant. This is similar
checking. I didn’t write everything down. Let me check that.
∅
//
X
∆
1
k
//
Y
1
2 GABRIEL C. DRUMMOND-COLE
So I have y
0
and y
1
and I want to lift everything. I again us that X → Y
is a trivial fibration. I can assume that there exists x
0
and x
1
which map
to each of these, by the same argument, and by fullness I’m surjective on
the homotopy class for this morphism, and this gives a way to define this
map. I should confess that I didn’t check every detail of that.
(3) These are about non-cofibrant objects. He claims that by exercise 7, you
can show that the dual numbers k[] is not a cofibrant dg category. Let me
briefly solve this exercise here. Let me explain this exercise. This says that,
let B be a commutative k-dg algebra whose underlying graded k-algebra is
a graded commutative polynomial algebra k[X, Y ] with X degree 0 and Y
degree −1. Let the differential of Y be X
2
. This is a dg category with one
object. Then the first claim is to show that there exists a natural quasi-
equivalence p from B → k[] = k[X]/X
2
. The second claim is to show that
there is no section in dg categories. Then this gives you that k[] is not
cofibrant. Because I can write
∅
//
B
p
k[]
id
//
k[]
We can show that B → k[] is a weak equivalence and the lack of a section
indicates that there is no lift so k[] is not cofibrant. So I have k[X, Y ],
and a differential, and in degree 0, I have k[X]. In −1 degree I have k[X]Y ,
and in degree −2 I have Y
2
= 0 as long as 2 is invertible. Then d
2
(Y ) = 0
so d(X
2
) = 0 so XdX + dX ⋅ X = 0. Then I have f (X)Y which goes to
f
′
(X)dX ⋅ Y + f(X)dY = f (X)X
2
. So in here the kernel is k[X] and the
image is k[X]X
2
so the homology is k[X]/X
2
. If this is zero then f is
zero in this ring. So I computed the cohomology, and H
0
(B) = k[X]/X
2
and H
−1
(B) = 0, et cetera. So I can naturally define k[X, Y ] → K[X]/X
2
naturally. This is a natural quasi-equivalence. This is an isomorphism in
homology. The first condition is satisfied. It does not admit a section. If p
admits a section, then this is k[X]+k[X]Y , it goes to some (f (X), g(X)Y ),
and it should square to zero, so f(X) = 0. Since it is a section, then it can
never go back to X.
(4) Let me finish by proving exercise 4, again very simple, I hope. There is
a non-cofibrant object and I want to show one more example, T is a dg
category with four objects, x, x
′
, y, y
′
, and
x
u
f
//
x
′
u
′
y
g
′
//
y
′
From x to y
′
I have k⟨u
′
f⟩ ⊕ k⟨g
′
u in degree 0. I put in another k⟨h⟩ with
h in degree −1, with boundary u
′
f − g
′
u.
DERIVED SEMINAR 3
There is a category ∆
1
k
⊗ ∆
1
k
. This has
0 ⊗ 0
//
1 ⊗ 0
0 ⊗ 1
//
1 ⊗ 1
with diagonal morphism just k. Again the argument says that there is a
natural fibration T → ∆
1
k
⊗ ∆
1
k
with no section, so then ∆
1
k
⊗ ∆
1
k
cannot be
cofibrant. There is a natural functor, just take everything to the thing it
looks like, and if I construct a functor between them, the only thing I have to
check is that Hom(x, y
′
) → Hom(0⊗0, 1⊗1) which takes ⟨u
′
f⟩⊕⟨g
′
u⟩⊕⟨h⟩ →
k⊗k is a quasi-isomorphism. This takes h to 0 and the other two generators
to 1 ⊗ 1. So this is a chain map. It’s also a quasi-isomorphism. You can
check that this is 0 → k → 0 in both cases. So this is a trivial fibration, and
there is no section, let me claim, that’s the only claim I want to say, this
is, this involves a lift
