Week 1
Main examples The first and most natural examples of large categories typically contain
sets with additional structures as objects and morphisms that are compatible
with the given structures as morphisms. So for instance we have
Set, Grp, Ab, Met, Ring, Top etc...
Examples Not all categories have this look though. As a mild counterexample, we
look at two large categories that have sets as objects but whose morphisms
are not functions between them, although somewhat related.
Rel is thus constituted: the objects are sets; the morphisms are defined
by Rel(A, B) = P(A × B), so they are precisely binary relations from A
to B. The identity on A is simply ∆A ⊆ A×A and composition is defined
as follows: for R ⊆ A × B and S ⊆ B × C, we have
S ◦ R = {(a, c) ∈ A × C|∃b ∈ B such that (a, b) ∈ R and (b, c) ∈ S}.
It’s a simple check that this is in fact a category.
Part is thus constituted: the objects are again sets; a morphism from
A to B is a pair (A0, f) where A0 ⊆ A and f : A0 → B (think of a
partially defined function A → B. The identity on A is (A, idA) and the
composition is
(B0, g) ◦ (A0, f) = (f−1
(B0), g ◦ f|f−1(B0)).
Exercise Show that i : N → Z is monic and epic in Mon but it is not an isomor-
phism.
Solution It is obviously monic, and obviously not surjective so not an isomorphism.
In order to prove that it is epic, consider two monoid homomorphisms
f, g : Z → M such that fi = gi. Since the i(n) = n for n ≥ 0, in order to
show that f = g it only remains to prove that f(−n) = g(−n). For this,
compute
f(−n) = −f(n) = −fi(n) = −gi(n) = −g(n) = g(−n)
where we use the condition on f and g and the fact that they are homo-
morphisms.
Exercise What is initial in Rel? What is terminal?
Solution The initial object is ∅. Indeed, for any set B we have ∅ × B = ∅ so it has
a unique subset, that is, ∅ itself. For the same reason, ∅ is also terminal.
Definition A zero object is an object that is both initial and terminal. A category is
called pointed if it has a zero object.
Exercise Is Set pointed? What about Top, Cat, Grp, Ring.
1
Solution Set is not pointed, becase it has an initial ∅ and a terminal ∗ and they
are not isomorphic. Similarly for Top and Cat, where the initial and
terminal objects have the same underlying sets. Grp is pointed, because
0 is both initial and terminal. In Ring, 0 is terminal but not initial: there
is a homomorphism 0 → R if and only if R = 0. On the other hand, Z is
initial, because 0 must be sent to 0, 1 to 1, and the rest is determined by
the homomorphism property.
Definition A pointed set is a pair (X, x) where x ∈ X. The category Set∗ of pointed
sets has pointed sets as objects, and a morphism (X, x) → (Y, y) is a
function f : X → Y such that f(x) = y.
Similarly, we define the category Top∗ of pointed spaces and Cat∗ of
pointed small categories (not pointed as categories, but rather having a
selected object that must be preserved by functors).
Definition Given a category C and an object C therein, the category C ↓ C has as
objects pairs (A, f) where A ∈ C and f : C → A. A morphism from (A, f)
to (B, g) is a morphism h : A → B such that hf = g. Dually we define
the category C ↓ C with objects being of the form A → C.
Remark Oftentimes, when we have a terminal object ∗ in a non-pointed category
C, we can define the category of pointed objects as the category ∗ ↓ C.
This is the case, for instance, of Set, Top, Cat etc...
A function between sets ∗ → X precisely selects an element of X, and a
function that preserves this selection of points is precisely one that makes
the diagram
∗
X Y
commutative.
ˆ Can we define something like pointed objects also in pointed categories?
This depends on the specific properties of the categories.
Example We know that for a group G there is an isomorphism Grp(Z, G) ∼= G, so
the datum of a morphism Z → G is precisely an arbitrary choice of an
element g ∈ G, and a homomorphism preserving these choices is precisely
a homomorphism G → H that makes the diagram
Z
G H
commutative. In other words, the category Z ↓ Grp is something like the
category of pointed groups.
Similarly, we have an isomorphism Ring(Z[x], R) ∼= R, because Z is initial
and x can be sent anywhere, thus determining a full homomorphism
Z[x] → R. So the category Z[x] ↓ Ring is something like the category of
pointed rings.
2
Products and coproducts Let us examine products and coproducts in Grp. For two groups G and
H, the product is G × H with the pointwise group structure and the
natural projections p1 : G × H → G and p2 : G × H → H. Given
another group A with homomorphisms f : A → G and f′
: A → H, define
(f, f′
) : A → G×H as a → (f(a), f′
(a)). This is clearly a homomorphism,
and clearly p1 ◦ (f, f′
) = f and p2 ◦ (f, f′
) = f′
. Moreover, it is unique,
in that if we have another morphism q : A → G × H such that p1q = f
and p2q = f′
, then compute the element q(a) for a ∈ A. It must be of the
form (g, h) ∈ G × H. Now we have
g = p1(g, h) = p1q(a) = f(a)
and similarly h = f′
(a).
The coproduct is the so called free product G ∗ H. Its elements are equivalence
classes of words (s1, . . . , sn) where for each i = 1, . . . , n we have
either si ∈ G or si ∈ H. The equivalence relation is that generated by
(. . . , si, si+1, . . .) ∼ (. . . , sisi+1, . . .)
whenever si and si+1 belong to the same group. The product operation is
given by concatenation. The inclusions i1 : G → G∗H and i2 : H → G∗H
are given by i1(g) = [g] and i2(h) = [h].
Now, given two homomorphisms f : G → A and f′
: H → A, we define
{f, f′
} : G ∗ H → A as [s1, . . . , sn] → f1(s1) · · · fn(sn) where fi = f if
si ∈ G and fi = f′
if si ∈ H. This is well-defined by the very definition of
the equivalence relation and the fact that f and f′
are homomorphisms.
It is also clearly a homomorphism, and we have {f, f′
} ◦ i1 = f and
{f, f′
} ◦ i2 = f′
. Moreover, this is unique, because given another q :
G ∗ H → A such that qi1 = f and qi2 = f′
, this condition says that
q([g]) = f(g) and q([h]) = f′
(h), so the homomorphism property implies
q([si, . . . , sn]) = q([s1] · · · [sn]) = q([s1]) · · · q([sn]) = f1(s1) · · · fn(sn).
Week 2
Review In predicate logic, a language is defined to be a set of symbols for variables,
a set of symbols for relations, a set of symbols for functions, and
the fixed set {≈, ∧, ¬, ∀}.
Starting with these, one can expand the dictionary with shortcuts, such
as p ∨ q := ¬(¬p ∧ ¬q), or p → q := ¬p ∨ q, or yet ∃xp(x) := ¬∀x¬p(x).
On the other hand, in equational logic we only have symbols for variables,
for functions and the set {≈, ∧, ∀}. The absence of ¬ in particular implies
that we don’t have ∨, →, ∃ among others.
Example Some objects can be axiomatizable in a certain context using the full
power of predicate logic but also admit an equivalent axiomatization in
3
equational logic.
A lattice can be defined to be a poset with all finite joins and meets. In
this case, the axioms saying that the relation is an order relation use the
→ connector, moreover the definition of joins and meets in a poset also
requires the use of →, because it says that for every element that is bigger
than something, then it is also bigger than the join, etc...
A lattice can also be formulated as having two binary operations ∨ and ∧
subject to the axioms:
Commutativity ∀x∀yx ∨ y ≈ y ∨ x;
Associativity ∀x∀y∀z(x ∨ y) ∨ z ≈ x ∨ (y ∨ z);
Idempotency ∀xx ∨ x ≈ x;
Absorption ∀x∀yx ≈ x ∨ (x ∧ y)
and their duals obtained inverting the symbols ∨ and ∧.
It is clear that a lattice in the former definition gives rise to one in the
latter. For the opposite direction, observe that you can define the binary
relation ≤ as x ≤ y := x ≈ x∧y and then verify that this is a poset having
∨ and ∧ as joins and meets respectively.
Exercise For an algebra A and a subset X ⊆ A, define ⟨X⟩ to be the smallest
subalgebra of A that contains X or, alternatively, the intersection of all
the subalgebras of A that contain X. How can this be written explicitly?
Solution Define the operator
EX := X ∪ {f(x1, . . . , xn)|x1, . . . , xn ∈ X, f is an n-ary operation of A}.
Then we have ⟨X⟩ = n∈N En
X. Indeed, if X ⊆ A0 and A0 is a subalgebra
of A, then EX ⊆ A0 because it is closed under all operations,
and by iterating this we obtain ⟨X⟩ ⊆ A0. Moreover, ⟨X⟩ is a subalgebra
of A because if x1, . . . , xn ∈ ⟨X⟩, then there is an n such that they are
in En
X. If f is an n-ary function symbol, then by definition we have
f(x1, . . . , xn) ∈ En+1
X ⊆ ⟨X⟩.
Normal subgroups In groups, there is a bijective correspondence between congruences and
normal subgroups.
Given a congruence ∼, define N = [1]. This a subgroup because if a ∼
1 and b ∼ 1 then ab ∼ 1 because ∼ is compatible with the product.
Moreover, if a ∼ 1 then a−1
∼ 1−1
= 1. Moreover, this subgroup is
normal because if a ∈ N then for h ∈ G we have hah−1
∼ hh−1
= 1,
which follows by a ∼ 1 and ∼ being compatible with the product.
Viceversa, let N be a normal subgroup, and define a ∼ b if and only if
ab−1
∈ N. Now we have
Reflexivity aa−1
= 1 ∈ N because N is a subgroup;
Symmetricity ab−1
∈ N then ba−1
= (ab−1
)−1
∈ N similarly;
4
Transitivity ab−1
∈ N and bc−1
∈ N then ac−1
= ab−1
bc−1
∈ N similarly;
Compatibility ac−1
∈ N and bd−1
∈ N then by normality abd−1
a−1
∈ N so
abd−1
a−1
ac−1
∈ N but this is abd−1
c−1
= (ab)(cd)−1
.
