E X A M P L E S O F M O D E L

S T R U C T U R E S

damien lejay

April 2017

Abstract. — We give an introduion to the andard model ruures on the

category of unbounded chain complexes by deﬁning ﬁr model ruures on

the category of Abelian groups.

contents

1 Reminder on why we like model categories 1

2 The father of all examples 2

3 Model ruures on Abelian groups 4

3.1 The andard projeive model ruure 5

3.2 The andard injeive model ruure 7

4 The model ruure on chain complexes over a ﬁeld 9

5 Model ruures on chain complexes of Abelian groups 12

5.1 The andard projeive model ruure 12

5.2 The andard injeive model ruure 13

References 14

1 reminder on why we like model categories

The derived category

D(Z)

is by deﬁnition the localisation of the category of

chain complexes where the set of maps we would like to invert is the set of all

quasi-isomorphisms Q:

D(Z) ≃ Q

−1

Ch(Z)

In fa, it is deﬁned as the model of localisation whose set of objes is the

set of all chain complexes (the localisation of a category is only deﬁned up to

equivalences of categories).

Given a category

and a set

of arrows of

, if

is known to be a

-small

category, then any model of the localisation

−1

is garanted to be equivalent

to a

-small category. Apart from that, essentially nothing is know on the

concrete description of the localisation. If we choose the model which has

the same objes, then the diﬃculty is to underand the set of arrows. In

full generality, the be we know is that it can be described as some huge

‘zig-zags’.

The situation is much better in the following two cases.

— The set S admits a calculus of fraions;

— The category C can be enhanced into a model category.

In the ﬁr case, the morphisms

X → Y

of the localisation may be described

by ‘fraions’ of one of the two types:

←− Z

−→ Y || X

−→ Z

←− Y

These notes were written for the ‘derived seminar’ I organised at IBS Centre for geometry and

physics in Pohang. Copyright (C) Damien Lejay 2017.

with

s ∈ S

, depending on whether

admits a right or left calculus of frac-

tions [1].

In the second case, a new model for the localisation may be proposed. The

ruure of model category allows us to isolate the subcategory of ﬁbrant-

coﬁbrant objes. The localisation — called the homotopy category in this

context — is then given by deﬁning maps between ﬁbrant-coﬁbrant objes as

homotopy equivalence classes of morphisms in the category C.

The ﬁr case is less convenient to work with because before arting to

compute the equivalence classes of arrows up to homotopy, one ﬁr has to

replace objes by equivalent ﬁbrant-coﬁbrant objes. Alas, the set

of all

quasi-isomorphisms does not have a calculus of fraions.

Proposition. — The set

of quasi-isomorphisms in

Ch(Z)

does not admit a

calculus of fraions.

Proof. — We will show that

does not satisfy the — left or right — cancellab-

ility condition.

Let

be the chain complex

−→ Z

and let

f : A → Z

be the projeion

on the ﬁr term. Then

s : 0 → A

is a quasi-isomorphism such that

f ◦ s = 0

but it is impossible to ﬁnd a quasi-isomorphism

′

: Z → B

such that

′

◦f = 0

(since f is surjeive, that would mean s

′

= 0).

In the other direion let

i : Z → A

be the injeion to the second term.

Then if

is the quasi-isomorphism

A → 0

, we have

t ◦ i = 0

. But since

injeive, it is impossible to have a quasi-ismorphism

′

: B → Z

such that

i ◦ t

′

= 0.

Hence, we are doomed to ﬁnd a model ruure on the category of chain

complexes. The minimum requirement to be able to ﬁnd such ruure with

weak equivalences the set

is that

be able under retras and satisfy the

2 out of 3 property.

Proposition. — The set

of quasi-isomrophisms of

Ch(Z)

is able under retras

and satisﬁes the 2 out of 3 property.

Proof. — The 2 out of 3 property is an immediate consequence of the funtori-

ality of the homology funor C 7→ H

∗

(C).

The same is true for the ability under retras, if

is a retra of a

given chain map

then,

∗

(g)

is a retra of

∗

(f )

and any retra of an

isomorphism is an isomorphism.

