Tony Geramita’s notes for his Homological Algebra course, Fall 1994 Books I am consulting

1). Charles A. Weibel - An Introduction to Homological Algebra, Cambridge Studies in Advanced Math., vol. 38 2). P.J. Hilton, U. Stammbach - A Course in Homological Algebra - Spring-Verlag (Graduate Texts in Math.) 3). A. Geramita, C. Small - An Introduction to Homological Methods in Commutative Rings, QPPAM, No. 43 4). D.G. Northcott - An Introduction to Homological Algebra, Cambridge University Press

Sect. 1A

Homological Algebra Preliminaries: Let R be an associative ring with 1 and let U, V and W be three (left) R-modules.

Let f : U → V and g : V → W be two R-module homomorphisms.

If g ∘ f = 0 then we say that

U⟶^{f}V ⟶^{g}W  (∗)

is a complex of R-modules.

Notice that g ∘ f = 0 ⇔ the image of f, (Im(f)), ⊆ Ker(g). In this case,

Ker(g)/Im(f) = H

is an R-module. H is called the homology module of the complex (*) above.

Definition: (*) is called an exact complex if and only if H = 0.

Thus, a homology module “measures ” the extent to which a complex is not exact.

We might as well now make the complete definition.

Definition: A sequence C_{•} = {C_{n}}_{n∈ℤ} of (left) R-modules and R-module homomorphisms d_{•} = {d_{n}}_{n∈ℤ}, where d_{n} : C_{n} → C_{n−1}, is called a chain complex of R-modules if d_{n−1}∘d_{n} = 0 for every n.

The maps d_{n} are called the differentials of the chain complex.

Much of the terminology here comes from homology theory in Algebraic Topology. Thus the elements of the modules C_{n} are referred to as n-chains, the elements of Ker(d_{n}) as n-cycles and the elements of Im(d_{n+1}) as n-boundaries and

H_{n} = Ker(d_{n})/Im(d_{n+1})

is called the nth homology module of the chain complex.

The complex C_{•} is called exact at C_{n} if H_{n} = 0 and (simply) an exact complex if all of its homology modules are 0.

Note: A complex like (∗) above can be thought of as a chain complex in many ways by letting U = C_{n}, V = C_{n−1} and W = C_{n−2} (for any n) and letting C_{i} = 0 otherwise.

Exercise 1.A Let C_{n} = ℤ/8 for n ≥ 0 and let C_{n} = 0 for n < 0. Let d_{n} : C_{n} → C_{n−1} be defined by x̄ → \overline{4x} for each n.

Show {C_{n}, d_{n}} is a chain complex and compute all the homology modules of this complex.

Definition: A sequence of R-modules C^{•} = {C^{n}}_{n∈ℤ} and R-module homomorphisms {∂^{n}} = ∂^{•}, where

∂^{n} : C^{n} → C^{n+1}

is called a cochain complex if ∂^{n+1} ∘ ∂^{n} = 0 for all n.

This time the elements of Im(∂^{n}) are call coboundaries and the elements of Ker(∂^{n+1}) are called cochains. The quotient module H^{n} = Ker(∂^{n})/Im(∂^{n−1}) is called the nth cohomology module of the cochain complex.

Note: If {C_{n}} is a chain complex and we let C^{n} = C_{−n} and ∂^{n} : C^{n} → C^{n+1} be defined by ∂^{n} = d_{−n}, then {C^{n}, ∂^{n}} is a cochain complex.

We say that the complex C_{•} is bounded if almost all the modules C_{n} are = 0. C_{•} is bounded above if there is an integer b such that C_{n} = 0 for all n ≥ b. C_{•} is bounded below if there is an integer c such that C_{n} = 0 for all n ≤ c.

If C_{•} is a bounded complex we usually do not write in all the zero modules.

Consider the complex

0 → A⟶^{f}B → 0 .

This complex is exact at A if and only if f is 1 − 1. The complex is exact at B if and only if f is onto. The complex is exact if and only if f is an isomorphism. So, it is possible to speak of properties of functions using the idea of exactness.

Definition: A complex is called a short exact sequence if, for some n, it is of the form

0 → C_{n} ⟶^{d_{n}} C_{n−1} ⟶^{d_{n−1}} C_{n−2} → 0  (†)

and (†) is an exact complex.

It is easy to see that the definition is equivalent to saying: d_{n} is 1 − 1, d_{n−1} is onto and H_{n−1} = 0.

If f : M → N is any R-module homomorphism then we have an exact complex:

0 → Ker(f)⟶^{i}M⟶^{f}N⟶^{η}N/Im(f) → 0

where i is the inclusion map and η is the canonical map onto a quotient.

Definition: The module N/Im(f) is called the cokernel of the map f.

The Snake Lemma:

Given a commutative diagram with exact rows

\begin{aligned}& & M′ & →^{α} & M & →^{α′} & M′′ & → & 0 \\ & & f′ ↓ & & f ↓ & & f′′ ↓ \\ 0 & → & N′ & →^{β′} & N & →^{β} & N′′ \end{aligned} .

Then, there is a complex:

Ker(f′)⟶^{\overline{α}}Ker(f)⟶^{\overline{α′}}Ker(f′′)⟶^{γ}Coker(f′)⟶^{\overline{β′}}Coker(f)⟶^{\overline{β}}Coker(f′′) (∗∗)

which is exact. Moreover:

i) If α is 1 − 1 then \overline{α} is 1 − 1,

ii) If β is onto then \overline{β} is onto.

Proof: (Mostly left as an exercise - 1A′) The only nontrivial part of this is the definition of the map γ.

To see how it is defined, let t′′ ∈ Ker(f′′), i.e. f′′(t′′) = 0. Then, since α′ is onto, t′′ = α′(t) for some t ∈ M. Thus, f′′α′(t) = 0 = βf(t). Thus, f(t) ∈ Ker(β) = Im(β′). I.e. there is a t′ ∈ N′ such that β′(t′) = f(t). We define

γ(t′′) = t′ ∈ Coker(f′) .

One must show that γ is well-defined and a homomorphism before showing that the sequence (∗∗) is exact.

Corollary: (5-Lemma)

Let

0 → M′ → M → M′′ → 0
0 ↓ f′ ↓ f ↓ f′′ ↓ 0 ↓
0 → N′ → N → N′′ → 0

be a commutative diagram with exact rows. Then

1) If f′ and f′′ are 1 − 1 then so is f.

2) If f′ and f′′ are onto then so is f.

3) If f′ and f′′ are isomorphisms then so is f.

Proof: I’ll leave this proof as an (easy) exercise - 1A′′.

The Group of Homomorphisms

Let M and N be (left) R-modules. We write Hom_{R}(M, N) to denote the set of all R-module homomorphisms from M to N. It is easy to see that this set is an abelian group if we let:

(f + g)(m) := f(m) + g(m) .