Ideals In rings, we have a similar correspondence between congruences and ideals.
Given a congruence ∼ we can define I = [0]. It is closed under sums by
compatibility, and moreover if a ∈ I, that is, a ∼ 0, for any b ∈ R we have
ab ∼ 0b = 0, so ab ∈ I and I is an ideal.
Conversely, given an ideal I, we define a ∼ b if and only if a − b ∈ I.
As above, this is an equivalence relation. We need to prove that it is
compatible with both sum and product.
Sum If a − c ∈ I and b − d ∈ I then a + b − c − d ∈ I;
Product Multiplying a − c on the right and knowing that I is an ideal, we
obtain ab − cb ∈ I, similarly we obtain cb − cd ∈ I. Adding these two
elements, we have ab − cb + cb − cd = ab − cd ∈ I.
Kernels For groups, rings and a number of other commonly studied algebraic structure,
whenever we have a homomorphism f : A → B, the kernel kerf ⊆ A
corresponds to a congruence. Following the above and calling this congruence
also kerf, we have
a ∼ b ⇔ ab−1
∈ kerf ⇔ f(ab−1
) = 1 ⇔ f(a)f(b)−1
= 1 ⇔ f(a) = f(b).
This allows to generalize the concept of kernel to any type of algebras.
Definition kerf = {(a, b) ∈ A2
|f(a) = f(b)}.
Theorem Just as in the known cases of groups and rings, we have for general algebras
the isomorphism theorem: given a homomorphism f : A → B, this induces
a congruence on A, therefore a quotient π : A → A/kerf so that f factors
uniquely as
A B
A/kerf.
π
f
t
Of course, t is defined as [a] → f(a).
Definition An element of a lattice x ∈ X is called compact if for every subset A ⊆ X
such that x ≤ A there is a finite subset A0 ⊆ A such that x ≤ A0.
Definition A lattice is called compactly generated if every element is a supremum of
compact elements.
Definition A lattice is called algebraic if it is complete and compactly generated.
5
Lattices of algebras Given an algebra A, we can take the set of its subalgebras S → A. These
are ordered by S ≤ S′
if and only if S can be embedded into S′
as subalgebras
of A. We have formed a poset SubA.
Similarly, we can take the set of all congruences on A (or quotients A ↠
A/ ∼ and order it by ∼≤∼′
if and only if a ∼ b implies a ∼′
b, in other
words when there is a factorization
A A/ ∼′
A/ ∼ .
This also forms a poset, that we are going to call ConA.
Theorem Given an algebra A, then both posets SubA and ConA are algebraic lattices.
Moreover, given an algebraic lattice L there is a set of operations Ω
and an Ω-algebra A such that L ∼= SubA.
Week 3
We want to prove the part about subalgebras in the last stated theorem.
Definition A closure operator on a set A is a function C : P(A) → P(A) such that
C1 ∀X ⊆ A, X ⊆ C(X);
C2 ∀X ⊆ A, C2
(X) = C(X);
C3 ∀X, Y ⊆ A, if X ⊆ Y then C(X) ⊆ C(Y ).
Definition Given a closure operator C, a closed subset is a subset X such that C(X) =
X. We define LC to be the poset of closed subsets under inclusion.
Proposition LC is a complete lattice.
Proof It suffice to observe that, for a family C(Xi)i∈I of closed subsets, we can
define
i∈I C(Xi) = C( i∈I Xi) and i∈I C(Xi) = i∈I C(Xi)
and observe that they are both closed subsets and play the roles of join
and meet respectively.
Proposition Given L a complete lattice, then there is a closure operator C such that
L ∼= LC.
Proof For X ⊆ L, define
C(X) := {a ∈ L|a ≤ X}.
This is a closure operator, and its closed subsets are precisely those containing
their join. So we have a bijection L ↔ LC given by b → {a ∈
L|a ≤ b} to the right, and by X → X to the left.
6
Definition A closure operator C is algebraic if
C4 ∀X ⊆ A, C(X) = {C(Y ), Y ⊆ X finite }.
Proposition Given an algebraic closure operator, then LC is an algebraic lattice.
Proof We only need to show that LC is compactly generated. Given C4, this
follows by proving that compact elements are precisely those of the form
C(X) for a finite set X, so we are reduced to proving that.
Suppose X = {a1, . . . , ak}, and suppose that
C(X) ⊆ i∈I C(Ai) = C( i∈I Ai).
For aj ∈ X, by C1 and C4 there is a finite Xj ⊆ i∈I Ai such that
aj ∈ C(Xj). By finiteness, Xj ⊆ Aj1 ∪ · · · ∪ Ajnj
, so that aj ∈ C(Aj1 ∪
· · · ∪ Ajnj ).
Therefore, we have X ⊆ 1≤j≤k C( 1≤i≤nj
Aji) ⊆ C( 1≤j≤k
1≤i≤nj
Aji. Applying
C and using C3 and C2, we obtain C(X) ⊆ C( Aji) so C(X)
factors through a finite join and it is compact.
Conversely, assume that C(X) is compact. By C4, we have C(X) ⊆
{C(Yi), Yi ⊆ X finite }. Using compactness, there is a finite set of indices
I0 such that C(X) ⊆ i∈I0
C(Yi) ⊆ C( i∈I0
Yi). Moreover, i∈I0
⊆ X,
therefore applying C3 we obtain C( i∈I0
⊆ C(X). In conclusion, we just
proved that C(X) = C( Yi) and the union on the right-hand side is a
finite union of finite sets, so it is finite and we are finished.
Proposition If L is an algebraic lattice, then L ∼= LC for some algebraic closure operator
C.
Proof Define A to be the set of all compact elements of L, and for every X ⊆ A
C(X) := {a ∈ A|a ≤ X}.
This is a closure operator. Let us show that it is algebraic. If a ∈ C(X),
then a ≤ X, and since a is compact we must have a ≤ Y for some
finite Y ⊆ X. By definition, this means that a ∈ C(Y ), which proves C4.
An isomorphism L ∼= LC is given by a → {b ≤ a} to the right, and
X → X to the left. They are mutually inverse, using that L is compactly
generated.
Proposition Given an algebra A, then SubA is an algebraic lattice.
Proof The operator ⟨−⟩ on subsets of A is clearly a closure operator. Since
L⟨−⟩ = SubA, we only need to show that ⟨−⟩ is algebraic. If E is the
operator that closes only once under all the operations of A, then we
have, for every X ⊆ A,
⟨X⟩ = n∈N En
X.
7
If a ∈ ⟨X⟩, then for some n we have a ∈ En
X. This means that it is of
the form a = f(a1, . . . , ak) for some k-ary operation f, and a1, . . . , ak ∈
En−1
X. The point is that the ai’s are in finite amount, and we can
reproduce the reasoning for each one of them, obtaining a finite set of
elements of En−2
X. Iterating this procedure until we reach E0
X = X,
we have obtained a finite set Y ⊆ X, such that a ∈ En
Y , so a ∈ ⟨Y ⟩.
This proves C4.
Proposition If L is an algebraic lattice, then L ∼= SubA for some algebra A.
Proof We know that L ∼= LC for some algebraic closure operator C on a set A.
We will define an algebra structure on A in such a way that SubA ∼= LC.
For each finite B ⊆ A and each b ∈ C(B), calling n := |B| we define an
n-ary operation
fB,b(a1, . . . , an)
b if B = {a1, . . . , an}
a1 otherwise.
Analyzing all possible values of such an fB,b, this implies that, for a subset
A ⊆ A and elements a1, . . . , an ∈ X then
fB,b(a1, . . . , an) ∈ C({a1, . . . , an}) ⊆ C(X).
By definition of EX, this means precisely that EX ⊆ C(X). By iterating
this, we see that for each n ∈ N we have En
X ⊆ C(X) so that ⟨X⟩ ⊆
C(X).
On the other hand, for a finite B ⊆ X given by B = {a1, . . . , an we have
C(B) = {fB,b(a1, . . . , an|b ∈ C(B)} ⊆ EB ⊆ ⟨B⟩ ⊆ ⟨X⟩. Using C4, this
implies C(X) ⊆ ⟨X⟩.
In conclusion, we have proven that ⟨−⟩ = C, so LC = L⟨−⟩, but the closed
subsets under ⟨−⟩ are precisely subalgebras of A, so this is precisely SubA.
HSP theorem Given a class of objects that are structures for some signature, then this is
precisely the class of all algebras for some equational theory if and only if it
is closed under subalgebras, homomorphic images and arbitrary products.
Remark The first part of the statement means that when we look at such a class of
objects, we must already know that they are sets with certain operations
defined on them. The question is whether these operations satisfy the
axioms of some equational theory.
Exercise Are lattices algebras for an equational theory? What about complete
lattices? Integral domains? Fields? Vector spaces?
Solution Using the HSP theorem, we see that lattices are algebras for an equational
theory. Complete lattices are not, in that they are not closed under subalgebras,
think for example of N ⊂ N∪{∞}. This reflects the fact that we
need implication symbols in order to define infinite joins and meets, but
the theorem tells us much more: it says that there is no way of axiomatizing
complete lattice that happens to be equational. Integral domains
also are not, because they’re not closed under products. Indeed, if R and
8
S are integral domains, take (0, 1), (1, 0) ∈ R × S, so (0, 1)(1, 0) = (0, 0).
Therefore, fields are likewise not algebras for an equational theory. Vector
spaces on a fixed base field are, which reflects the fact that, although it is
impossible to equationally define multiplicative inversion for fields, we can
circumvent this issue in defining vector spaces by defining a unary operation
for each element of the field, and then subjecting those to equational
axioms.