Remark. — More generally, given any funor

F : C → D

, the set of arrows

deﬁned by

s ∈ S

if and only if

F(s)

is an isomorphism, will satisfy the 2 out of

3 property and be able under retras.

This means aually that the two condition imposed on a set to be weak

equivalences of a model ruure are faily weak: for any category

and set

of arrows

, the set of arrows

that become and isomorphism in the

localisation

−1

is able under retra and satisﬁes the 2 out of 3 property.

And of course we have S

−1

C ≃ W

−1

2 the father of all examples

Before inveigating the possible model ruures on the category of chain

complexes, we shall ﬁr focus on more elementary Abelian categories.

Let

be your prefered ﬁeld. Let

Vect

denote the category of veor

spaces over

. We shall classify all the possible model ruures on

Vect

and we shall discover that only two sets of arrows can be taken as set of weak

equivalences for a model ruure.

The key to classify model ruures on

Vect

is to ﬁr classify lifting

syems (L, R). We shall see that there are only a handful of them.

So let

(L, R)

be a lifting syem i.e the elements of

are the maps

such

that the following diagram can be ﬁlled up

A C

B D

for every arrow

r ∈ R

. In the same fashion, the set

is required to be deﬁned

as the maps

that lift again all maps of

. Hence by deﬁnition,

contains

all isomorphisms.

Case 1. — L = {iso}. In this case, R = {all} and we have a ﬁr pair

(iso, all)

Case 2. —

contains a map

that not injeive. Any map

retras to the

zero map

Ker f → 0

; in this case, since the kernel of

is not zero,

retras

to the zero map

π : F → 0.

So by the ability properties of lifting syems, we know that

is in

Computing the set of maps that can be lifted again π, we get

= {injections}

Let ι be the zero injeion

ι : 0 → F

By computation we have

ι = {surjections}.

Hence, we get that

{surjections} =

ι ⊂

) ⊂ L

Case 3. —

contains a map

that is not surjeive. Then in this case, it is

possible to retra f to ι. By computation,

= {surjections}

Since we also have

π = {injections},

we have

{injections} =

π ⊂

) ⊂ L

Conclusion. — Since

is also able by composition, we have only four possible

choices for L and they correspond to the following lifting pairs

(iso, all) (inj, surj) (surj, inj) (all, iso)

Notice that all those four lifting pairs are also weak faorisation syems.

Indeed, the faorisation for the pair (surj, inj) is the image faorisation

A B

Im(f )

and it is the graph faorisation for the pair (inj, surj):

A B

A ⊕ B

Getting back to ﬁnding all model ruures, we are looking for three sets

of maps

(W, C, F)

such that

(W ∩ C, F)

and

(C, W ∩ F)

are weak faoriion

syems and

satisﬁes the 2 out of 3 property. Since there are only four

lifting syems, we can deduce that any set

has to contain either

iso, inj, surj

or be the set of all arrows.

Suppose

contains the set of all injeive maps. Then give any map

f : A → B, since we have the following commutative diagram

0 B

by the 2 out of 3 property, f is in W and W = {all}.

Suppose now that

contains the set of all surjeive maps. Then for any

f : A → B, the following diagram

A 0

and the 2 out of 3 property means that f is in W and W = {all}.

We can know li all the model ruures on the category Vect

— (iso, all, all), it is the trivial model ruure;

— (all, iso, all)

, it is also a trivial model ruure but the homotopy cat-

egory in this case is the point;

— (all, all, iso), analogous to the previous one;

— (all, surj, inj);

— (all, inj, surj).

Remark. — The same inveigations can be made for

Set

the category of sets.

However, since it is usually impossible to retra to the empty set, there

are a total of 9 diﬀerent model ruures on

Set

with 3 diﬀerent homotopy

categories.

As we can see, only the la two model ruures are not trivial. But the

very la one is far more useful than the other. We shall call it the andard

model ruure on

Vect

. The reason behing this is that both the set of ﬁbrant

objes and coﬁbrant objes are reduced to

for the model ruure given

by the triplet

(all, surj, inj)

. Hence, there is no space for tweaking: if we

would like to change some parts of the ruure and keep the ﬁbrations for

examples, then whatever we do, the set of ﬁbrant-coﬁbrant objes is ill

going to be reduced to a point. The same happen if we would like to keep the

coﬁbrations. On the opposite, in the andard model ruure, every obje is

both ﬁbrant and coﬁbrant!