If, moreover, R is a commutative ring, then we can even make this group of homomorphisms into an R-module by:

if ϕ ∈ Hom_{R}(M, N) and r ∈ R then (rϕ)(m) := r(ϕ(m)) .

Exercise 1.B Show that this makes Hom_{R}(M, N) into an R-module. Also, show that this is not necessarily true if R is not commutative, i.e. give a specific example.

If β : N → N′ is an R-module homomorphism, then there is an induced homomorphism of abelian groups, β_{∗} : Hom_{R}(M, N) → Hom_{R}(M, N′) by

N

↗
M β_{∗} ↓
↘
N′
β_{∗}(ϕ) = β ∘ ϕ .

Exercise 1.C Check that β_{∗} is a homomorphism of abelian groups.

Note further that if:

β : N → N′ and β′ : N′ → N′′

then

(β′β)_{∗} : Hom_{R}(M, N) ⟶ Hom_{R}(M, N′′)

satisfies

(β′β)_{∗} = (β′)_{∗} ∘ β_{∗} .

Also, if β : N → N is the identity R-module homomorphism then

β_{∗} : Hom_{R}(M, N) → Hom_{R}(M, N)

is the identity group homomorphism.

We think of the discussion above in the following way: Fix an R-module M and consider the symbol Hom_{R}(M, −). This symbol defines a function from the set of (left) R-modules to the set of Abelian groups by

Hom_{R}(M, −) : N ⟶ Hom_{R}(M, N) .

But, this does more! The “function” also acts on homomorphisms between R-modules i.e.

Hom_{R}(M, −) : \begin{matrix}N & & Hom_{R}(M, N) \\ β ↓ & ⟶ & β_{∗} ↓ \\ N′ & & Hom_{R}(M, N′) \end{matrix}

Moreover, this “function” respects composition and identities (the two most important things about functions). Moreover, it maintains the order of composition.

When we make the formal definition later you will that see that we have just shown that Hom_{R}(M, −) is a covariant functor from the category of (left) R-modules to the category of Abelian groups.

Sect.1B

In a completely analogous way (and I’ll leave you to fill in all the details) Hom_{R}(−, N) is also a functor from R-modules to Abelian groups, with one important difference. If we let β^{∗} be the result of applying this functor to β : M → M′ then

β^{∗} : Hom_{R}(M′, N) ⟶ Hom_{R}(M, N)

is defined by: β^{∗}(ϕ) = ϕ ∘ β, and

(β_{1} ∘ β)^{∗} = β^{∗} ∘ β_{1}^{∗}

i.e. the order of composition is reversed. I.e. (as we shall see later) Hom_{R}(−, N) is a contravariant functor from (left) R-modules to Abelian groups.

We can say more about Hom_{R}(M, −).

Proposition: Let

0 → A⟶^{f}B⟶^{g}C → 0

be a short exact sequence of R-modules.

If M is any R-module then

0 → Hom_{R}(M, A)⟶^{f_{∗}}Hom_{R}(M, B)⟶^{g_{∗}}Hom_{R}(M, C)  (†)

is also exact.

Proof: (See also exercise 10A) \underline{Exactness at Hom_{R}(M, B) :}

It is easy to see that if 0 denotes the 0-homomorphism then 0_{∗} = 0. Since (g ∘ f)_{∗} = g_{∗} ∘ f_{∗} we get that (†) is a complex. So, it suffices to prove that:

Ker(g_{∗}) ⊆ Im(f_{∗}) .

So, let ϕ ∈ Ker(g_{∗}), i.e. ϕ ∈ Hom_{R}(M, B) and g_{∗}(ϕ) = 0, i.e. g ∘ ϕ = 0. Thus, Im(ϕ) ⊆ Ker(g).

However, since the original sequence is exact we have Ker(g) = Im(f) and, since f is 1 − 1, f gives an isomorphism from A to the Im(f). Thus χ = f^{−1} ∘ ϕ is an element of Hom_{R}(M, A) and f_{∗}(χ) = f ∘ (f^{−1} ∘ ϕ) = ϕ, i.e. ϕ ∈ Im(f_{∗}) as we wanted to show.

\underline{Exactness at Hom_{R}(M, A) :} It suffices to show that f_{∗} is 1−1. So, suppose that f_{∗}(ϕ) = 0, ϕ ∈ Hom_{R}(M, A), i.e. f ∘ ϕ = 0. Since f is 1 − 1, ϕ must be the zero map. ⊓ ⊔

Exercise 2A: Consider the short exact sequence of ℤ modules:

0 → ℤ⟶^{×n}ℤ⟶^{η}ℤ/nℤ → 0 .

Apply the functor Hom_{ℤ}(ℤ_{n}, −) to this sequence and show that η_{∗} is not onto.

In the light of this proposition we say that the functor Hom_{R}(M, −) is a left exact functor. The exercise shows that, in general, Hom_{R}(M, −) is not an exact functor.

In a completely analogous way we have:

Proposition: Let

0 → A⟶^{f}B⟶^{g}C → 0

be a short exact sequence of R-modules. If N is any R-module then

0 → Hom_{R}(C, N)⟶^{g^{∗}}Hom_{R}(B, N)⟶^{f^{∗}}Hom_{R}(A, N)

is exact.

Exercise 2.B Prove this proposition and then, using the example of Exercise 2.A, show that f^{∗} need not be surjective. This will show that the contravariant functor Hom_{R}(−, N) is left exact but not necessarily exact.

Direct Sums and Direct Products

Let R be a ring, M, N both R-modules. The Direct Sum of M and N is

M ⊕ N = {(m, n) | m ∈ M and n ∈ N}

where

(m, n) + (m_{1}, n_{1}) = (m + m_{1}, n + n_{1}) and λ(m, n) = (λn, λn) .

There are also two distinguished R-module homomorphisms

ι_{M} : M → M ⊕ N , ι_{N} : N → M ⊕ N

by ι_{M}(m) = (m, 0) and ι_{N}(n) = (0, n).

Note that ι_{M} and ι_{N} are 1 − 1 and that ι_{M}(M) + ι_{N}(N) = M ⊕ N.

Proposition: The triple {M ⊕ N, ι_{M}, ι_{N}} satisfies the following universal mapping property: if V is any R-module and there are maps ϕ_{M} : M → V and ϕ_{N} : N → V then there is a unique R-module homomorphism ϕ : M ⊕ N → V such that the following diagram commutes:

M
↙ ι_{M}  ↘ ϕ_{M}
M ⊕ N −− −^{ϕ} − → V
↖ ι_{N}  ↗ ϕ_{N}
N

Proof: Once stated, the proof is obvious if we let

ϕ(m, n) := ϕ_{M}(m) + ϕ_{N}(n)

Then ϕ satisfies the conditions and the result is immediate. ⊓ ⊔

Notice that the UMP says that an R-module homomorphism out of M ⊕ N is completely determined by its values on Im(ι_{M}) and Im(ι_{N}).