Week 4
Our next purpose will be to create a connection between the language of universal
algebra and category theory. We will first explore the relation between
two ubiquitous category-theoretic concepts, adjunctions and monads.
Definition An adjunction between two categories is a pair of functors L : C ⇆ D : R
such that, for every pair of objects C ∈ C and D ∈ D we have bijections
D(LC, D) ∼= C(C, RD)
that are natural in the sense that, for two morphisms f : C′
→ C and
g : D → D′
in the respective categories, the square
D(LC, D) C(C, RD)
D(LC′
, D′
) C(C′
, RD′
)
D(Lf,g)
∼=
C(f,Rg)
∼=
is commutative. We also denote adjunctions by L ⊣ R.
Definition In the situation above, consider the bijection D(LC, LC) ∼= C(C, RLC).
There is a unique morphism ηC : C → RLC corresponding to idLC via
this bijection. By the naturality conditions on an adjunction, all the morphisms
ηC aggregate into a natural transformation idC → RL, which is
called the unit of the adjunction.
Analogously we obtain morphisms ε : LRD → D which aggregate into a
natural transformation ε : LR → idD, known as the counit of the adjunc-
tion.
Example The functors F : Set ⇆ Grp : U are one of the archetypical examples
of adjunction. Indeed, by the universal property of adjunctions says that
whenever we have a set X and a group G with a function f : X → UG,
then there is a unique group homomorphism k : FX → G such that the
triangle
X UG
UFX
f
Uk
9
commutes.
For a set X, the unit X → UFX is the natural inclusion of elements of X
as words of length 1 in the free group. For a set G, the counit FUG → G
takes a finite word of elements of G and takes it to the product of all of
them.
Definition A monad on a category C is a functor T : C → C, with two natural
transformations µ : T2
→ T and η : id → T, called the multiplication and
unit of the monad respectively, such that the diagrams
T3
T2
T2
T
T µ
µT µ
µ
T T2
T
T
T η
idT
µ
ηT
idT
are commutative. The first is known as the associativity condition and
the second as the unitality conditions.
Definition Given a monad (T, µ, η), a T-algebra is an object C ∈ C with a morphism
h : TC → C satisfying the condition that the diagrams
T2
C TC
TC C
T h
µC h
h
C TC
C
ηC
idC
h
are commutative. This says that the “algebraic” structure that we put on
C is compatible with the requirements dictated by the monad T.
A T-homomorphism between T-algebras (C, h) and (D, k) is a morphism
f : C → D such that the square
TC TD
C D
h
T f
k
f
is commutative. In other words, it is compatible with the “algebraic”
structures of C and D.
Example Given a monad T, there is a category Alg(T) whose objects are T-algebras
and whose morphisms are T-homomorphisms. Clearly, there is a functor
UT
: Alg(T) → C which simply forgets the additional structure. Moreover,
there is a functor FT
: C → Alg(T) which takes an object C to the “free Talgebra
on C”, that is, the pair (TC, µC). This is a T-algebra by definition
of monad. It is called free because, similarly to other definitions of freeness,
it satisfies no conditions other than those declared by the monad itself.
There is an adjunction FT
⊣ UT
(this is left as an exercise).
10
Example Given an adjunction L : C ⇆ D : R, we can construct a monad on C
as follows: set T := RL. As a multiplication µ : T2
→ T, we take
RεL : RLRL → RL, and as a unit η : id → T we simply take the unit of
the adjunction η : id → RL. The conditions of a monad are satisfied by
the naturality of an adjunction. We also say that T is the monad induced
by the adjunction L ⊣ R.
Definition A fork is a diagram a b c
f0
f1
e
in a category. It is a coequalizer
if it is a colimit diagram, i.e. ef0 = ef1 and for every other map t : b → d
such that tf0 = tf1, then there exists a unique q : c → d such that qe = t.
Intuitively, the morphism e : b → c is the minimal thing that equalizes f0
and f1 by postcomposition.
Definition An absolute coequalizer is a fork whose image along any functor is a
coequalizer. In particular, it is obviously a coequalizer itself after pulling
it through the identity functor.
Definition A split fork is a fork as above such that there exist maps a b ct s
satisfying ef0 = ef1, es = idc, f0t = idb and f1t = se.
Example Every split fork is in particular an absolute coequalizer (left as an exercise).
Therefore, we will in the following also say split coequalizer for split fork.
Example If T is a monad and (x, h) is a T-algebra, then the diagram T2
x Tx x
µx
T h
h
is a split fork. Indeed, if we consider the arrows T2
x Tx x
µT x ηx
,
we have h ◦ µx = h ◦ Th and h ◦ ηx = idx by definition of T-algebra,
µx ◦ ηT x = idT x by definition of monad, and Th ◦ ηT x = ηx ◦ h by naturality
of η. As a consequence, for every T-algebra there is a split coequalizer
diagram as above.
ˆ We only need one last ingredient before embarking into our journey which
will eventually relate adjunctions and monads in a precise way. This
relation will then be applied to constructions from universal algebra.
Definition A functor F : C → D creates coequalizers if, whenever we have a diagram
a ⇒ b in C such that Fa ⇒ Fb admits a coequalizer e : Fb → d, then there
exists a unique k : b → c such that Fk = e and, moreover, the diagram
a ⇒ b → c is itself a coequalizer.
Lemma Assume we are given a solid diagram
D′
D
C
K
R′
RL′
L
where the arrows form two adjunctions inducing the same monad T. Moreover,
assume that R creates coequalizers for all pairs f0, f1 such that
Rf0, Rf1 have a split coequalizer.
Then, there exists a unique comparison functor K : D′
→ D such that
KL′
= L and RK = R′
.
11
Proof If K exists, observe that we have LRK = KL′
R′
and Kε′
= εK. We
need to prove that K is unique, and in the follwing we will freely use the
equalities just mentioned. For an object d′
∈ D′
, consider the fork
L′
R′
L′
R′
d′
L′
R′
d′
d′
.
ε′
L′R′d′
L′
R′
ε′
d′
ε′
d′
Applying K to shis fork, we obtain
LRLR′
d′
LR′
d′
Kd′
.
εLR′d′
LR′
ε′
d′
Kε′
d′
Applying R to this second fork, we obtain
RLRLR′
d′
RLR′
d′
R′
d′
RεLR′d′
RLR′
ε′
d′
R′
ε′
d′
which is a particular case of the split fork in the example above, applied
to the free T-algebra on R′
d′
. Since R creates split coequalizers, then
there is a unique k : LR′
d′
→ d such that Rk = R′
εd′ . But we already
know by the above discussion that Kε′
d′ : LR′
d′
→ Kd′
has this property,
so it must be Kε′
d′ = k. In particular, we have Kd = k, so that K is
uniquely determined on objects. Moreover, we know that k is a coequalizer
(again by the assumption on the creation of coequalizers). For a morphism
f : d′
→ d′′
, we have an induced diagram
LRLR′
d′
LR′
d′
Kd′
LRLR′
d′
LR′
d′′
Kd′′
εLR′d′
LR′
ε′
d′
LRLR′
f
Kε′
d′
LR′
f Kf
εLR′d′′
LR′
ε′
d′′
Kε′
d′′
where the horizontal arrows constitutes forks as the second above. Since
as we have shown they are coequalizers, there is a unique factorization
q : Kd′
→ Kd′′
which makes the right square commute, but we know
that Kf already does the job, so Kf = q. Therefore K is also uniquely
determined on morphisms, and we are done.
Now we need to show that such a K exists. We already know that, if it
does, it must have the form above, so let us define it as a coequalizer in
the first line of the diagram
LR′
L′
R′
d′
LR′
d′
Kd′
LR′
L′
R′
d′′
LR′
d′′
Kd′′
.
LR′
L′
R′
f LR′
f
12
If we have a morphism f : d′
→ d′′
, this induces the diagram above which,
by universal property of the coequalizer Kd′
, induces a unique morphism
Kd′
→ Kd′′
which makes the right square commute. This will be the
value of Kf, so we have defined a functor K : D′
→ D.
Now there is a fork of the form
RLR′
L′
R′
d′
⇒ RLR′
d′
→ R′
d′
in C, and we can see that it is a split fork. Since R creates coequalizers
for split forks, it must be RKd′
= R′
d′
, because we already know that
LR′
L′
R′
d′
⇒ LR′
d′
→ Kd′
is a coequalizer. An analogous argument shows that for a morphism f we
have RKf = R′
f, so that we have RK = R′
. Moreover, if we perform
the construction above on an object of the form L′
c, we are taking the
coequalizer of a diagram which is of the form L(R′
L′
R′
L′
c ⇒ R′
L′
c), so
we have a coequalizer
LR′
L′
R′
L′
c ⇒ LR′
L′
c → KL′
c.
Moreover, we have a fork
LR′
L′
R′
L′
c ⇒ LR′
L′
c → Lc.
Applying R to both these forks, we obtain the same diagram
RLR′
L′
R′
L′
c ⇒ RLR′
L′
c → R′
L′
c
and since RL = R′
L′
(because they induce the same monad), this is a
special case of the split fork in the example above, applied to the free
algebra on c. Since R creates split coequalizers, the two forks above must
be the same, so KL′
c = Lc. An analogous argument will show that this
also holds on morphisms, so KL′
= L, as we desired.
Week 5
Last week, we introduced adjunctions and monads, and we showed a way to
associate certain adjunctions to monads, and monads to adjunctions. Namely,
given a monad T on C, we have an adjunction
FT
: C ⇆ Alg(T) : UT
and given an adjunction L ⊣ R, we can generate a monad (RL, RεL, η). In
general, there are many adjunctions inducing the same monad. In particular,
the adjunction FT
⊣ UT
is one of these. We recall the proven Lemma.
Lemma 1 We are given a solid diagram
13
D′
D
C
K
R′
RL′
L
where the diagonal arrows constitute two adjunctions inducing the same
monad. Assume that R creates coequalizers for all pairs f0, f1 such that
Rf0, Rf1 have a split coequalizers. Then there exists a unique comparison
functor K : D′
→ D such that KL′
= L and RK = R′
.