Theorem (Experimental fa). — All useful model categories are based on the

andard model ruure.

3 model structures on abelian groups

The ﬁr main diﬀerence between

the category of Abelian groups and

Vect

is that we ill do not have a complete classiﬁcation of objes in

[

As a consequence, it is impossible to compute all the model ruures on

. What we shall do inead, is to build model ruures that mimic the

andard model ruure on Vect

Fa. — The pair (inj, surj) is not a lifting pair in Ab.

To see this, have a look to the following commutative square,

0 Z

mod

3.1 The andard projeive model ruure

Idea 1. — Keep the ﬁbration and look for the corresponding left part of the

pair.

Fir let us look at the corresponding coﬁbrant objes.

Definition 3.1. — An abelian group

is called projeive if the map

0 → P

satisﬁes the left lifting property again surjeions. That is, we can always ﬁnd a

lift for the following types of diagrams,

P Y

Proposition 3.2. — The projeive Abelian groups are the free Abelian groups.

Proof. — If

is a free Abelian group then it is projeive as any map with

domain P is only determined by the image of elements of a basis of P.

Suppose P is projeive and let L(P) be the free Abelian group generated

by the elements of

. Then because

is projeive we get the following

retraion

L(P)

P P

can be seen as a subgroup of a free Abelian group and since

is a PID,

it means P itself is a free Abelian group.

We can now generalise this setting to describe all the maps that can be

lifted again surjeions.

Proposition. — A map

f : A → B

is inside

surj

if and only if it is a (split)

injeion with a free cokernel.

Proof. — If

is an injeion with a free cokernel, then it has to be split (thanks

to the lifting property of free modules), so we can write

B ≃ A ⊕ Coker f

and

is the identity of

. Hence, we can split the problem into a lifting

problem for

Coker f

and

. We have already deﬁned the former to be

objes that solve that very problem and isomorphisms can always be lifted.

Suppose now that

satisﬁes the left lifting property. Denote by

the

projeion

π : B −→ Coker f

To show that the cokernel is free, we will show that it is projeive. Let us be

given any diagram

0 X

Cocker f Y

and we wish to lift

. Composing,

with

gives us a map

B → Y

whose

composition with

is equal to zero. Hence we can complete the former

diagram into

A 0 X

B Cokerf Y

Using the lifting property of f , we can complete the square in

A X

B Y

g◦π

But since

composed with

is zero, by the universal property of the cokernel,

we get a new map ϕ : Coker f → X such that ϕ = ϕ ◦ π. So we now have

p ◦ ϕ ◦ π = g ◦ π

and since π is an epimorphism we get the following lifting

0 X

Coker f Y

This proves that the cokernel has to be projeive. Since maps with a projeive

cokernel split, we can then rewrite B as the sum

B ≃ Im f ⊕ Coker f

And we can now reri our attention to the case where the cokernel is zero.

In this case the following diagram

A A

B B

f f

tells us that we can split A into

A ≃ Ker f ⊕ B

This la diagram will tell us that the kernel of f has to be zero:

Ker f ⊕ B Ker f

B 0

Ker f

Corollary. — The pair made of injeive maps with free cokernels on one side

and surjeive maps on the other side, is a lifting pair.

Proof. — All we need to prove is that any surjeion has the right lifting

property again any injeive map

f : A → B

with free cokernel. But as we

have already said, such a map can be split into the sum of an isomorphism

(which can be lifted) and the zero map

0 → Coker f

which by deﬁnition of

projeiveness can also be lifted.

Theorem. — There exis a model ruure (W, C, F) on Ab where:

— W is the set of all maps;

— F is the set of all surjeive maps;

— C is the set of all injeive maps with free cokernel.

This model category is called the andard projeive model ruure.

Proof. — Since

W ∩ F = F

and

W ∩ C = C

, the only thing that is left to

prove is the faoriion property. Let

f : A → B

be any map between Abelian

groups. Let

L(B)

be the free Abelian group with basis the elements of

the canonical map

π : L(B) → B

is surjeive. We obtain the sought for

faorisation as

A B

A ⊕ L(B)

f +π

where

is the inclusion on the ﬁr faor. The cokernel of

L(B)

and it

is free.