Notation: In view of the way the map ϕ was constructed we often write ϕ = (ϕ_{M}, ϕ_{N}).

There is nothing stopping us from making this definiton in general. Let {M_{j}}, j ∈ J be any family of R-modules indexed by the set J.

Definition: The direct sum of the modules {M_{j}}, j ∈ J is the set

⊕_{j∈J} M_{j} = {(m_{j})_{j∈J} | m_{j} = 0 for almost all j}

(where addition and the R-action are defined componentwise) and a family of 1 − 1 maps,

ι_{M_{j}} : M_{j} → ⊕_{j∈J} M_{j}

defined by ι_{M_{j}}(m_{j}) is the infinite tuple which is 0 everywhere except in position j where it is m_{j}.

This set and the family of maps so defined satisfy a universal mapping property. To wit:

Proposition: Let V be an R-module and suppose there are homomorphisms ϕ_{j} : M_{j} → V for each j.

Then there is a unique map ϕ : ⊕_{j∈J} M_{j} → V such that ϕ ∘ ι_{M_{j}} = ϕ_{j} for all j ∈ J.

Proof: We define ϕ as follows

ϕ((m_{j})) = ∑_{j} ϕ_{j}(m_{j}) .

Notice that the sum makes sense because it is finite. The uniqueness follows from the fact that

⊕_{j∈J} M_{j} = ∑_{j∈J} Im(ϕ_{j})

Notation: As with the case above, we denote the uniquely determined map ϕ = (ϕ_{j})_{j∈J}.

There is a similar construction for what is called the direct product. To be more precise,

Definition: Let {M_{j}}_{j∈J} be a family of R-modules. The direct product of this family is the R-module

Π_{j∈J} M_{j} = {(m_{j})_{j∈J} | m_{j} ∈ M_{j}}

(with addition and the R action defined componentwise) along with a family of onto mappings

π_{j} : Π_{j∈J} M_{j} → M_{j}

(called projections) where π_{j}((m_{j})_{j∈J}) = m_{j} .

Note: For J a finite set the direct sum and the direct product are the same modules, but that is obviously not the case if J is infinite. One need think only of the difference between polynomials and powerseries!

The direct product also satisfies a Universal Mapping Property.

Proposition: Let V be any R-module and suppose there are R-module homomorphisms ρ_{j} : V → M_{j} for every j.

Then there is a unique map ρ : V → Π_{j∈J} M_{j} such that π_{j} ∘ ρ = ρ_{j}.

Proof: Let ρ(v) = (ρ_{j}(v))_{j∈J}. It is an easy matter to verify that this works and is unique. ⊓ ⊔

Note that this UMP says that a map to the Direct Product is determined if we know what it is when followed by all the projections.

Exercise 2D: Let M = ⊕_{i=1}^{m}M_{i} and N = ⊕_{j=1}^{n}N_{j} and let Φ : M → N be any R-module homomorphism. Show that Φ may be represented by a matrix (ϕ_{ij}) where ϕ_{ij} : M_{j} → N_{i}.

The notions of direct sum and direct product for modules plays a very fundamental role in our understanding of modules and we shall have occasion to return to these ideas later when we discuss more general categories.

We continue our discussion of some basic ideas about R-modules.

Definition-Proposition: Let M be an R-module and let S ⊆ M be any subset of M. Then

< S >= {∑_{s∈S} r_{s}s | r_{s} ∈ R and s ∈ S, r_{s} ≠ 0 for only finitely many s}

is a submodule of M and is called the submodule of M generated by S.

Notes-Definitions: 1). < S > is the smallest submodule of M containing S. 2). If < S >= M, then S is said to generate M and S is called a set of generators for M. 3). If M has a finite set of generators then M is called a finitely generated R-module.

4). A set of generators S of M is called a basis for M if every element of M can be uniquely expressed as a combination of the elements of S.

Exercise 2C: Not every R-module with a finite generating set has a basis. Show that the ideal I = (x, y) ⊂ k[x, y] does not have a basis.

Definition: The R-module M is called free if it has a basis.

Proposition: M is a free R-module with basis S, ⇔

M ≃ ⊕_{s∈S}R_{s} where R_{s} ≃ R .

Proof: ⇒: Let m ∈ M then m = ∑_{s∈S} r_{s}s uniquely with all but a finite number of the r_{s} ≠ 0.

Let ϕ : M → ⊕_{s∈S}R_{s} be given by ϕ(m) = (r_{s})_{s∈S} where the r_{s} are the coefficients of m above. By the uniqueness it is easy to see that ϕ is an R-module homomorphism.

Define ϕ_{s} : R_{s} → M by ϕ_{s}(r_{s}) = r_{s}s. By the UMP for direct sums there is a map χ : ⊕R_{s} → M such that χ ∘ ι_{s} = ϕ_{s}.

Claim: χ and ϕ are inverse to each other. Proof: ϕχ((r_{s})) = ϕ((ϕ_{s})((r_{s})) = ϕ(∑_{s∈S} r_{s}s) = (r_{s}).

Also χϕ(∑ r_{s}s) = χ((r_{s})) = ∑_{s} r_{s}s.

⇐ I’ll leave this as a simple exercise.

Sect.2A

Free modules have a nice universal property.

Proposition: Let F be a free R-module on the set S and let M be an R-module.

If f is any \underline{set} map from S to M, then there is a unique R-module homomorphism \overline{f} : F → M such that \overline{f} restricted to S is f.

Proof: If we set f(s) = m_{s} and let

\overline{f}(∑ r_{s}s) = ∑ r_{s}m_{s}

then it is easy to see that \overline{f} is a homomorphism with the desired property and is the only homomorphism extending f to F. ⊓ ⊔ Proposition: Any R-module M is a quotient of a free module.

Proof: Let F be the free module F = ⊕_{m∈M}R_{m}. Let S be the standard basis for F and let f : S → M be defined by:

f(0, ⋯, 0, 1_{m}, 0, ⋯) = m .

Clearly f is onto and thus extends to an onto homomorphism \overline{f} : F → M and so M ≃ F/Ker(\overline{f}). ⊓ ⊔

There is another interesting property that free modules have.

Proposition: Let F be a free module, and consider the following diagram

F
↓ ϕ
M ⟶^{β} N → 0

where β : M → N is onto and ϕ : F → N is any homomorphism.

Then, one can find a map χ_{ϕ} = χ : F → M such that β ∘ χ = ϕ.