Lemma 2 Given a monad T, then the forgetful functor UT
: Alg(T) → C creates
coequalizers for all pairs f0, f1 such that UT
f0, UT
f1 have an absolute
coequalizer.
Proof Take a pair (x, h) (y, k)
f0
f1
of T-homomorphisms, and assume that
there is an absolute coequalizer x y z
f0
f1
e
in C. We need to
show that there is a unique algebra structure on z such that e is a Thomomorphism.
Moreover, we need to show that the resulting diagram is
also a coequalizer in Alg(T). Let us look at the solid diagram
Tx Ty, Tz
x y z.
T f0
T f1
h
T e
k m
f0
f1
e
Since our coequalizer is absolute, the first line of horizontal arrows is a
coequalizer diagram. Therefore, there is a unique m : Tz → z such that
the right square commutes. In particular, if m is a T-algebra structure
the same square automatically proves that e is a T-homomorphism. To
see that m is a T-algebra structure, we need to show that the outer square
in the diagram
T2
z Tz
T2
y Ty
Ty y
Tz z
T m
µz m
T k
µy
T 2
e
k
T e
k
T e e
m
commutes. We already know that all the other squares are commutative,
some by naturality of µ, some by the diagram above, some because y is a
T-algebra. Therefore, we know that
m ◦ Tm ◦ T2
e = m ◦ µz ◦ T2
e.
14
Since our first coequalizer was absolute, T2
e is a coequalizer, and since
both m ◦ Tm and m ◦ µz are factorizations of e ◦ k ◦ Tk(= e ◦ k ◦ µy), the
universal property of coequalizers says that they are equal. Similarly, we
prove that m ◦ ηz = idz, so (z, m) is a T-algebra.
It remains to show that x ⇒ y → z is also a coequalizer of T-algebras, so
consider a diagram in Alg(T)
(x, h) (y, k) (z, m)
(w, n)
f0
f1
e
g
where gf0 = gf1. Since the horizontal arrows constitute a coequalizer in
C, there is a unique factorization q : z → w with qe = g. We only need to
show that q is a T-homomorphism, that is, that the outer square in the
diagram
Tz Tw
Ty
y
z w
T q
m nk
T e T g
e g
q
commutes. All other squares and triangles there commute, so n◦Tq◦Te =
q◦m◦Te. Again since our first coequalizer was absolute, we know that Te
is a coequalizer, so reasoning as above we conclude that m ◦ Tq = q ◦ m,
which is what we wanted.
Comparison theorem We are given a diagram
D Alg(T)
C
R
K
UTL
F T
where the adjunction L ⊣ induces the monad T, there is a unique comparison
functor K : D → Alg(T) such that KL = FT
and UT
K = R.
Proof Since all split coequalizers are in particular absolute, Lemma 2 says that
the adjunction on the right satisfies the hypotheses of Lemma 1. Therefore,
we may use the latter and obtain the comparison functor.
Beck’s monadicity theorem Given a diagram
15
D Alg(T)
C
R
K
UTL
F T
where the adjunction L ⊣ R induces the monad T and K is a the comparison
functor, the following conditions are equivalent:
1. there is a comparison functor K which is an isomorphism of cate-
gories;
2. R creates coequalizers for all pairs f0, f1 such that Rf0, Rf1 have an
absolute coequalizer;
3. R creates coequalizers for all pairs f0, f1 such that Rf0, Rf1 have a
split coequalizer.
Proof If there is a comparison isomorphism K, we may reduce to the case where
the adjunction L ⊣ R is precisely FT
⊣ UT
, so the implication 1 ⇒ 2
is precisely Lemma 2. The implication 2 ⇒ 3 is obvious since all split
coequalizers are absolute.
For the implication 3 ⇒ 1, observe that by Lemma 1 we obtain both our
K : D → Alg(T) and a comparison functor M : Alg(T) → D. Moreover,
it is clear that the composite MK : D → D is a comparison functor as
well. Since idD is another comparison functor, by its uniqueness we know
that MK is the identity. Analogously, KM is the identity on Alg(T), so
K has an inverse and is therefore an isomorphism, which proves 1.
Corollary A variety of algebras is monadic over Set.
Proof Given a variety of algebras V with signature Ω, there is a forgetful functor
U : V → Set by definition, and in the course we have seen that free
algebras exist, so U has a left adjoint F : Set → V. By Beck’s theorem,
we only need to show that U creates coequalizers for pairs f0, f1 such that
Uf0, Uf1 have an absolute coequalizer. Let us take a pair of homomorphisms
(A, (αi)i∈Ω) (B, (βi)i∈Ω)
f0
f1
such that there is an absolute
coequalizer
A B C
f0
f1
e
in Set. Given i ∈ Ω of arity ni, let us consider the solid diagram
Ani
Bni
Cni
A B C
f
ni
0
f
ni
1
αi
eni
βi γi
f0
f1
e
16
Since our coequalizer was absolute, the first line is a coequalizer, therefore
there is a unique γi : Cni
→ C such that the right square commutes. This
is an ni-ary operation on C which is compatible with the corresponding
operation on B. Doing this for every i ∈ Ω, we have endowed C with
a unique structure of Ω-algebra such that e is an Ω-homomorphism. We
need to prove that the resulting fork is a coequalizer of Ω-algebras, so
consider a diagram of Ω-algebras
(A, (αi)i∈Ω) (B, (βi)i∈Ω) (C, (γi)i∈Ω)
(D, (δi)i∈Ω)
f0
f1
e
g
where gf0 = gf1. Since we have a coequalizer of sets, we get q : C → D
such that qe = g. We need to show that q is a Ω-homomorphism. For
this, take i ∈ Ω and consider the diagram
Cni
Dni
Bni
B
C D.
qni
γi δi
eni gni
βi
e g
q
We know that everything commutes except perhaps for the outer square,
so δi ◦ qni
◦ eni
= q ◦ γi ◦ eni
and, since eni
is a coequalizer, as in Lemma
2 we conclude that δi ◦ qni
= q ◦ γi. Since this can be done with every
i ∈ Ω, it means that q is an Ω-homomorphism.
Week 6
We would like to establish a converse of the last corollary. In other words, we
would like to have a way of detecting varieties of algebras that only makes use of
purely categorical means. We remind the previous result, whose proof is based
on the powerful Beck’s monadicity theorem.
Theorem Given a variety of algebras V, this is isomorphic to a category of the form
Alg(T), for some monad T on Set.
Remark In order to have a full converse to the above theorem, we need to make
some allowances on our definition of variety of algebras. Namely, we will
allow the signature Ω to be a proper class rather than just a set, and we
will allow our operations to be infinitary, that is, of the form Aλ
→ A, for
λ a not necessarily finite cardinal.
17
Lemma Given a monomorphism m : X → Y , and two morphisms x : TX → X, y :
TY → Y such that m ◦ x = y ◦ Tm, if (Y, y) is a T-algebra then (X, x) is
also a T-algebra.
Proof Consider the diagram
T2
X TX
T2
Y TY
TY Y
TX X.
µX
T x
T 2
m
x
T m
µX
T y y
y
x
T m
m
We already know that every square except maybe for the outer one commutes,
so m ◦ x ◦ µX = m ◦ x ◦ Tx. Since m is a monomorphism, this
implies x ◦ µX = x ◦ Tx, which is the associativity condition. Similarly we
prove the commutativity of the unitality triangles.
Theorem Given a canonical adjunction F : Set ⇆ Alg(T) : U, then Alg(T) is a
variety of algebras.
Proof Remember that a cardinal λ can be regarded as a set (namely, the set of
all cardinals < λ, so the monad T can be applied to cardinals. We define
a signature by
Ω := {(λ, i)|λ is a cardinal and i ∈ Tλ}
and we set the arity of (λ, i) to be λ. Given a T-algebra (X, x) and a
symbol (λ, i), we define an operation ωx
(λ,i) : Xλ
→ X as follows: remembering
that an element g ∈ Xλ
can be regarded as a function λ → X, our
definition is g → x ◦ Tg(i). Doing this for every element of Ω gives an
Ω-algebra E(X, x) = (X, (ωx
(λ,i))(λ,i)∈Ω).
Our first claim is that a morphism f : X → Y between two T-algebras is
a T-homomorphism if and only if it is an Ω-homomorphism. So assume
that it is a T-homomorphism, then for every element Xλ
we have
f ◦ ωx
(λ,i)(g) = f ◦ x ◦ Tg(i) = y ◦ Tf ◦ Tg(i) = y ◦ T(fg)(i) = ωy
(λ,i)(fg) =
ωy
(λ,i) ◦ fλ
(g)
which means that f ◦ ωx
(λ,i) = ωy
(λ,i) ◦ fλ
, and this for every (λ, i), so f is
an Ω-algebra.
Conversely, if f is an Ω-algebra, define λ = |X| and choose an isomorphism
g : λ → X. Now, for every element i ∈ Tλ, compute
f ◦ x ◦ Tg(i) = f ◦ ωx
(λ,i)(g) = ωy
(λ,i) ◦ fλ
(g) = ωy
(λ,i)(f ◦ g) =
y ◦ T(fg)(i) = y ◦ Tf ◦ Tg(i)
18
which means f ◦ x ◦ Tg = y ◦ Tf ◦ Tg. Since g is an isomorphism, then so
is Tg, so we obtain f ◦ x = y ◦ Tf, and f is a T-algebra.
What we have just proven is that the image E(Alg(T)) is isomorphic to a
full subcategory of Alg(Ω). By Birkhoff’s HPS theorem, we now only need
to verify that it is closed under products, subalgebras and surjections.
An Ω-algebra structure on a product is defined componentwise, and it
corresponds to the componentwise Ω-algebra structure defined by E, so
we have closure under products.