Remark. — The andard projeive model ruure will exis for every

category of modules over a commutative ring. Ju replace free Abelian

groups by the corresponding projeive modules over the given ring (if the

ring is not a PID).

Remark. — In this model ruure every obje is ﬁbrant but the only coﬁbrant

objes are the free Abelian groups. Many times in the proves, we have used

the fa that for every Abelian group

, there exis a surjeion

L(B) ↠ B

with

L(B)

a free Abelian group. This is nothing else than the coﬁbrant replacement

of B.

3.2 The andard injeive model ruure

Idea 2. — Keep the weak equivalences and the coﬁbrations.

In this case let us give a name to the new ﬁbrant objes.

Definition 3.3. — An Abelian group

, is injeive if it satisﬁes the right lifting

property again all injeions, i.e if every such diagrams can be ﬁlled,

X I

Y 0

Proposition. — The injeive Abelian groups are the divisible Abelian groups,

that is the Abelian groups

such that for every

a ∈ I

and

n ∈ Z

, there exis

b ∈ I

such that a = nb.

Proof. — The divisibility condition for an Abelian group

is equivalent to

being able to ﬁll the following diagrams

Z I

Z 0

×n

Hence every injeive Abelian group is divisible. Conversely, given a divisible

Abelian group

, a map

f : X → I

and an inclusion

X ⊂ Y

, we will show

that

can be extended to

. Let

be an element of

that is not in

. Then

either

is free relative to

and

may be extended to

X ⊕ Zy

by choosing

any element r ∈ I and letting f (x + ny) = f (x) + n f (y) be the extension.

Either there exis an element

x ∈ X

and a prime number

p ∈ Z

such that

py = x

in which case, thanks to the divisibility of

, there is an element

r ∈ I

such

that

pr = f (x)

. Then to any element

x + my ∈ X + Zy

, the extended map

is given by

f (x + my) = f (x) + m r

. The extended map is correly deﬁned

since f (py) = pr = f (x).

This extension can be performed on any submodule of

on which

has previously been extended. The results then comes from (transﬁnite)

induion.

Definition 3.4. — For a prime number

, let

∞

be the Abelian group colimit of

the diagram of inclusions Z

,→ Z

n+1

Remark. — Since every cyclic group Z

can be represented as the quotient

≃ Z ⟨1/p

⟩/Z

the group Z

∞

may be seen as

∞

≃ Z[1/p]/Z.

where

⟨.⟩

denotes group generation and

[.]

denotes ring generation. It can also

be described as

∞

≃ Q/pZ.

Proposition (Classiﬁcation of injeive Abelian groups). — Every injeive

Abelian group is a sum of the groups Q and Z

∞

for every prime p.

Proof. — Fir

and

∞

are injeive because they are divisible. This is

obvious for the ﬁr one and a consequence of Bezout theorem for the second

one.

Let

be any injeive Abelian group, we will show that

splits as the sum

of its injeive subgroups ‘divisibly generated’ by one element. Let

be an

element of

. If

is not a torsion element, since

is divisible, it deﬁnes an

injeion

: Q → I

and as

is injeive,

splits into a sum

I ≃ Q ⊕ J

where

is also divisible (every quotient of a divisible group is divisible).

Hence we can now suppose that

is equal to its torsion subgroup. For

a given prime number

, let

be an element of

-torsion of

. Since

divisible, for every k ≥ 1, there exis an element x

∈ I such that

x = p

k−1

The subgroup of

generated by the

for

k ≥ 1

is isomorphic to

∞

. The

map sends the generators

−k

and all we need to show is that every

linear equation

k=1

= 0

with all

integer numbers not divisible by

, image of a element

k=1

∞

, is in fa impossible. For every

, by conruion

= 0

. It follows

that

N−1

= 0

. But since

= 0

it means that

divides

which is

impossible unless a

= 0. This shows that the built map

∞

−→ I

is injeive. The end of the proof consis in a usual (transﬁnite) induion.