Proof: Let S be a set which generates F. Then ϕ is determined by its values on S. Since β is onto we can choose t = β^{−1}(ϕ(s)) for every s ∈ S. Define g(s) = t. Then β ∘ g(s) = ϕ(s) for every s ∈ S. If we let χ be the homomorphism which extends g to F, then χ works. ⊓ ⊔

Remark: Coupling this proposition with our earlier remarks we see that if

0 → A → B → C → 0

is an exact sequence of R-modules and F is a free R-module then

0 → Hom_{R}(F, A) → Hom_{R}(F, B) → Hom_{R}(F, C) → 0

is also exact.

Thus, if F is a free R-module we have that Hom_{R}(F, −) is an exact functor.

In the light of this result we make the following definition.

Definition: An R-module P is called projective if Hom_{R}(P, −) is an exact functor.

From what we have already seen, free modules are all projective.

Proposition: {P_{i}}_{i∈I} are all projective R-modules if and only if ⊕_{i∈I}P_{i} is a projective R-module.

Proof: To keep the notation down, I’ll only do this for the case of two summands.

⇒: We just need to show that if

M⟶^{β}N → 0

is exact and

ϕ : P_{1} ⊕ P_{2} ⟶ N

then we can “lift ” ϕ to M.

Let ι_{P_{j}}, be the injections given as part of the definition of the direct sum. Then, ϕ ∘ ι_{P_{1}} can be lifted since P_{1} is projective. I.e. we have

χ_{1} : P_{1} → M such that β ∘ χ_{1} = ϕ ∘ ι_{P_{1}} .

Similarly we have

χ_{2} : P_{2} → M such that β ∘ χ_{2} = ϕ ∘ ι_{P_{2}} .

Then χ = (χ_{1}, χ_{2}) : P_{1} ⊕ P_{2} → M is the R-module homomorphism guaranteed by the UMP such that χ ∘ ι_{P_{j}} = χ_{j}.

We rewrite

β ∘ χ_{j} = β ∘ (χ ∘ ι_{P_{j}}) = ϕ ∘ ι_{P_{j}} .

I.e. β ∘ χ and ϕ are both R-module homomorphisms out of P_{1} ⊕ P_{2} which agree after composition with the ι_{P_{j}} for j = 1, 2. By the UMP for direct sums we obtain that β ∘χ = ϕ as we wanted to show. ⊓ ⊔

Definition: The short exact sequence

0 → A⟶^{f}B⟶^{g}C → 0

is called a split exact sequence if there is an R-module homomorphism h : C → B such that g ∘ h = 1_{C}. If it exists, we call h a “splitting”.

Example: The exact sequence

0 → A⟶^{ι_{A}}A ⊕ C⟶^{π_{C}}C → 0

is a split exact sequence with splitting C⟶^{ι_{C}}A ⊕ C.

Proposition: Let

0 → A⟶^{f}B⟶^{g}C → 0

be a split exact sequence with splitting σ. Then, there is an isomorphism

ϕ : A ⊕ C ⟶ B

such that we have a commutative diagram

B
f ↗  ↘ g
0 → A ϕ ↑ C → 0
ι_{A} ↘  ↗ π_{C}
A ⊕ C

Proof: We have f : A → B and σ : C → B thus, by the UMP for direct sums, there is a unique map ϕ : A ⊕ C → B such that

i) f = ϕ ∘ ι_{A} and ii) σ = ϕ ∘ ι_{C} .

From i) we get the commutativity of the left hand triangle. To see the commutativity of the right hand triangle, consider

g ∘ ϕ(a, c) = g(f(a) + σ(c)) = (gf)(a) + (g ∘ σ)(c) = 0 + id(c) = c = π_{C}(a, c)

as we wanted to show.

Finally, using the 5-Lemma, we see that the map ϕ is an isomorphism. ⊓ ⊔

We can put all these observations together to get a very nice characterization of projective R-modules.

Theorem: The following are equivalent:

1). P is a projective R-module; 2). The functor Hom_{R}(P, −) is an exact functor; 3). If B→^{f}P → 0 is exact then there is an R-module homomorphism g : P → B such

that f ∘ g = 1_{P}; 4). Every short exact sequence ending in P splits; 5). P is a direct summand of every R-module of which it is a quotient; 6). P is a direct summand of a free R-module.

Proof: 1) ⇔ 2): This is the definition.

2) ⇒ 3): We have:

0 → Ker(f)→^{i}B→^{f}P → 0

exact, so f_{∗} : Hom_{R}(P, B) → Hom_{R}(P, P) is onto. Since 1_{P} ∈ Hom_{R}(P, P) there is a g ∈ Hom_{R}(P, B) such that f_{∗}(g) = 1_{P}. That is what we wanted.

3) ⇒ 4): This is immediate from the definition.

4) ⇒ 5): If M⟶^{f}P → 0 is exact then

0 → Ker(f) → M⟶^{f}P → 0

is exact. From 4) we get that P is a direct summand of M.

5) ⇒ 6): Since every module is a homomorphic image of a free module, this is immediate from 5).

6) ⇒ 1): P is a direct summand of a free (and hence projective) R-module and hence, by an earlier proposition, P is projective.

⊓ ⊔

Exercise 3A:

i). Prove that ℤ_{n} is \underline{not} a projective ℤ-module. ii). Using the structure theorem for finitely generated abelian groups show that a f.g.

projective ℤ-module must be free. Hence, for f.g. ℤ-modules the notions of “projective” and “free” are the same.

iii). Show that if n is not a prime power there always exist f.g. projective ℤ_{n}-modules which are not free.

(We’ll see later that if n is a prime power then every f.g. projective ℤ_{n}-module is free.)

Sect2B

We made the notion of a projective R-module from the functor Hom_{R}(M, −). With that in mind, we make the following “co-definition”.

Definition: An R-module N is called injective if Hom_{R}(−, N) is an exact functor.

From what we have already seen, N is an injective R-module if and only if whenever

0 → A⟶^{f}B

is exact, then

Hom_{R}(B, N)⟶^{f^{∗}}Hom_{R}(A, N) → 0

is also exact.

I.e. given a diagram

0 → A ⟶^{f} B
ϕ ↓
N

there is a map χ : B → N making it commute.

To explore this idea we need to extend a theorem from the previous lecture.

Lemma: The short exact sequence

0 → A⟶^{f}B⟶^{g}C → 0

splits if and only if there is a map j : B → A such that j ∘ f = 1_{A}.

Proof: ⇒: By the previous proposition about split exact sequences we have: B

f ↗  σ ↖ g ↘
0 → A ϕ ↑ C → 0
ι_{A} ↘  ↗ π_{C}

A ⊕ C

-17- 9/7/100 where ϕ is an isomorphism, g ∘ σ = 1_{C}, ι_{A} and π_{C} are the maps associated with the direct sum and direct product, and all diagrams commute.