Now take a subalgebra embedding m : (X, (ω(λ,i))) → E(Y, y). Choose
an isomorphism g : λ → X and define a function f(λ,g) : Tλ → X to be
i → ω(λ,i)(g). We compute, for i ∈ Tλ,
m◦f(λ,g)(i) = m◦ω(λ,i)(g) = ωy
(λ,i) ◦mλ
(g) = ωy
(λ,i)(mg) = y ◦Tm◦Tg(i),
so that m ◦ f(λ,g) = y ◦ Tm ◦ Tg. Consider the solid square
Tλ TX
X Y.
f(λ,g)
T g
y◦T m
x
m
Since Tg is an isomorphism, there exists a diagonal arrow x : TX → X
making the two triangles commute. By the previous lemma, we know
that (X, x) is a T-algebra. Moreover, E(X, x) and (X, (ω(λ,i))) are both
obtained from E(Y, y) by restricting all the operations therein, so they
must coincide, and (X, (ω(λ,i))) is in the image of E.
Finally, let us consider a surjection q : E(X, x) ↠ (Y, (ω(λ,i))). By a direct
check, we know that this is the coequalizer of the fork P ⇒ E(X, x), where
P = {(x0, x1) ∈ X × X|q(x0) = q(x1)} ⊆ X × X, and the two arrows are
the projections on the two components. By the previous discussion, we
know that P is in the image of E, so there is a T-algebra (Z, z) such that
E(Z, z) = P. Applying UT
to this coequalizer, we obtain a diagram
P ⇒ X ↠ Y
in Set. Now by choosing a section of q, we obtain a map Y → X. Moreover,
we can consider the diagonal X → P. These two make the fork
under observation into a split fork, a since UT
creates coequalizers for
these, there is a unique T-algebra structure y on Y for which q is a Thomomorphism.
Now, E(Y, y) and (Y, (ω(λ,i))) are obtained from E(X, x)
via the same quotient, so they must coincide, and (Y, (ω(λ,i))) is in the
image of E. This concludes the proof.
Definition A cardinal λ is regular if, whenever (µi)i∈I is a family of cardinals such
that |I| < λ and ∀i ∈ I, µi < λ, then i∈I µi < λ.
Example ω is regular. Indeed, a finite sum of finite cardinals is finite.
Any infinite successor cardinal, i.e. of the form µ+
is regular. Indeed,
being < µ+
is equivalent to being ≤ µ. Therefore, if we have a family
(µi)i∈I with |I| ≤ µ and µi ≤ µ, then i∈I µi ≤ i∈I µ = µ · µ = µ.
19
Definition Given a regular cardinal λ, a poset is called λ directed if every λ-small
subposet has an upper bound.
Example The typical example of this is, given a set S, the poset Pλ
(S) = {X ⊆ S
such that |X| < λ}. It is a poset by inclusion, and λ-directed because,
by regularity of λ, a λ-small union of elements of Pλ
(S) is an element of
Pλ
(S).
Definition A functor T : Set → Set is called λ-accessible if it preserves colimits
indexed by λ-directed posets.
To illustrate this, for every set we have S = Pλ
(S). The condition for
T of being λ-accessible means that this union also holds after applying T,
that is, TS can be recovered as the union of the images of all sets TX, for
X ∈ Pλ
(S).
Theorem Given a canonical adjunction F : Set ⇆ Alg(T) : U, where T is λaccessible,
then Alg(T) is a variety of algebras for a small signature of
operations whose arities are < λ.
Proof We can retrace the proof of the previous theorem, defining the signature
Ω = {(µ, i)|µ < λ and i ∈ Tµ}.
The fact that we had arbitrarily large cardinals in our signature was only
used in two places in the previous proof, the rest can be replayed as it
is. Namely, we need to tweak our proof that an Ω-homomorphism is a
T-homomorphism, and that the image of E is closed under subobjects.
For the first, the set X can be written as X = {Yj ⊆ X with |Yj| < λ}.
Let us call gj the inclusion Yj → X. For every such subset Yj of cardinality
µ < λ, and every i ∈ Tµ, we compute, as in the original proof,
f ◦ x ◦ Tgj = y ◦ Tf ◦ Tgj.
Since TX is the union of all the images of Tgj, these are jointly surjective,
which lets us cancel them on the right, so we obtain f ◦x = y◦Tf anyway.
In the discussion of subobjects, we similarly obtain commutative squares
of the form
Tµ TX
X Y.
f(µ,gj )
T gj
x
y◦T m
m
Now the arrows Tgj present TX as a colimit of the diagram (TYj), so that
there exists a unique factorization x : TX → X. Now, again by universal
property of colimits, we have y ◦ Tm = m ◦ x, then we proceed as in the
original proof.
Corollary A category of the form Alg(T), where T is a finitely accessible monad, is
a variety of algebras for a small signature of finitary operations.
20
Proof Just take λ = ω in the previous theorem.
Example Let us show in with purely categorical methods that monoids can be obtained
by a set of finitary operations as an equational theory. By the
previous corollary, all we have to do is verify that the free monoids monad
is finitely accessible.
For a set X, the set TX contains finite sequences of elements of X. Given
such a word (a1, . . . , an), we obviously have
(a1, . . . , an) ∈ T({a1, . . . , an})
which is a finite subset of X. Since this is true for every word, we may
conclude that TX = {TY |Y ⊆ X finite }.
Example The functor (−)N
can be given the structure of a monad with unit X → XN
taking constant sequences, and multiplication XN×N
→ XN
sending a
double sequences (xij)i,j∈N to (xii)i∈N. This is ℵ1-accessible but not ℵ0-
accessible.
Indeed, given a set X, this can be written as {Y ⊆ X|Y is at most
countable}. A sequence (xi) ∈ XN
is in particular contained in {xi}N
,
which is a countable subset of X, and thus we recover all sequences, therefore
(−)N
is ℵ1-accessible.
To see that it is not ℵ0-accessible, take a set X with |X| ≥ |N|. Therefore,
there exist a sequence (xi) in XN
such that xi ̸= xj for i ̸= j. Now, this
sequence cannot be contained in any Y N
with Y finite. We conclude that
XN
is not equal to the union of all sets Y N
for Y ⊂ X finite, so that (−)N
is not ℵ0-accessible.
Week 7
Exercise Decide whether R is finitely generated as a Q-module.
Solution No, it is not. Indeed, a map Qn
→ R can never be surjective, because the
domain is countable but the codomain is not.
Exercise Decide whether Q is finitely generated as a Z-module.
Solution It is not. Indeed, we can prove that there is no surjective homomorphism
Zn
→ Q. Consider a homomorphism as indicated. The domain contains n
elements of the form ei = (0, . . . , 1, . . . , 0) having all components 0 except
for the i-th, which is 1. The image of ei is of the form pi
qi
. Notice that the
denominators of linear combinations of these elements are finite products
of qi’s, and the function is completely determined by the values of ei by
taking linear combinations. Now, if r is a prime number > max{qi}, by
this discussion we know that 1
r cannot be in the image. Therefore, our
function is not surjective.
Proposition If M is a torsion Z-module, then Q ⊗Z M = 0.
Proof We prove that every element of the tensor product is 0. It suffices to show
it for elementary tensors, that is, elements of the form p
q ⊗ m. Since M is
a torsion module, there is an integer n such that nm = 0, so we have
21
p
q ⊗ m = np
nq ⊗ m = p
nq ⊗ nm = p
nq ⊗ 0 = 0.
Observation If M is a Z-module, then an explicit computation shows that Q ⊗Z M is
the module of fractions obtained by inverting all scalars. The previous
proposition says that the torsion component of M, if it has one, doesn’t
contribute to the process.
Remark The same exact proof shows that, if R is an integral domain and F its
field of fractions, then tensoring with F over R is equivalent to inverting
formally all scalars. Moreover, torsion components are annihilated in the
process. Here, an element m ∈ M is torsion if there is r ∈ R such that
rm = 0.
Exercise Show that a sequence 0 A B Cu v
of R-modules is exact
if and only if for any other module N the sequence
0 HomR(N, A) HomR(N, B) HomR(N, C)
u∗ v∗
is exact.
Solution If the first sequence is exact, take two maps f, g : N → A such that
u∗(f) = u∗(g), i.e. uf = ug. Since u is injective, it is a monomorphism,
therefore f = g, and u∗ is injective. Now take f : N → B in ker(v∗),
that is, vf = 0. Then im(f) ⊆ ker(v) = im(u) ∼= A because u is injective.
So we can compose with an inverse map im(u) → A and obtain a map
g : N → A such that f = ug. This means that f ∈ im(u∗). Lastly, if
f ∈ im(u∗), then it is of the form f = ug. Since we know that vu = 0,
then vf = vug = 0, and f ∈ ker(v∗).
Conversely, if the shown induced sequence is exact, for every N, the we
just choose N = R to reobtain the first sequence.
Remark As assigned in the exercise sheet, we also know that A → B → C → 0 is
exact if and only if for each other R-module N the sequence
0 → HomR(C, N) → HomR(B, N) → HomR(A, N)
is exact.
Observation For every three R-modules A, B and C, we have an isomorphism of mod-
ules
HomR(A ⊗R B, C) ∼= HomR(A, HomR(B, C))
which is natural in all three variables. Therefore, for every module M
there is an adjunction − ⊗R M ⊣ HomR(M, −).
Exercise Show that a sequence A → B → C → 0 is exact if and only if for every
module M the induced sequence
M ⊗R A → M ⊗R B → M ⊗R C → 0
22
is exact.
Solution By the previous exercises, the first sequence is exact if and only if for each
N the sequence
0 → HomR(C, N) → HomR(B, N) → Hom(A, N)
is exact, if and only if for each N and M the sequence
0 → HomR(M, HomR(C, N)) → HomR(M, HomR(B, N)) →
HomR(M, HomR(A, N))
is exact, if and only for each N and M if the sequence
0 → HomR(M ⊗R C, N) → HomR(M ⊗R B, N) → HomR(M ⊗R A, N)
is exact, if and only if for each M the sequence
M ⊗R A → M ⊗R B → M ⊗R C → 0
is exact.