Remark. — By the classiﬁcation of divisible Abelian groups, we see that they

are exaly the Abelian groups where the ruure of

(Z, +)

-module can be

extended to a ruure of (Q, +)-module.

Proposition (Fibrant replacement). — The induion funor

Q ⊗

(Z,+)

: Ab −→ Mod

(Q,+)

is left adjoint to the forgetful funor and is such that the unit of the adjunion is

a monomorphism i.e for every Abelian group A, the canonical map

A ,→ Q ⊗

(Z,+)

is an injeion.

In particular, we get the following replacements:

Z ,→ Q ; Z

,→ Q/p

Z ≃ Z

∞

Proof. — The adjunion is a andard result in the theory of modules over

monoids. The only thing we have to prove is that the unit is a monomorphism.

Let

be any element of

, then its image by the unit is

(0, a)

which is equal

to 0 if and only if a is equal to zero in A.

Now that we have dealt with the necessary prerequisites, we can describe

the new model ruure.

Theorem. — There exis a model ruure on the category of Abelian groups

where

— any map is a weak equivalence;

— a map is a coﬁbration if and only if it is injeive;

—

a map is a ﬁbration if and only if it is a (split) surjeion with divisible

kernel.

This model ruure is called the andard injeive model ruure.

The proof of this theorem is a complete mirror of the proof of the exience

of the projeive model ruure (once you know how to build a ﬁbrant

replacement). Indeed, it is the projeive model ruure on the category

Remark. — As we see, the key fa about the category of Abelian groups is

that we are always able to ﬁnd, for every group

, a surjeive map to

with

free domain and an injeive map from

to a divisible group. We say that

the category of Abelian groups has both enough projeives and injeives.

4 the model structure on chain complexes over a field

Let us move on to the category

Ch(F)

of unbounded chain complexes over

your cheriched ﬁeld

. This time we will build a model ruure with non-

trivial weak equivalences, namely the set Q of quasi-isomorphisms.

But ﬁr we may warn the reader of the following fa.

Proposition. — Any chain complex in Ch(F) retras to its homology.

Proof. — Let

... C

−1

...



−1

be any chain complex over

. Let us denote by

the image of the map



n+1

and let

be the kernel of



inside

. Finally let

be the quotient

. Since on a ﬁeld, every submodule is a dire summand, we get that

for every n,

≃ B

⊕ H

⊕ B

n−1

Hence it is very raighforward to build a retraion map

... C

−1

...

... H

−1

...



−1

0 0 0 0

with raighforward seion

... C

−1

...

... H

−1

...



−1

0 0

−1

Remark. — Although every chain complex retras to its homology, it is not

true that every map of chain complexes f retras to H(f ).

As a consequence, we have the following description of the derived cat-

egory of chain complexes over F.

Theorem. — The homotopy category

D(F)

is equivalent to the category of graded

veor spaces Mod

Proof. — Thanks to the previous proposition, the homology funor

H : Ch(F) → Vect

is left adjoint to the inclusion of graded veor spaces inside chain complexes

Vect

Ch(F)

with π being the unit of the adjunion and ι being the counit.

Moreover, a chain map is an isomorphism precisely if its image by

an equivalence of graded veor spaces. Thus, the derived category is even a

reﬂeive localisation of Ch(F).

As a consequence, we do not need any model ruure on the category of

chain complexes to be able to underand its homotopy category. Nonethe-

less, underanding the andard model ruure on the category of chain

complexes over a ﬁeld will be an important ep toward the model ruures

we are going to build on the category of chain complexes of Abelian groups.

Theorem. — There exis a model ruure on the category of chain complexes

over F where

— the weak equivalences are the quasi-isomorphism;

— the ﬁbrations are the epimorphism;

— the coﬁbrations are the monomorphisms.

This model rure is called the andard model ruure.

Proof. — In view of the total symmetry of the given ruure, we shall

only prove that the pair made of trivial coﬁbrations and ﬁbrations is a weak

faorisation syem. The same proof applied to

Ch(F)

will prove that

coﬁbrations and trivial ﬁbrations also form a weak faorisation syem.

Let us ﬁr setup some notations. We will call

the following trivial

coﬁbration

... 0 0 0 ...