Let j = π_{A} ∘ ϕ^{−1}. Then j ∘ f = π_{A} ∘ ϕ^{−1} ∘ f. But, f = ϕ ∘ ι_{A}, so

j ∘ f = π_{A} ∘ ϕ^{−1} ∘ (ϕ ∘ ι_{A}) = π_{A} ∘ ι_{A} = 1_{A}

as we wanted to show.

⇐: Given j : B → A, g : B → C then, by the UMP for the direct product, we have

Φ : B → A ⊕ C such that i) π_{A} ∘ ϕ = j and ii) π_{C} ∘ Φ = g .

Consider

B
f ↗ j ↙  g ↘
0 → A Φ ↓ C → 0
ι_{A} ↘  ↗ π_{C}
A ⊕ C

From ii) above, we get the commutativity of the left hand triangle.

Now,

(Φ ∘ f)(a) = Φ(f(a)) := (j(f(a)), g(f(a))) = (a, 0) = ι_{A}(a)

which, from the previous proposition about split exact sequences, is all we have to show. ⊓ ⊔

We know give several characterizations of injective modules, similar (in the “co” sense) to those we gave for projective modules.

Theorem: The following statements are equivalent:

1) N is an injective R-module;

2) If

0 → A⟶^{f}B⟶^{g}C → 0

is a short exact sequence, then

0 → Hom_{R}(C, N)⟶^{g^{∗}}Hom_{R}(B, N)⟶^{f^{∗}}Hom_{R}(A, N) → 0

is also a short exact sequence.

3) If 0 → N⟶^{f}B is exact then there is an R-module homomorphism h : B → N such that h ∘ f = 1_{N}.

4) N is a direct summand of every module M which contains it.

Proof:

1) ⇔ 2): This is the definition.

2) ⇒ 3): An easy exercise (Exercise 4A(5)).

3) ⇔ 4): trivial

4) ⇒ 1): This is the difficult part of the proof and will take quite a while to prove. In order to do this, we will need two lemmas, one easy, one difficult. Let me state the easy one first.

Lemma 1: Let N = Π_{i∈I}N_{i}. Then

N is an injective R-module ⇔ each N_{i} is an injective R-module.

Proof: Exercise 4B (10)

Lemma 2: Every R-module is contained in an injective R-module.

Proof: Postponed for a bit.

Assuming that the two lemmas are proved, we shall now show how to prove 4) ⇒ 1).

Given N, an R-module which is a direct summand of any R-module which contains it, include it in an injective module ℑ, as guaranteed by Lemma 2. Then ℑ = N ⊕ N′ by assumption. But, by Lemma 1, ℑ injective implies that N (and N′) is injective and we are done. ⊓ ⊔

We now return to the problem of finding a proof of Lemma 2. In order to get started on this we first try to get some examples of injective modules. This is non-trivial! We will start by considering injective modules over Principal Ideal Domains (PID’s). This includes ℤ and k[x], k a field. We start a bit more generally.

Definition: Let R be an integral domain (this means, among other things, commutative with an identity). An R-module, D, is called divisible if, for every d ∈ D and every 0 ≠ r, r ∈ R, there is a c ∈ D such that rc = d (i.e. we can “divide” by r in D).

Examples:

1) If R = ℤ and D = ℚ, the rationals (under addition), then ℚ is a divisible ℤ-module.

2) If R = ℤ and D = ℚ/ℤ then D is divisible.

3) ℤ is not a divisible ℤ-module and neither is ℤ_{n}.

Exercise 4C: i). (10) No f.g. abelian group is a divisible ℤ-module. ii). (10) The direct product of divisible R-modules is again divisible.

We are now ready for our main result about injectives over a PID.

Theorem: Let R be a PID. The R-module M is injective if and only if it is a divisible R-module.

Proof: ⇒: (Notice that this direction does not use the fact that R is a PID, only that R is a domain. i.e. This proves: R a domain and M injective ⇒ M divisible.)

Suppose that M is injective and let m ∈ M and let 0 ≠ r ∈ R. We want to find m′ such that rm′ = m.

Let ϕ : R → M by ϕ(1) = m. Since R is a free R-module this is enough to well-define ϕ. Similarly, let χ : R → R be defined by χ(1) = r. Since R is an integral domain, ker(χ) = 0, i.e. have the following diagram:

0 → R ⟶^{χ} R
ϕ ↓
M

Since M is injective, there is a map δ : R → M such that δ ∘ χ = ϕ.

Then, m = ϕ(1) = δ ∘ χ(1) = δ(r) = rδ(1). Then m′ = δ(1) ∈ M is the required element.

⇐: As for the converse, suppose that M is divisible and that we have a diagram

0 → A ⟶^{μ} B
α ↓
M

-20- 9/7/100 where μ is 1 − 1. We need a map ϕ : B → M such that ϕ ∘ μ = α.

Consider pairs (A_{j}, α_{j}) where A ⊆ A_{j} ⊆ B (we are thinking of “μ” as embedding A in B) and such that α_{j} : A_{j} → M and α_{j} restricts to α on A. Let Φ be the set of all such pairs.

Φ is non-empty since (A, α) ∈ Φ .

Define (A_{j}, α_{j}) ≤ (A_{j′}, α_{j′}) if A_{j} ⊆ A_{j′} and α_{j′} restricted to A_{j} is α_{j}. This ordering is inductive, i.e. every chain has an upper bound (namely (∪A_{j}, ∪α_{j}),

where ∪α_{j} is defined by:

if a ∈ ∪A_{j} ( then a ∈ A_{l} for some l), and (∪α_{j})(a) = α_{l}(a) .

Using the definition of the ordering, this map is well-defined.

By Zorn’s Lemma, Φ contains maximal elements. Let (\overline{A}, \overline{α}) ∈ Φ be a maximal element of Φ. If we can show that \overline{A} = B then we are done. So, suppose that \overline{A} ⊈ B and let b ∈ B, b ∉ \overline{A}.

Consider

\{r | rb ∈ \overline{A}, r ∈ R\} = I .

It is easy to see that I is an ideal of R, and since R is a PID, then I = (λ), λ ∈ R. We now try to extend the homomorphism \overline{α} to the module < \overline{A}, b >.

Case 1: If λ ≠ 0.

Since λb ∈ \overline{A} we must have \overline{α}(λb) = m ∈ M. Since M is divisible, m = λm′ i.e. \overline{α}(λb) = λm′.

We extend \overline{α} to < \overline{A}, b > by:

\overline{α}(\overline{a} + sb) = \overline{α}(\overline{a}) + sm′ .