Week 8
Definition A pullback in a category is the limiting cone of a diagram of shape
•
• •
. More explicitly, it is a commutative square
C′
D′
C D
f′
g′ g
f
such that, whenever there is an object K with morphisms as in the solid
diagram
K
C′
D′
C D
p
q
t
f′
g′ g
f
with gp = fq, then there exists a unique factorization t : K → C such
that f′
t = p and g′
t = q.
The dual notion is that of pushout. It is therefore an (inner) commutative
square as in the diagram
23
C D
C′
D′
K
g
f
g′
p
f′
q
t
such that whenever there is a commutative outer square, then there is a
unique factorization t : D′
→ K.
Lemma In a pullback square as above, if f is monic, then so is f′
. Dually, in a
pushout square as above, if f is epic, then so is f′
.
Proof Consider a diagram
K C′
D′
C D
p
q
g′
f′
g
f
with f′
p = f′
q, this being the upper bent arrow. The two lower bent
arrows are g′
p and g′
q respectively. Now we have
fg′
p = gf′
p = gf′
q = fg′
q
and since f is monic by assumption, this implies that g′
p = g′
q, so that the
two lower bent arrows are in fact the same. This means that we are looking
at a commutative outer square having K as an upper left vertex, and by
universal property of pullbacks there is a unique factorization K → C′
.
But now both p and q are such factorizations, therefore it must be p = q.
Proposition A sequence of R-modules 0 → A → B → C is exact if and only if the
square
A B
0 C
u
v
is a pullback.
A sequence A → B → C → 0 is exact if and only if the square above is a
pushout.
Proof We will prove the former statement, the latter being simply the dual.
However, it will be included in the exercise sheet.
Assume that the sequence is exact, and consider a solid diagram
24
M
A B
0 C.
p
k
u
v
The commutativity of the outer square just means that vp = 0, so im(p) ⊆
ker(v) = im(u) ∼= A, where we are using both that the sequence is exact at
B and at A. Therefore, just use this to define the factorization k : M → A,
which is moreover unique because it is given by composition with the
inverse isomorphism im(u) → A.
Conversely, if the square above is a pullback, then we know by the previous
lemma that u : A → B is monic, because 0 → C of course is. Moreover,
vu = 0 implies that im(u) ⊆ ker(v). It remains to show that ker(v) ⊆
im(u). To this end, consider a solid diagram
ker(v)
A B
0 C.
i
k
u
v
Since obviously we have vi = 0, the outer square is commutative, therefore
there is a unique factorization k : ker(v) → A. Since i is monic, then so
must be k. In other words, ker(v) may be regarded as a submodule of A,
which is isomorphic to im(u). This concludes the proof.
Remark A sequence M0 M1 · · · Mn
f1 f2 fn
is exact if and
only if all the induced sequences of the form
0 → Ni → Mi → Ni+1 → 0
are exact, where Ni = im(fi).
Notice that the arrow Ni → Mi is an injection and Mi → Ni+1 is a
surjection by definition. If the long sequence is exact, then Ni = ker(fi+1),
which means that the short sequence above is exact. Conversely, if all
those short sequences are exact, then Ni = ker(Mi → Ni+1) = ker(fi+1),
but since by definition Ni = im(fi) we are done.
Definition A functor F : R-Mod → R-Mod is exact if it preserves long exact se-
quences.
Proposition A functor F is exact if and only if it preserves short exact sequences.
Proof It F preserves long exact sequences, then it also preserves short ones, because
they are a particular case.
Suppose that F preserves short exact sequences. Given a long exact sequence
as above, let us split it into short exact sequences. At each node
Mi, we have the following diagram
25
Mi−1 Mi+1
Ni Mi Ni+1
fi fi+1
where the lower horizontal arrows constitute a short exact sequence. Let
us plug the whole diagram through F, observing that, since it preserves
SES’s, in particular it preserves both monomorphisms and epimorphisms.
The resulting diagram therefore will be
FMi−1 FMi+1
FNi FMi FNi+1
F fi
j
F fi+1
π
and the lower horizontal row remains a short exact sequence. Now com-
pute
im(Ffi) = im(j) = ker(π)
the first equality by surjectivity of FMi−1 ↠ FMi, the second by exactness
of the horizontal row. Now, since FNi+1 → FMi+1, an element
of FMi is sent to 0 by Ffi+1 if and only if it is sent to zero by π, so
ker(π) = ker(Ffi+1), and we conclude.
Lemma A left adjoint preserves colimits, a right adjoint preserves limits.
Proof Given a colimit diagram (Ai → A), consider the induced diagram (LAi →
A) and another cocone (LAi → K). By adjunction, this cone corresponds
to one of the form (Ai → RK), therefore by universal property this yields
a unique factorization A → RK, which again by adjunction corresponds
to a unique factorization LA → K. Similarly in the dual case.
Proposition If a functor R-Mod → R-Mod preserves finite limits, then it is left exact.
If it preserves finite colimits, then it is right exact.
Proof This follows from the characterization of left exact sequences given above
in terms of pullbacks of pushouts, and by the observation that final and
initial objects are limits and colimits respectively of the empty diagram,
and in R-Mod they are always zero objects.
Corollary A left adjoint is right exact, a right adjoint is left exact.
ˆ Now we want to be able to give alternative definitions of flatness, some of
which may be more usefule in some contexts, and some in other contexts.
We need a preliminary lemma.
Lemma Given xi ∈ M and yi ∈ N such that xi ⊗ yi = 0 ∈ M ⊗ N, then there
exist finitely generated M0 ⊆ M and N0 ⊆ N such that xi ⊗ yi = 0 ∈
M0 ⊗ N0.
26
Remark The fact that this is not true for all finitely generated submodules reflects
the fact that not all modules are flat, so tensoring does not always preserve
injectivity. However, even when our modules are not flat, the lemma says
that there are some finitely generated submodules for which injectivity is
preserved.
Proof Remember that a tensor product can be constructed as a quotient module
F(M × N)/Q, where the numerator is the free module on the underlying
set of M × N, and Q is generated by all expressions of the form
(x + x′
, y) − (x, y) − (x′
, y)
(x, y + y′
) − (x, y) − (x, y′
)
(ax, y) − a(x, y)
(x, ay) − a(x, y).
Now since xi⊗yi = 0, this means that the formal expression (xi, yi) ∈
Q. By definition of Q, this means that it can also be expressed as a finite
formal sul gj, where every gj is an expression of the form above. Now
take as M0 the submodule generated by all the xi’s and all the elements
appearing as first coordinates of the gj’s. Define N0 similarly.
It is clear then that (xi, yi) ∈ Q0, this being the corresponding module
over which we quotient in order to obtain M0 ⊗ N0. In other words, our
tensor is 0 in M0 ⊗ N0.
Remark If xi ⊗ yi = 0 ∈ M0 ∈ N0, a fortiori this implies that it is 0 in M0 ⊗ N.
This means that the result is true even if we replace only one of the two
modules with a finitely generated one.
Theorem Given a module M, the following conditions are equivalent:
1. the functor M ⊗ − preserves long exact sequences;
2. the functor M ⊗ − preserves short exact sequences;
3. given an injection A → B, then M ⊗ A → M ⊗ B is an injection;
4. given an injection A → B between finitely generated modules, then
M ⊗ A → M ⊗ B is an injection.
Proof The equivalence 1 ⇔ 2 has been discussed. The equivalence 2 ⇔ 3 follows
since all functors M ⊗ − are right exact, so the only thing left to preserve
short exact sequences is the first map remaining an injection after tensoring.
Moreover, 3 ⇒ 4 is obvious.
It remains to show 4 ⇒ 3. Given an injection f : A → B, and an element
mi ⊗ ai in the kernel of M ⊗ f, we want to see that this is 0. Since it
is in the kernel, we have mi ⊗ f(xi) = 0 ∈ M ⊗ B. Now take A0 ⊆ A
to be generated by all ai. By the previous lemma (and remark) we can
choose a finitely generated B0 ⊆ B that moreover contains f(A0) (by enlarging
it if necessary) in which mi ⊗ f(ai) = 0 ∈ M ⊗ B0. By our
hypothesis, this implies mi ⊗ ai = 0 ∈ M ⊗ A0. A fortiori, this means
mi ⊗ ai = 0 ∈ M ⊗ A.
27
Week 9
ˆ This week’s topic is a very brief introduction to homological algebra, which
incidentally also includes some interesting constructions which is possible
to perform with projective and injective modules.
Definition A chain complex is a Z-indexed sequence of modules and morphisms of
the form
. . . C1 C0 C−1 . . .
d2 d1 d0 d−1
with the property that for every i we have di−1 ◦ di = 0. The maps di’s
are called boundary operators.
We denote moreover Bi(C) := im(di+1) and Zi(C) := ker(di). Elements
of Ci are called i-chains, elements of Bi(C) are i-boundaries and elements
of Zi(C) are i-cycles.
Definition A cochain complex is the dual of a chain complex, i.e. the maps are going
upward. Conventionally, all the indexes are usually put in superscript
instead of subscript. The di
’s are known as differentials, elements of Ci
are i-cochains, elements of Bi
(C•) are i-coboundaries and elements of
Zi
(C•) are i-cocycles.
Remark Sometimes, boundary operators of a chain complex (but not differentials
of a cochain complex) are denoted witih ∂ instead of d. These differences
in terminology are due to the fact that in the motivating applications one
or the other form is preferred in different context, e.g. chain complexes
are often used in algebraic topology, cochain complexes in differential geometry,
so the terms are borrowed from the respective branches of mathematics.
On a purely categorical perspective, the two notions are nothing
more than dual to one another.
