CGP DERIVED SEMINAR
GABRIEL C. DRUMMOND-COLE
1. June 27: Tae-su Kim
I’m going to talk about the Yoneda embedding, which was part of Youngjin’s
talk last time, but he skipped it. So k will be our ring, a commutative unital ring,
and T a dg category defined over k. Then a T -dg module is nothing but a dg
functor F ∶ T → C(k). What we’re going to do is construct a dg functor from T to
another category which is quasi-fully faithful, and this will be Int(T
op
− mod ).
Let me construct this functor. I’ll denote a functor h ∶ T → T
op
− mod which
will give rise to the one I want later. So we have x ↦ h
x
∶ T
op
→ C(k). This
functor h
x
takes y to T (y, x). Moreover, a morphism α ∶ x → x
′
gives a natural
transformation given using the composition map, from h
x
to h
x
′
. So for each z this
defines a map T (z, x) → T (z, x
′
) and it’s just composition with α.
I want to show that this is a fibrant and cofibrant object in T
op
− mod to say
that it lands in the interior. To talk about this I need to define the model structure
on T
op
− mod . If I have F and F
′
objects in T
op
− mod , then morphisms between
them are natural transformations between functors. This F → F
′
is a fibration if
F
x
→ F
x
′
is a fibration if it’s a fibration in C(k) and similarly for weak equivalences.
This is our model category structure on T
op
− mod . We can also find an initial
and terminal object, which is a 0 object, which to any object in x assigns the zero
chain complex. My claim is that h
x
→ 0 is a fibration. This is easy to show because
what we have to show is that (h
x
)
y
→ 0 is a degreewise surjection. But with 0 as
a target it will be surjective, so this is a fibrant object.
The more tricky part is to show that h
x
is cofibrant. What we have to show is,
we need this kind of diagram
0
//
F
′
p
h
x
>>
//
F
for F, F
′
objects of T
op
− mod and p a trivial fibration. So to each object y I need
to assign maps
T (x, y) → F
′
y
→ F
y
What do I get for y = x? I can assign id
x
to something in F
′
x
because F
′
x
→ F
x
is
a trivial fibration so a degreewise surjection. Now for α in T (y, x), I want to give
something in F
′
y
. But α defines a map F
′
x
→ F
′
y
. Since F is a T
op
-module, there is
a map φ
′
α
which takes a to φ
′
α
(a). Then my lift will be α ↦ φ
′
α
(a).
This data will define a natural transformation as I desired. Let me check. I
have p
y
⋅ g
y
= f
y
. So p
y
⋅ g
y
(α) = p
y
⋅ φ
′
α
(a). By the definition, since p is a natural
1
2 GABRIEL C. DRUMMOND-COLE
transformation, I have
F
′
x
φ
′
α
//
F
′
y
F
x
φ
α
//
F
y
Then
p
y
⋅ φ
′
α
(a) = φ
α
p
x
(a) = φ
α
⋅ f
x
(id
x
) = f
y
φ
h
α
(id
x
) = f
y
(α).
Here φ
h
is a transformation between f
x
and f
y
.
So we have checked that h
∶ T → T
op
− mod is fibrant and cofibrant in the
category, so instead we can write it as landing in Int(T
op
− mod ). Our next claim
is that this is a quasi-fully faithful functor between these two dg categories.
What does this mean? It means that the morphism level map T (x, y) → Hom(h
x
, h
y
)
is a quasi-isomorphism. Let me check this. I’ll directly construct a map ψ
z
for every
z in the objects of T from
Z(T (x, y))/B(T (x, y)) → Z(Hom(T (z, x), T (z, y))/B(Hom(T (z, x), T (z, y))).
So suppose I have [u] in the domain, then Ψ
z
([u])
∶
= [φ
z
(u)] where φ ∶ T (x, y) →
{Hom(T (z, x), T (z, y))} is defined by composition with u. Let me check that this
Ψ
z
is an injective map. So φ
z
(u)(b) = (dα, b) = d(α(b)) ± α(db). If for all b we have
this equality, for z = x we get φ
x
(u)(id
x
)dα(id
x
) ± α(did
x
) = dα(id
x
). This is u on
the left, so u is zero in the domain. So the map is injective. [Some problems with
the setup]
Surjectivity is more tricky, I’ll explain after that talk.
Assume we have defined this, then we’ll call this the Yoneda embedding.
The next topic I’m going to talk about is about [T, Int(M)]. I want to state a
proposition here and Gabriel will prove it later. This is isomorphic to Iso Ho(M
T
).
Here M is a C(k)-model category and T is our dg category, and we’ll assume
two conditions, M is cofibrantly generated. This means that there is some set of
cofibrations and trivial cofibrations, suitably small, that generates the cofibrations.
The second thing is, if E → E
′
is a quasi-isomorphism in C(k) and X is cofibrant
in M . Then we demand that E ⊗ X → E
′
⊗ X is an equivalence. Under these two
conditions, we can show this, we can prove this. I want to talk about a lemma,
for f a quasi-equivalence from T to T
′
, quasi-fully faithful and quasi-essentially
surjective, then, under the conditions of the hypothesis, the homotopy categories
of Ho(M
T
) and Ho(M
T
′
) are equivalent, witnessed by f
∗
and f
!
.
Let me quickly mention how to prove this lemma and then I’ll stop. The first one
is about any object in the homotopy category. Any object in M
T
can be written as
a homotopy colimit of something. Let I be a category and let c ∶ M
T
→ (M
T
)
I
be
the constant functor. Then this induces a map on the homotopy level Ho(M
T
) →
Ho((M
T
)
I
) which has a left adjoint, the homotopy colimit functor. Any object in
M
T
can be written hocolim(F ) with F (i)
∶
= h
x
i
⊗ X
i
where X
i
are cofibrant.
The second fact is that hocolim and Lf
i
(and f
∗
) commute. So then it’s enough
to prove it for just cofibrant elements h
x
i
⊗X
i
. Then he proves the statement in this
case. We have to show there is a natural isomorphism Lf
!
f
∗
→ id and f
∗
Lf
!
→ id.
These follow similar logic so let me show one case. So what are we trying to do?
We’re trying to show an isomorphism Lf
!
f
∗
(h
x
i
⊗ X
i
) ≅ h
x
i
⊗ X
i
.
DERIVED SEMINAR 3
So I want to say that this is isomorphic in the homotopy category for some x
′
to Lf
!
f
∗
(h
f(x
′
)
⊗ X) by quasi-essential surjectivity. Then this is isomorphic to
Lf
!
(h
x
′
⊗ X), from quasi-full faithfulness. Then the object we have is coifbrant
and then we can erase L, and then f
!
(h
x
′
⊗ X), by adjointness, this is the same as
h
f(x)
⊗ X ≅ h
x
⊗ X.
The opposite direction can be shown in a similar way. Then we have the equiv-
alence of categories between Ho(M
T
) and Ho(M
T
′
). I should have gone into more
detail but I didn’t have much time, I think this is where I should stop.