... F[n + 1] F[n] 0 ...

A map of complexes

has the right lifting property again

if and only if

n+1

is surjeive, hence

{ι

}

n∈Z

= {ﬁbrations}

Let

denote the surjeion

: F[n] −→ 0

. A map

has the left lifting

property again

if and only if

(f )

is injeive. Finally let us denote by

the following ﬁbration

... F[n + 1] F[n] 0 ...

... F[n + 1] 0 0 ...

A map of chain complexes

has the left lifting property again

if and

only if H

n+1

(f ) is surjeive and f

is injeive. As a consequence

{π

, p

}

n∈Z

= {trivial coﬁbrations}

Laly we need to show that it is aually possible to lift any of our deﬁned

trivial coﬁbrations again a ﬁbration. Let

f : X → Y

be a trivial coﬁbration

and g : R → S be a ﬁbration and let be given a commutative square

X R

Y S

∼

We will conru a map

ϕ : Y → R

to ﬁll up this diagram. A andard rategy

would be to art lifting

and then induively build the other

but since

we are dealing with unbounded chain complexes there isn’t really a right

place to art and we will see that being able to lift is in fa some kind of a

global property of a complex. Let us adopt another rategy.

Fir we may split Y as a sum

Y ≃ X ⊕ Coker f

and since

is a quasi-equivalence, its cokernel is aually quasi-isomorphic to

zero. As it is always possible to lift an isomorphism, we can suppose that

zero and that

is quasi-isomorphic to zero. Hence what we shall prove is that

trivial coﬁbrant objes lift againts ﬁbrations. We wish to lift the following

diagram

0 R

Y S

∼

Since

is a surjeion for every

, it is possible to ﬁnd a map

: Y

→ R

lifting s

. This gives us a map ϕ of graded veor spaces but it may fail to be

a chain map. Let ψ be deﬁned as the failure of ϕ to be a chain map:

= 

n+1

◦ ϕ

n+1

− ϕ

◦ 

n+1

By conruion

is a chain map from

R[1]

and since

is a lift again

the image of ψ is contained in (Ker g)[1],

(Ker g)[1]

Y R[1]

g[1]

Now thanks to the exaness of the complex

, it is aually homotopic to zero

(since exa complexes are split over a ﬁeld). Hence, we can ﬁnd a homotopy

h between ψ and 0:

= 

n+1

◦ h

n+1

− h

◦ 

n+1

We now deﬁne a new map

τ = ϕ − h

By conruion

is a chain map from

. By conruion also, since

has its values in Ker g, we have

g ◦ τ = t

And the diagram is ﬁlled.

We now approch the very ﬁnal ep: showing that any map can be faored

into a trivial coﬁbration followed by a ﬁbration. Let

f : X → Y

be any map of

chain complexes. Let also

be the chain complex quasi-isomorphic to zero

such that

= Y

−1

= Y

and the only non-zero map is the identity of

. Let

π :

Y → Y

be the ﬁbration deﬁned by projeion on the

faor. Then we can

faor the map f as follows,

X Y

X ⊕

∼

f ⊕π

where ι is the inclusion on the ﬁr faor.

5 model structures on chain complexes of abelian groups

5.1 The andard projeive model ruure

The situation for the andard projeive model ruure on chain complexes

of modules over a ring is aually a bit worse than for the projeive model

ruure on the category of modules or on chain complexes over a ﬁeld.

Indeed, it is not true for example that for any ring, a chain complex of free

modules is going to have the left lifting property again all epimorphisms.

Here is a famous counter example: the ground ring is

and the following

chain complex is only made of free modules,

. . . Z

. . .

×2 ×2 ×2 ×2

It is an exa complex but as it does not split, it is not homotopic to zero! Hence

it can’t be a coﬁbrant obje and as a consequence, ﬁnding free resolutions is

no longer enough to build coﬁbrant replacement... On a PID, the situation is

aually much better.

Proposition. — Any exa chain complex of free Abelian groups is homotopic to

zero.

Proof. — Let

be an exa chain complex of free Abelian groups. For every

n ∈ Z

⊂ C

is a subgroup of a free Abelian groups, hence it is also free.