The only question is: Is this extension well-defined, and is it a homomorphism? Well-defined: I.e. suppose \overline{a} + sb = \overline{a}_{1} + s_{1}b. Is it true that:

\overline{α}(\overline{a}) + sm′ = \overline{α}(\overline{a}_{1} + s_{1}m′) ?

But, \overline{a} + sb = \overline{a}_{1} + s_{1}b implies that \overline{a} − \overline{a}_{1} = (s_{1} − s)b. Therefore, s_{1} − s = uλ for some u ∈ R. I.e. \overline{a} − \overline{a}_{1} = uλb.

Thus, \overline{α}(\overline{a} − \overline{a}_{1}) = \overline{α}(uλb) = u\overline{α}(λb). But, \overline{α}(\overline{a} − \overline{a}_{1}) = \overline{α}(\overline{a}) − \overline{α}(\overline{a}_{1}) = uλm′ = (s_{1} − s)m′, which is what we wanted to show.

Thus, the map \overline{α} is well-defined. It is easy to see that it is an R-module homomorphism. So, this contradicts the maximality of (\overline{A}, \overline{α}), unless \overline{A} = B.

Case 2: If λ = 0.

In this case we do exactly as above, but we just choose c ∈ M arbitrarily and define: \overline{α}(\overline{a} + λb) = \overline{α}(\overline{a}) + λc.

Is this well-defined? Suppose that \overline{a} + λb = \overline{a}_{1} + λ′b, do they have the same image under this extended \overline{α}?

But, as before, \overline{a} − \overline{α}_{1} = (λ′ − λ)b. Since I = (0) we must have λ = λ′. But that, in turn, implies that \overline{a} = \overline{a}_{1}, i.e. the representation is unique and so there is nothing to showing that the extension of \overline{α} is well-defined. It’s an easy exercise to show that it is also an R-module homomorphism. ⊓ ⊔

Exercise 4D: 1). (10) Prove that every quotient of a divisible R-module is again a divisible R-module. 2). (5) If R is a PID then every quotient of an injective R-module is again an injective R-module. 3). (10) The R-module ℑ is injective if and only if for every (left) ideal J ⊂ R and any R-module homomorphism ϕ : J → ℑ, the following diagram

0 → J ⟶^{ι} R
ϕ ↓
ℑ

can be completed with a map χ : R → I such that χ ∘ ι = ϕ. 4). (10) Let R = ℤ_{m}. Prove that R is an injective R-module.

Sect.3A

Theorem: Every abelian group G can be embedded into an injective abelian group.

Proof: If G is free then G ≃ ⊕ℤ. Since ⊕ℚ is divisible (hence injective) and clearly contains ⊕ℤ we are done in this case.

If now G is an arbitrary abelian group then we know that we can find a free abelian group F such that:

0 → K → F → G → 0

is exact.

Now G ≃ F/K ⊆ D/K, where F ⊆ D, D a divisible abelian group. By a previous exercise, D/K is divisible and hence injective as a ℤ-module. ⊓ ⊔

Let R be any ring and let C be a right R-module and let G be an abelian group. We can form the abelian group Hom_{ℤ}(C, G). But, it is possible to view this abelian group as a left R-module as follows:

if f ∈ Hom_{ℤ}(C, G) and r ∈ R then (rf)(c) = f(cr) .

(I’ll leave the verification that i). rf is a ℤ-module homomorphism and ii). the action of R on Hom_{ℤ}(C, G) makes Hom_{ℤ}(C, G) into a left R-module , as a (simple) exercise.)

It is also easy to see that if ϕ : C → D is a homomorphism of R-modules then the induced map on abelian groups

ϕ^{∗} : Hom_{ℤ}(D, G) → Hom_{ℤ}(C, G)

is an R-module homomorphism. Similarly, if χ : G_{1} → G_{2} is a homomorphism of abelian groups then the induced homomorphism of abelian groups

χ_{∗} : Hom_{ℤ}(C, G_{1}) → Hom_{ℤ}(C, G_{2})

is an R-module homomorphism. (I suggest you fill in the details of this verification.)

The reason for introducing this R-module is apparent from the next Lemma.

Lemma: If D is an injective abelian group then Hom_{ℤ}(R, D) is an injective (left) Rmodule.

Proof: By an earlier exercise, it is enough to show: if L is any left ideal of R and we are given a diagram:

0 → L ⟶^{j} R
f ↓
Hom_{ℤ}(R, D)

then f can be lifted to R.

So, if x ∈ L then f(x) : R → D is a ℤ-module homomorphism. In particular, (f(x))(r) ∈ D.

We can think of f as a function of two variables, f(x, y), where x ∈ L and y ∈ R. Thought of this way we have:

f(x, y_{1} + y_{2}) = f(x, y_{1}) + f(x, y_{2}) ,
f(x_{1} + x_{2}, y) = f(x_{1}, y) + f(x_{2}, y) ,
f(rx, y) = f(x, yr) .

Let g(x) = f(x, 1) for x ∈ L. This is a map from L → D. It is a ℤ-module homomorphism. Since D is an injective abelian group, this map can be extended to a homomorphism of abelian groups, g′, g′ : R → D. We use this map g′ to extend f to f′, as follows:

f′(x, y) = g′(yx) where x ∈ R and y ∈ R .

Claim: f′(x, y) satisfies the following properties, for x_{1}, x_{2}, y_{1}, y_{2}, x, y ∈ R:

i) f′(x_{1} + x_{2}, y) = f′(x_{1}, y) + f′(x_{2}, y) .
ii) f′(x, y_{1} + y_{2}) = f′(x, y_{1}) + f′(x, y_{2}) ,
iii) f′(rx, y) = f′(x, yr) for r ∈ R ,
iv) f′(x, y) = f(x, y) = f(yx, 1) = g(yx) for x ∈ L and y ∈ R.

Proof: Combining i)−iii) gives that we have an R-module homomorphism from R to Hom_{ℤ}(R, D) and iv) shows that this homomorphism extends f.

These are all very simple verifications and I leave them to you as an easy exercise to complete the proof of the Lemma. ⊓ ⊔

Theorem: If M is a left R-module then M can be embedded in an injective (left) Rmodule.

Proof: First think of M as an abelian group. Then, we have an exact sequence of abelian groups

0 ⟶ M⟶^{ι}D

where D is an injective abelian group.

It follows that

0 → Hom_{ℤ}(R, M)⟶^{ι_{∗}}Hom_{ℤ}(R, D)

is also exact.

Claim: M ≃ Hom_{R}(R, M) ⊆ Hom_{ℤ}(R, M), where the isomorphism is of R-modules.

Proof: Clearly, any homomorphism of R-modules is a homomorphism of abelian groups so Hom_{R}(R, M) ⊆ Hom_{ℤ}(R, M).

