CGP DERIVED SEMINAR
GABRIEL C. DRUMMOND-COLE
1. April 11: Cheolgyu Lee: Model categories I
So you have seen the definition, so now we will give one hour of just examples.
Let C be a category with objects non-negatively graded chain complexes on
Mod
R
, for R a ring with identity and morphisms chain maps. Then we can define
weak equivalences to be quasi-isomorphisms, cofibrations to be chain maps which
is injective in each degree with projective cokernels. Let fibrations be surjections in
positive degrees. Then we can check that for any 0 → M
∗
, then there is a projective
resolution 0
ι
Ð→ P
π
Ð→ M , a projective resolution. We can easily see that π is in
F ∩ W , it’s an acyclic fibration and ι is a cofibration.
Suppose we’re given a chain map N
∗
→ M
∗
, then we have a complex N
∗
⊕ P
∗
,
and a factorization N → N ⊕ P → M. So now we have P sitting inside N ⊕ P and
also P projecting to M . We can define a homotopy category Ho C, where we invert
weak equivalences, but this constructs a quiver, arrows in C where the arrows in
W are inverted, this has index set objects in C and we need a bigger universe to
construct it. With the structure of a model category, we can define the homotopy
relation in morphisms in C. Let C
cf
be the full subcategory of objects that are both
cofibrant and fibrant.
I need to define fibrant and cofibrant. We call X cofibrant if the unique morphism
from the initial object is a cofibration and fibrant if the unique morphism to the
final object is a fibration. I want to assume the existence of two functors Q and R
which are called cofibrant replacement and fibrant replacement functors. I might
need some version of the axiom of choice to define it. This is a functor from C to
itself so that QX is a cofibrant replacement and RX is a fibrant replacement of X.
I want to explain a first lemma, Ken Brown’s lemma.
Lemma 1.1. Let C be a model category with structure (W, C, F ) and D a category,
not necessarily a model category, with some class of weak equivalences satisfying
the two out of three property.
Suppose a functor G sends acyclic cofibrations between cofibrant objects to weak
equivalences. Then G sends weak equivalences between cofibrant objects to weak
equivalences.
There is also a dual version that I won’t state.
Proof. Let f ∶ A → B be a cofibration between cofibrant objects. Then I can
factorize A ⊔ B → B into a cofibration b followed by an acyclic fibration a. Then
the identity is a weak equivalence. Because A and B are cofibrant, then c
1
and c
2
are cofibrations, then because pushout preserves cofibrations, then the inclusions
of A and B into A ⊔ B are cofibrations. Then b ○ c
′
1
and b ○ c
′
2
are both acyclic
cofibrations. Then F (b ○ c
′
1
) and F (b ○ c
′
2
) are weak equivalences. So F (a) is in W
′
1
2 GABRIEL C. DRUMMOND-COLE
by the two out of three property. Then F (a) ○ F (b ○ c
′
2
) is in W
′
; that’s the same
as F (f ).
Let me give a definition. A cylinder object Cyl(X) is an object of C such that
X ⊔ X
c∈C
ÐÐ→ Cyl(X)
w∈W
ÐÐÐ→ X
where the composition is id ⊔ id and a path object is dual:
Y
w∈W
ÐÐÐ→ Path(Y )
f∈F
ÐÐ→ Y × Y
where the composition is the diagonal id × id.
A left homotopy from f to g is a map H ∶ Cyl(X) → Y satisfying that Hi
0
= f
and Hi
1
= g where i
0
⊔ i
1
is the map from X ⊔ X → Cyl(X).
If there is a left homotopy from f to g then we write f
`
∼ g.
If objects are cofibrant and fibrant then left homotopy implies right homotopy
and vice versa. We say that f ∼ g if f is both right and left homotopic to g. We
say that f is a homotopy equivalence if there is a “homotopy inverse” so that both
compositions are homotopic to the respective identities.
From now on I will construct an equivalence relation on the space of morphisms
between two objects. We can check the following
Lemma 1.2. f
`
∼ g implies that h ○ f
`
∼ h ○ g and f
r
g
implies that f ○ h
r
∼ g ○ h.
Lemma 1.3. If Y is fibrant then f
`
g
implies f ○ h
`
∼ g ○ h. Dually if X is cofibrant
and f
r
∼ g then h ○ f
r
∼ h ○ g.
Let me give a partial proof. So you have X⊔X → Cyl(X) → X and H ∶ Cyl(X) →
Y . We can assume that w is an acyclic fibration because Y is fibrant, you can form
a lift
Cyl(X)
//
%%
Y
Cyl
′
(X)
;;
yy
X
//
∗
Then we can use this to lift.
I didn’t prove that this is an equivalence relation. I didn’t prove transitivity.
Lemma 1.4. If X is cofibrant then
`
∼ is an equivalence relation.
It suffices to show transitivity. So assume f
`
∼ g and g
`
∼ h. So we have Cyl(X)
0
→
Y and Cyl(X)
1
→ Y . Then we can take a colimit of Cyl(X)
0
⊔Cyl(X)
1
over X with
respect to i
′
0
and i
1
. We have a map to X and both maps are weak equivalences.
Since X is cofibrant, the maps i and i
′
are cofibrations, in fact acyclic cofibrations
(by an easy diagram chase).
Then i
0
and i
1
are trivial cofibrations. We can’t deduce that X ⊔ X → Z is a
cofibration, but if we factorize this into a cofibration followed by a trivial fibration,
then we get a cylinder obect K, as desired.
So we get a homotopy relation.
DERIVED SEMINAR 3
Lemma 1.5. Suppose X is cofibrant and Y and Z are fibrant. Then a weak equiv-
alence Y → Z induces a bijection Hom
C
(X, Y )/
`
∼→ Hom
C
(X, Z)/
`
∼
I should also have assumed that the morphism spaces are sets.
Lemma 1.6. If X is cofibrant then f
`
∼ g implies f
r
∼ g.
By these five lemmas we can deduce that
Theorem 1.1. The category C
cf
/ ∼ exists.
But we didn’t see how we can invert the weak equivalences. I will state it.
Theorem 1.2. A map of C
cf
is a weak equivalence if and only if it is a homotopy
equivalence.
So this is how we invert weak equivalences in that category. Actually,
Theorem 1.3. Let δ ∶ C → C
cf
/ ∼ and Q and R be cofibrant and fibrant replacement
functors. Then δQR satisfies the universal property for the homotopy category
Ho(C).
Let me just give a proof. Let D be a category and F a functor from C to
D sending weak equivalences to isomorphisms. Then there is a “unique” functor
that I will construct G ∶ C
cf
/ ∼→ D. Then G(δQRX) = F (δQRX) and [f] ∈
Hom
C
cf
(δQRX, δQRY ) ≅ Hom
C
(δQRX, δQRY )/ ∼, whith G([f ]) = F (f).
We should check athat it’s well-defined. Suppose f ∼ g, then there exists a
cyclinder on δQRX with
δQRX ⊔ δQRX → Cyl(δQRX) → δQRY
and because δQRX is cofibrant, then i
0
and i
1
are cofibrations and using the
identity wi
0
= id
δQRX
= wi
1
, and F (w) is an isomorphism, so F (i
0
) and F (i
1
)
have the same image. So F (f) = F (Hi
0
) = F (Hi
1
) = F (g). Then we can check
composition. Now I can say that I defined the homotopy category for a model
category.