Moreover, since

is exa, the quotient

is isomorphic to the image

n−1

which is free (as a subgroup of a free Abelian group) so that

can

aually be split

≃ B

⊕ B

n−1

And split exa chain complexes are homotopic to zero.

Thanks to this proposition we can now deﬁne the andard model ruure

on the category of chain complexes of Abelian groups.

Theorem. — There exis a model ruure on the category

Ch(Z)

of chain

complexes of Abelian groups where:

— the weak equivalences are the quasi-isomorphisms;

— the ﬁbrations are the epimorphisms;

—

the coﬁbrations are the split monomorphisms with cokernel made of free

Abelian groups.

This model ruure is called the andard projeive model ruure.

Remark. — In the case of the category of chain complexes over a non PID ring,

the andard model ruure ill exisits but the coﬁbrant objes are a bit

diﬀerent: there are the complexes

made of projeive modules such that

every map

P −→ K

with

quasi-isomorphic to zero, is homotopic to the zero map. We shall call

such a chain complex dg-projeive.

The proof of the exience of the projeive model ruure is essentially

the same as the one on chain complexes over a ﬁeld; the previous proposition

guaranties that chain complexes of free Abelian group are coﬁbrant.

In order to prove the two faorisations some extra work is needed and in

particular, the following propoion mu be used.

Proposition. — Given a chain complex of Abelian groups

there exis a chain

complex of free Abelian groups L and a trivial ﬁbration

L C

∼

Proof. — We are going to give the reciepe to build a dg-projeive resolution

(the fa that the ring is a PID is not relevant here). Fir truncate on the right

the chain complex and then build induively a projeive resolution. This is

always possible since there are enough projeives. We then have

≥0

∼

Then funorially complete the projeive resolution into a projeive res-

olution

≥n

→ C

≥n

for

n ≤ 0

. I.e for every negative

we want to have the

following diagram

≥n

≥n−1

≥n

≥n−1

∼

Then take the ﬁltered colimit indexed by

to get a dg-projeive resolu-

tion. This happens because both weak equivalences and ﬁbrations are able

under ﬁltered colimits. As such, any ﬁltered colimit of dg-projeive chain

complexes is again dg-projeive and every ﬁltered colimit of resolutions is a

resolution. Laly, every right bounded chain complex of projeive modules is

dg-projeive

5.2 The andard injeive model ruure

For injeive modules, the same problems appear as with the projeive mod-

ules. Namely, some chain complexes of injeive modules may not be ﬁbrant

in the andard model ruure. As before, the situation on a PID is much

better.

Proposition. — Any exa chain complex of divisible Abelian groups is homotopic

to zero.

Proof. — The proof boils down to the following fa: every quotient of a

divisible group is again divisible. This means that given any complex of

divisible groups

, for every

the boundaries subgroup

⊂ C

is divisible

and then

may be split into

≃ B

⊕ B

n−1

were exa. This means

that C is homotopic to zero.

Theorem. — There exis a model ruure on the category of chain complexes of

Abelian groups where:

— the weak equivalences are the quasi-isomorphisms;

— the coﬁbrations are the monomorphisms;

—

the ﬁbrations are the epimorphisms with kernel made of divisible Abelian

groups.

This model ruure is called the andard injeive model ruure.

In the same way that we reﬁne the deﬁnition of coﬁbrant objes, we may

deﬁne a dg-injeive chain complex as a complex

made of injeive modules

such that for every map

K −→ I

with K quasi-isomorphic to zero, is homotopy equivalent to zero.

But contrarily to the projeive case, we need to use one more assumption

to be able to build a dg-injeive resolution for every chain complex: we need

to use the fa that ﬁltered colimits in the Abelian category we are working

in are exa i.e that they preserve the monomorphisms. Apart from that the

proves are totally similar.

references

[1] Gabriel P. and Zisman M., Calculus of fraions and homotopy theory,

vol. 35. Springer Berlin Heidelberg, 1967.

[2] Fuchs L., Inﬁnite Abelian groups. Vol. I. Academic Press, 1970.

[3] Fuchs L., Inﬁnite Abelian groups. Vol. II. Academic Press, 1973.