Remark Often, when it is clear from the context, indices will be suppressed, for
the sake of notational simplicity.
Definition A chain map is a homomorphism in the category of chain complexes, i.e.
a sequence of maps fi : Ci → C′
i such that dfi+1 = fid.
Definition The condition d ◦ d = 0 can be rewritten as Bi(C) ⊆ Zi(C). Therefore,
given a chain complex Ci, we can define the i-th homology module as
Hi(C) := Zi/Bi.
Dually, for a cochain complex we can define the i-th cohomology module
as Hi
(C) := Zi
/Bi
.
Remark The operations H• are functorial, i.e. they are functors from the category
of chain complexes to that of modules. Moreover, they are additive, in the
sense that Hi(f + g) = Hi(f) + Hi(g) (this will be on the exercise sheet).
Definition Given two chain maps f, g : C• → C′
•, a chain homotopy from f to g is a
sequence of maps si : Ci → C′
i+1 such that fi − gi = dsi + si−1d.
28
. . . Ci+1 Ci Ci−1 . . .
. . . C′
i+1 C′
i C′
i−1 . . .
d d
fi+1 gi+1
d
fi gi
si
d
fi−1 gi−1
si−1
d d d d
Lemma If two chain maps f and g are chain homotopic, then they induce the same
maps on all homology modules.
Proof By definition of chain homotopy, there are maps s in every degree such
that f −g = ds+sd. Now, by definition of homology, all chains of the form
d(c) are collapsed to 0 in the quotient, so for every i, by the definitions
and additivity of the homology functors, we have
Hi(f) − Hi(g) = Hi(f − g) = Hi(ds + sd) = Hi(ds) = Hi(sd) = 0
which implies Hi(f) = Hi(g).
ˆ Next, we are going to use these notions in combinations with projectivity
and injectivity.
Definition A left resolution C• of a module M is an exact sequence of the form
. . . → C2 → C1 → C0 → M → 0.
A right resolution is the dual notion.
Definition A projective resolution of a module M is a left resolution P• where each
module Pi is projective.
An injective resolution is a right resolution I• where each module Ii is
injective.
Remark Every module admits both a projective and an injective resolution (this
also will be in the exercise sheet).
Remark Note that a left resolution of M can be regarded as a chain complex, having
Ci in non-negative degree, M in degree -1 and 0 in all other degrees. The
condition d ◦ d = 0 is verified, because the stronger condition of exactness
is true.
Dually, we can regard a right resolution as a chain complex which is 0 in
all degrees < −1.
Lemma Given a map f′
: M → N and two projective resolutions as in the diagram
. . . P2 P1 P0 M 0
. . . Q2 Q1 Q0 N 0
d d ε
f′
d d η
then we can lift f′
to a chain map f. Moreover, this chain map is unique
up to chain homotopy.
29
Proof We already have the chain map constructed in all degrees ≤ −1, this being
f′
in degree -1 and 0 in all other negative degrees. So we can proceed by
induction, assuming that we have defined our chain map up to fn, with
n ≥ −1. In this case, we have fn−1d = dfn. Taking an n-cycle p, i.e.
d(p) = 0, then we have dfn(p) = fn−1d(p) = 0, which means that fn is
also an n-cycle. In other words, fn restricts to a map Zn(P) → Zn(Q).
Let us represent this in the solid diagram
Pn+1 Zn(P) 0
Qn+1 Zn(Q) 0
d
fn+1 fn
d
where the two d’s are surjective by exactness of the rows. By projectivity
of Pn+1, there is a map fn+1 making the square commute, so the induction
step is complete. It only remains to prove that the obtained chain map f
is unique up to homotopy.
Suppose we have two chain maps f, g : P → Q, both lifting f′
. We want
to produce a chain homotopy between them. Define h := f − g. For all
degrees ≤ −2, there is nothing to prove. Moreover, h−1 = 0, so we can
just define s−1 = 0 : M → Q0 and the chain homotopy condition will
be trivially satisfied. For all non-negative degrees, we proceed again by
induction, so first assume n = 0. We have ηh0 = ηf − ηg = f′
ε − f′
ε = 0.
This means that h0 lands in Z0(Q). Now consider the diagram
P0 Z0(Q)
Q1.
h0
s0
d
Again d is surjective by exactness, so projectivity of P0 ensures the existence
of s0 such that ds0 = h0. Since s−1 = 0, this is already the required
condition.
Now assume we have defined our si’s up to some n ≥ 0. We have
hn = dsn + sn−1d. Consider the map hn+1 − snd : Pn+1 → Qn+1, and use
this to compute
d(hn+1 − snd) = dhn+1 − (hn − sn−1d)d = dhn+1 − hnd = 0
the last equality following from the fact that h is a chain map. So hn+1 −
snd lands in Zn+1(Q). Similarly as above, in the solid diagram
Pn+1 Zn+1(Q)
Qn+1
hn+1−snd
sn+1
d
there exists a lift sn+1, so that dsn+1 = hn+1 − snd, which is the chain
homotopy condition, so our induction step is complete.
30
Definition Given a right exact functor F on some category of modules, we can define
the left derived functors of F by choosing a projective resolution P for
each modules M, and then the n-th left derived functor will be computed
on M as LnF(M) = Hn(F(P)).
Proposition The left derived functors are well-defined. Moreover, they are indeed
functors.
Proof Choosing two different projective resolutions P and Q for M, the lemma
above says that they are connected by a map f. Since this map is unique
up to chain homotopy, it becomes strictly unique after passing to homology.
Analogously, we have a map g in the other direction which
is unique after passing to homology. Again by uniqueness, it must be
Hn(f) ◦ Hn(g) = Hn(id) = id and Hn(g) ◦ Hn(f) = Hn(id) = id, so that
Hn(P) ∼= Hn(Q).
By the same reasoning, this assignment is functorial.
Remark Dually, all maps can be completed to cochain maps between injective
resolution in a way that is unique after taking cohomology. Given a left
exact functor F, we can defined the n-th right derived functor of F by
choosing such an injective resolution I for a module M, and then defining
RnF(M) = Hn
(F(I)).
Definition Remember that the tensoring functors are right exact. The higher Tor
modules are defined as follows:
Torn
(A, B) = Ln(A ⊗ −)(B).
Similarly, remember that the hom-functors are left exact, then define the
higher Ext modules as:
Extn
(A, B) = Rn(Hom(A, −))(B).
ˆ We also have the two following results, allowing to compute Tor and Ext
modules in symmetric ways. They are non-trivial and require some more
advanced tools from homological algebra to be proven.
Theorem Extn
(A, B) ∼= Rn(Hom(A, −))(B) ∼= Rn(Hom(−, B))(A).
Torn
(A, B) ∼= Ln(A ⊗ −)(B) ∼= Ln(− ⊗ B)(A).
Proposition If R is a PID (see exercise sheet), then for all R-modules A and B and
n ≥ 2 we have Extn
(A, B) = 0.
Proof We can choose an injective resolution of B of the form
0 → M → I0 ↠ I1 → 0 → . . .
where I0 is an injective hull (see the course for its existence) and I1 is
the quotient I0/M, which is itself injective (see the exercise sheet). After
applying the functor Hom(A, −), we obtain a cochain complex which is 0
in degree ≥ 2, so its cohomology is 0.
31
Week 10
Construction Suppose we are given a short exact sequence 0 A• B• C• 0u v
.
For each n ∈ Z, we construct a homomorphism ∂ : Hn(C) → Hn−1(A) as
follows:
let us keep in mind a specific section of our short exact sequence of chain
complexes, that it, the diagram
0 An+1 Bn+1 Cn+1 0
0 An Bn Cn 0
0 An−1 Bn+1 Cn−1 0
0 An−2 Bn−2 Cn−2 0.
u
d
v
d d
u
d
v
d d
u
d
v
d d
u v
An element of Hn(C) is a homology class of the form [c], for some cycle
c ∈ Zn(C). By surjectivity of v, there is b ∈ Bn such that v(b) = c.
Now observe that vd(b) = dv(b) = d(c) = 0 because c is a cycle, so we
know that d(b) = ker(v) = im(u), and by injectivity of u there is a unique
a ∈ An−1 such that u(a) = d(b). Now we want to show that this a is a
cycle. For this we compute ud(a) = du(a) = dd(b) = 0, and since u is
injective we have d(a) = 0, so a ∈ Zn−1(A). Now we choose ∂([c]) to be
the homology class [a] ∈ Hn−1(A).
Proposition The construction above is well-defined and it is a module homomorphism.
Proof We need to show that our construction is independent of the representative
c of the homology class [c], and also of the preimage b ∈ Bn. We will
kill both birds with one stone. So let us assume that we have another
b′
∈ Bn such that v(b′
) ∈ [c]. By definition of homology, this means that
v(b′
) = c + d(c′′
) for some c′′
∈ Cn+1. Now choose b′′
∈ Bn+1 such that
v(b′′
) = c′′
, and compute
v(b′
) = v(b) + dv(b′′
) = v(b) + vd(b′′
) = v(b + d(b′′
))
which means that b′
− b − d(b′′
) ∈ ker(v) = im(u), and by injectivity of u
there is a unique a′′
∈ An such that b′
− b − d(b′′
) = u(a′′
). Now, we chose
above a to be the preimage of d(b), and analogously we choose a′
∈ An−1
to be the preimage of (b′
), which is similarly a cycle. Finally, we compute
u(a′
) = d(b′
) = d(b) + dd(b′′
) + du(a′′
) = u(a) + ud(a′′
) = u(a + d(a′′
))
which by injectivity of u implies a′
= a+d(a′′
). This means that [a′
] = [a],
so our construction is well-defined. With a similar argument we can show
that it is a homomorphism.