Define

χ : Hom_{R}(R, M) ⟶ M by χ(f) = f(1) .

Since R is a free R-module, χ is clearly an onto function. Also,

χ(f_{1} + f_{2}) = (f_{1} + f_{2})(1) = f_{1}(1) + f_{2}(1) = χ(f_{1}) + χ(f_{2}) .

Moreover,

χ(rf) = (rf)(1) = f(1r) = f(r) = rf(1)

this last since f is an R-module homomorphism. Since rf(1) = rχ(f) we are done with the claim.

The Theorem is an immediate consequence of the Claim. ⊓ ⊔

That ends, for the moment, our general discussion of modules over rings. We shall come back to a more specialized study of these things later in the term.

§2: Categories and Functors

Now that we have a few examples under our belt I would like to spend a little time to make the fundamental definitions from category theory that we shall be using. This is only a very cursory introduction to some of the ideas of category theory. A more detailed discussion would take us too far afield since, although we will use the general ideas, we won’t need (at first encounter) some of the more subtle and interesting aspects of duality from category theory.

We move to give the first (rambling) definition.

Definition: A category ℭ is:

i) a class of objects, usually denoted Obj(ℭ), and usually written as A, B, C, … ∈ Obj(ℭ);

ii) for every pair of objects A, B ∈ Obj(ℭ), we have a set, denoted ℭ(A, B). The elements of this set are called morphisms from A to B. Although, in general, these morphisms need not be functions (we’ll see plenty of examples below), we use the notation:

if f ∈ ℭ(A, B) then f : A → B

and call A the domain of f and B the codomain of f.

iii) to each triple A, B, C ∈ Obj(ℭ) there is a law of composition, i.e. a function

ℭ(A, B) × ℭ(B, C) → ℭ(A, C)

Again, following the “notation” of functions for the “not necessarily functions” of ℭ(−, −) we usually denote the composition law as “composition of morphisms” , i.e.

(f, g) ⟶ g ∘ f or, simply gf .

We require three axioms for an arrangement as above to be a category.

A1: The sets ℭ(A, B) and ℭ(C, D) are disjoint unless A = C and B = D.

A2: The composition law is associative, i.e.

h(gf) = (hg)f .

A3: There is, for each A ∈ Obj(ℭ), an element in ℭ(A, A), denoted 1_{A}, such that

f ∘ 1_{a} = f for every f ∈ ℭ(A, B) and 1_{A} ∘ f = f for every f ∈ ℭ(C, A) .

Remarks on the Axioms: 1). Although A1 is a familiar idea about functions (i.e. we don’t say two functions are equal unless they also have the same domain and codomain i.e. we distinguish between the function f(x) = 5 if we have domain all the reals and codomain either the reals, or the positive reals, or simply the number 5 !), in homological algebra the distinction is more than just pedantic.

Consider the following example from algebraic topology: let f_{1} : S^{1} → ℝ^{2} and f_{2} : S^{1} → ℝ^{2} −{(0, 0)}. I.e. the first map embeds S^{1} in the real plane with center at the origin and the second map embeds S^{1} in the real punctured plane, with the center of the circle still at the origin. It’s crucial to distinguish between these two functions in homotopy theory since the first is homotopic to zero while the second is not! 2). Note that composition is defined \underline{only} if codomain (f) = domain(g). 3). The element 1_{A} ∈ ℭ(A, A) is uniquely determined by A (an easy exercise).

Examples: There are many examples of categories all over mathematics. In each of the examples below I will describe, first, the objects of the category and then say what the morphisms are.

1). Obj(ℭ) consists of all sets. If A and B are any two sets then ℭ(A, B) consists of all the functions with domain A and codomain B.

2). This time Obj(ℭ) consists of all topological spaces and if A and B are topological spaces then ℭ(A, B) consists of all continuous maps from A to B. Notice that the same set would appear more than one time among the objects of this category if it came equipped with two different topologies.

3). Let Obj(ℭ) consist of all groups and if A and B are groups then ℭ(A, B) is all the group homomorphisms from A to B.

4). This time let Obj(ℭ) consist of all abelian groups and ℭ(A, B) the group homomorphisms from A to B.

The example of 4) is called a subcategory of the category in 3) since the objects of 4) are a subset of the objects of 3) and the morphisms between two objects of the category in 4) are a subset of the morphisms between those objects considered in the category 3).

This example is somewhat special in that whenever A and B are objects of 4), the morphisms between A and B are the same in both categories. When this is the case the smaller category is called a full subcategory of the larger one.

5). Fix a field F and let Obj(ℭ) consist of all F -vector spaces. If V and W are two F -vector spaces then we let ℭ(V, W) be all the linear transformations from V to W.

6). Let Obj(ℭ) consist of all rings, and if A and B are two rings let ℭ(A, B) be all the ring homomorphisms from A to B.

7). This time let Obj(ℭ) consist of rings with identity and if A and B are such rings then ℭ(A, B) should only consist of ring homomorphisms which preserve the identity.

8). Fix a ring R and let Obj(ℭ) consist of all the right (left) R-modules and let ℭ(A, B) be the R-module homomorphisms between A and B.

9) Fix a ring R and let Obj(ℭ) consist of all chain complexes of R-modules. We should define the morphisms in this category.

Definition: A collection of R-module homomorphisms {f_{n}}, n ∈ ℤ is a morphism between the chain complexes {C_{i}, d_{i}} and {C′_{i}, d′_{i}} of R-modules if

for each i ∈ ℤ, f_{i} : C_{i} → C′_{i}

and we have, for each i, the following commutative diagram

\begin{matrix}C_{i} & ⟶^{f_{i}} & C′_{i} \\ d_{i} ↓ & & ↓ d′_{i} \\ C_{i−1} & ⟶^{f_{i−1}} & C′_{i−1} \end{matrix} .

Exercise 5A: (10) Show that this is a category. Note that in this case the morphisms are not functions but sets of functions.

Sect.3B

10). In this example let the objects be topological spaces and the morphisms between two topological spaces to be homotopy classes of maps between them.

Using facts from algebraic topology one can see that this is indeed a category. Note also that in this example the morphisms are not functions but “homotopy classes” of functions!

11). (A strange example!)

A category with one object, denoted *, is the same thing as a monoid M, i.e. if we let ℭ(∗, ∗) = M.

In particular, any group G can be considered as a category with one object.