32
Snake lemma Given a short exact sequences of chain complexes 0 A• B• C• 0u v
,
there exists a long exact sequence of the form
. . . Hn(A) Hn(B) Hn(C) Hn−1(A) . . .
u∗ v∗ ∂
Proof First, observe that a cycle a ∈ Zn(A) yields a cycle u(a) = Zn(B), because
du(a) = ud(a) = u(0) = 0, so u restricts to a map u : Zn(A) → Zn(B).
Moreover, a boundary d(a) ∈ Bn(A) yields likewise a boundary, in that
ud(a) = du(a), so this induces a well-defined map u∗ : Hn(A) → Hn(B).
Analogously, we have a map v∗ : Hn(B) → Hn(C). We will leave it up to
the reader to show that the chain is exact at Hn(B), and prove explicitly
exactness at Hn(A) and Hn(C).
Let us show that ker(u∗) ⊆ im(∂). If [a] ∈ Hn(A) is in ker(u∗), it means
that [u(a)] = 0, that is, u(a) = d(b) for some b ∈ Bn+1. Now, observe
that dv(b) = vd(b) = vu(a) = 0, so v(b) is a cycle. By construction of ∂,
we have ∂([v(b)]) = [a].
Now let us show that im(∂) = ker(u∗). If [a] ∈ im(∂), there exists a cycle
c ∈ Zn+1(C) such that, chosen b with v(b) = c, we have [u−1
d(b)] = [a].
This means that u−1
d(b) = a + d(a′
) for some a′
∈ An+1. Now we have
u(a) = d(b) − ud(a′
) = d(b − u(a′
))
so it the boundary of b − u(a′
), which is 0 after passing to homology.
We have proven that [a] ∈ ker(u∗). The proof of exactness at Hn(A) is
complete.
Let us prove first that im(v∗) ⊆ ker(∂). For a homology class of the form
v∗([b]) ∈ Hn(C), reproduce the construction of ∂ to obtain
∂v∗([b]) = ∂([v(b)]) = [u−1
dv−1
v(b)] = u−1
d(b)] = [u−1
(0)] = [0]
where we slightly abuse notation in that v−1
is not uniquely determined,
but we have seen above that it doesn’t matter what element of the preimage
we choose. On the other hand, u−1
is not an abuse of notation, because
u is injective. Moreover, we are using d(b) = 0 because homology classes
are defined starting with cycles.
Now we prove ker(∂) = im(v∗). Assume that [c] ∈ ker(∂). If v(b) = c, this
means that [u−1
d(b)] = 0, so that u−1
d(b) = d(a′
) for some a′
∈ An. Let
us consider the element b − u(a′
) ∈ Bn, we have
d(b − u(a′
)) = d(b) − ud(a′
) = 0
so b − u(a′
) is a cycle, hence it defines a homology class. We conclude by
seeing that
v∗([b − u(a′
)]) = [v(b) − vu(a′
)] = [v(b)] = [c]
because vu = 0. So [c] is in the image of v∗, and our proof is complete.
33
Corollary For a short exact sequence of modules 0 → A → B → C → 0 and a right
exact functor F, there is a long exact sequence
. . . → L2F(C) → L1F(A) → L1F(b) → L1F(C) → F(A) → F(B) →
F(C) → 0.
(Remember, from the exercise sheet, that L0F ∼= F.)
Corollary For a short exact sequence as above and a module M, there are long exact
sequences
. . . → Tor1
(M, A) → Tor1
(M, B) → Tor1
(M, C) → M ⊗ A → M ⊗ B →
M ⊗ C → 0
. . . → Tor1
(A, M) → Tor1
(B, M) → Tor1
(C, M) → A ⊗ M → B ⊗ M →
C ⊗ M → 0.
Remark Dually, there is are cohomology long exact sequences going upward in
dimension, with homomorphisms ∂ : Hn
(C) → Hn+1
(A). In particular,
given a short exact sequence of modules and a left exact functor, there is
a long exact sequence of the form
0 → F(A) → F(B) → F(C) → R1F(A) → R1F(B) → R1F(C) →
R2F(A) → . . ..
Specializing even more, we get long exact sequences of the form
0 → Hom(M, A) → Hom(M, B) → Hom(M, C) → Ext1
(M, A) →
Ext1
(M, B) → Ext1
(M, C) → . . .
0 → Hom(C, M) → Hom(B, M) → Hom(A, M) → Ext1
(C, M) →
Ext1
(B, M) → Ext1
(A, M) → . . ..
Week 11
Question Why are Ext modules called Ext? The name Ext originated from extension
groups, that we will illustrate in a moment. It was then shown
that these groups could equivalently be constructed by the means of projective
or injective resolutions of abelian groups, and such methods are
quite naturally generalized to the context of modules, where the higher
constructions are possibly non-trivial, giving thus rise, by analogy, to the
nomenclature Extn
for the higher steps of the same constructions. We are
then reduced to the question of what extension groups are and how they
are related to Ext1
.
Definition Given two abelian groups A and B, an extension of A through B is a short
exact sequence of the form
0 → B → M → A → 0.
34
Two such sequences are called equivalent if there is a map M → N such
that the diagram
M
0 B A 0
N
is commutative. It can be checked that whenever this is the case the map
M → N is necessarily an isomorphism. The extension group Ext(A, B)
has equivalence classes of extensions of A through B as elements. The
group structure is depicted below.
Baer sum Given two extensions as above, with f : B → M and g : B → N respectively
as first arrow, their Baer sum is defined to be the equivalence class
of the short exact sequence
0 → B → E → A → 0
where E is the quotient of M ⊕ N on the subgroup generated by all
elements of the form (f(b), 0) − (0, −g(b)) for all elements b ∈ B.
The unit for this monoid structure is given by the splitting sequence, that
is, the sequence
0 → B → A ⊕ B → A → 0
and the inverse of a sequence given by the two maps u : B → M and
v : M → A is the sequence given by −u and v.
Theorem The above construction gives a group structure on Ext(A, B). Moreover,
there is an natural isomorphism Ext(A, B) ∼= Ext1
(A, B).
Idea of proof We construct a map Ext(A, B) → Ext1
(A, B) and omit the rest of the
proof. Choose an extension of A through B, and choose a projective
resolution of A, as in the diagram
0 P1 P0 A 0
0 B M A 0.
By projectivity of P0, there is an arrow P0 → M that makes the right
square commute, then since the lower row is exact, its universal property
yields the arrow P1 → M that makes the left square commute as well. Now
observe that Ext1
(A, B) is the quotient of Hom(P1, B) on the subgroup of
all maps that factor through P1 → P0. Our starting short exact sequence
will be associated to the equivalence class of the obtained map P1 → B.
35
Question Why are Tor modules called Tor? The answer to this also restricts to
the case of Tor1
on abelian groups. The reason is that, in that case, Tor
groups are always torsion groups. We need a couple of preliminaries to
prove that.
Proposition Let F be a right exact functor, and U an exact functor. Then there is a
natural isomorphism ULnF ∼= LnUF, where Ln denotes as usual the n-th
left derived functor.
Proof Exercise sheet.
Proposition Let I be a directed diagram, then the functor colim : R − ModI
→
R − Mod is exact.
Sketch of proof We know that colim is left adjoint to the diagonal functor δ : R−Mod →
R−ModI
sending an object M to the constant functor on M. Therefore,
as we have already seen in a past session, it is right exact. It only remains
to show that it preserves monomorphisms. This can be done explicitly,
keeping in mind that monomorphisms are precisely injections, and colimits
of directed diagrams behave like unions of all the involved modules.
Corollary Given a left adjoint (hence right exact) functor F and a directed diagram
of modules (Ai)i∈I, we have a natural isomorphism LnF(colim Ai) ∼=
colim LnF(Ai).
Proof Let us compute
LnF(colim Ai) ∼= Ln colim F(Ai) ∼= colim LnF(Ai)
where the first isomorphism is given by the fact that L is a left adjoint
and therefore it preserves all colimits, and the second isomorphism is a
combination of the two previous propositions.
Remark A dual result is obtained if we choose a left exact functor F and its right
derived functors.
Corollary For a directed diagram of modules (Ai)i∈I and a module B, we have
natural isomorphisms Torn
(colim Ai, B) ∼= colim Torn
(Ai, B).
Theorem For A and B abelian groups, Tor1
(A, B) is a torsion group.
Proof A can be expressed as the union of all its finitely generated subgroups
Ai. In particular, this union is the colimit of a directed diagram. Now we
compute
Tor1
(A, B) ∼= Tor1
(colim Ai, B) ∼= colim Tor1
(Ai, B)
by the corollary above. Now torsion groups are stable under colimits,
therefore we are reduced to the case where A is finitely generated. In
other words, it must be of the form
A ∼= Zm
⊕ Z/n1 ⊕ . . . ⊕ Z/nr
36
for appropriate natural numbers m, n1, . . . , nr. Next, we compute
Tor1
(A, B) ∼= Tor1
(Zm
, B) ⊕ Tor1
(Z/n1, B) ⊕ . . . ⊕ Tor1
(Z/nr, B)
since all functors Torn
preserve finite sums (see exercise sheet). Now, we
know that the first term is 0 because Zm
is projective, and all other terms
are B/niB, hence torsion. This concludes the proof.
Exercise Let us compute the group Ext1
(Q, Z/n).
Solution Let us choose an injective resolution of Z/n as follows:
0 Z/n Q/Z Q/Z 0
1
n n
where the first map is the injection that sends 1 ∈ Z/n to 1
n ∈ Q/Z. The
second map is then the quotient of this injection. We can picture it as
the rational circle wrapping around itself n times. Now apply the functor
Hom(Q, −) and consider the degree 1 of the resulting chain complex. This
is Hom(Q, Q/Z). Now Q has no torsion elements, and Q/Z has only torsion
elements. Since a homomorphic image of a torsion element is itself torsion,
we conclude that Hom(Q, Q/Z) ∼= 0. Since Ext1
(Q, Z/n) is a quotient of
that, this must also be 0.
37