12). Recall that a partially ordered set (i.e. a poset) is a set 𝔓 with a reflexive, transitive and antisymmetric relation “≤ ”. We can regard 𝔓 as a category in the following way:

Obj(𝔓) = {p | p ∈ 𝔓} ,
𝔓(p, q) = \{ \begin{matrix}∅ & unless p ≤ q; \\ \{ι_{pq}\} & one morphism from p to q if p ≤ q. \end{matrix}

Some of the categories I’ve described above are so fundamental that they even have their own names! E.g. 1), the “mother of all categories” is called Sets, while 3) is called Groups, 4) is called Ab and 6) is called Rings. If we restrict to left R-modules, then 8) is called R-mod, while if we consider only right R-modules, the category is called mod-R. Those are among the most important names. If we start using a category a lot it is often useful to give it a name. I’ll leave those baptisms until they are needed.

Further Categorical Terminology

An element f ∈ ℭ(A, B) is called invertible if there is an element g ∈ ℭ(B, A) such that g ∘ f = 1_{A} and f ∘ g = 1_{B}.

It’s easy to see that if f is invertible then g is uniquely determined by f and so we may call g = f^{−1}. As always, we have (f^{−1})^{−1} = f and (f ∘ g)^{−1} = g^{−1} ∘ f^{−1} when f and g are invertible.

If there is an invertible morphism in ℭ(A, B) we say that A is isomorphic to B in ℭ. This is easily seen to be an equivalence relation on the objects of ℭ, which has been given a different name in different categories.

E.g. 1−1 correspondence in the category of sets and set mappings, homeomorphism in the category of topological spaces and continuous maps, group isomorphism in the category of groups and group homomorphisms. etc.

Sometimes a category ℭ contains a special object, often denoted 0, with the property that:

for any object X ∈ Obj(ℭ) the sets ℭ(X, 0) and ℭ(0, X) contain a single element.

For example, the group with one element is such an object in the category of groups (or in the category of abelian groups). The module consisting only of the zero element of R is such an object in the category of left (right) R-modules. The ring with only the zero element works in the category of rings and ring homomorphisms.

An object like the one described above is called a zero object in the category ℭ. Clearly, any two zero objects in a category ℭ are isomorphic.

If ℭ has a zero object 0, then for any X, Y ∈ Obj(ℭ) we have a distinguished morphism in ℭ(X, Y) given by

X ⟶ 0 ⟶ Y .

This morphism is called the zero morphism of ℭ(X, Y) and denoted 0_{XY}. It has the property that if f ∈ ℭ(Y, Z) then f ∘ 0_{XY} = 0_{XZ} and if g ∈ ℭ(A, X) then 0_{XY} ∘ g = 0_{AY}.

Exercise 6A: (10) The map 0_{XY} ∈ ℭ(X, Y) does not depend on the zero object chosen in the category ℭ.

Definitions: An object I in a category ℭ is called an initial object if ℭ(I, X) consists of one element for every X ∈ Obj(ℭ); it’s called a terminal object (or coinitial object) if ℭ(X, I) consists of one element for every X ∈ Obj(ℭ).

Clearly, being a zero object in the category ℭ implies that the object is both an initial and terminal object. The converse is also obviously true.

Having said that, it is interesting to note that in the category SETS, the empty set, ∅, is an initial object and every one point set is a terminal object. (Explain about the initial object!) It’s easy to show that all initial objects in a category are isomorphic and all terminal objects are isomorphic.

Exercise 6B: (10) Prove that in the category SETS there is no zero object.

A morphism f : B → C in a category ℭ is called a monomorphism (or mono or monic) if:

whenever e_{1}, e_{2} ∈ ℭ(A, B) and f ∘ e_{1} = f ∘ e_{2} then e_{1} = e_{2} .

In SETS the monos are precisely the 1 − 1 maps (in fact, that’s also true in AB and R-mod and mod-R). In order to remember where the cancellation takes place, monos are often written f : A ↣ B where the extra “>” is on the left – the side on which we can cancel.

Exercise 6C: (10) Let ℭ be a category in which the objects are all sets and the morphisms are all functions. Show that the monomorphsims are precisely the 1 − 1 maps.

Analogously f : B → C ( f a morphism in ℭ(B, C)) is called an epimorphism (or epic or an epi) if

whenever e_{1}, e_{2} ∈ ℭ(C, D) and e_{1} ∘ f = e_{2} ∘ f then e_{1} = e_{2} .

As above, if f : B → C is an epi we write the arrow as ↠ to remember where the cancellation can take place.

If the underlying objects of the category ℭ are sets than any “onto” morphism is an epi but the converse is not true.

Exercise 6D: a) (10) Show that in Rings the inclusion of ℤ into ℚ is an epi. More generally, whenever R is commutative with identity and S is a multiplicatively closed set, not containing 0, then the natural map R → S^{−1}R (given by r → [r/1]) is an epi in Rings.

b) (5) Show that the inclusion of ℚ into ℝ is an epi in the category of Hausdorff topological spaces and continuous maps.

Definition: Let ℭ be a category with a zero object and let f ∈ ℭ(B, C). A kernel of f is: i) a morphism ι ∈ ℭ(A, B) such that f ∘ ι = 0 in ℭ(A, C); ii) which satisfies the following Universal Mapping Property: if g ∈ ℭ(A′, B) is any morphism such that f ∘ g = 0 then there is a unique morphism χ : A′ → A such that:

A′

g ↘
χ ↓  B →^{f} C
ι ↗
A

and the triangle commutes. In some sense, (A, ι) is the “biggest” pair with the property that f ∘ ι = 0.

Claim: ι is monic. Proof: Suppose we have h_{1}, h_{2} : D → A where ιh_{1} = ιh_{2}. Then

f(ιh_{1}) = f(ιh_{2}) = 0

i.e. (fι)h_{1} = 0 and (fι)h_{2} = 0, and we have a diagram:

A → B →^{f}
ι ↑
A
h_{1} ↑
D

So, we have a unique map χ : D → A such that χι = h_{1}ι. But χ = h_{1} = h_{2} both work! So, h_{1} = h_{2}. ⊓ ⊔

It is easy to see that any two kernels of f : B → C are isomorphic.

In like manner we can define a cokernel of a morphism f : B → C as

i) a morphism p : C → D such that pf = 0 and ii) satisfying the following Universal Mapping Property:

if g : C → D′ is any morphism such that gf = 0 then there is a unique morphism χ : D → D′ such that the following diagram commutes:

D

p ↗
B ⟶^{f} C  χ ↓
g ↘
D′

-32- 9/7/100 In a sense, the pair (D, p) is the “smallest” with the property that pf = 0.

It’s easy to show (using the UMP) that a coker is an epi and any two cokernels of f : B → C are isomorphic.

Continuing, we define a product of objects {C_{i}}_{i∈I} in the category ℭ, as an object (denoted Π_{i∈I}C_{i}) together with morphisms π_{j}, where

π_{j} : Π_{i∈I}C_{i} → C_{j}

satisfying the Universal Mapping property we described earlier in the lectures.

Likewise, a coproduct of objects {C_{i}}_{i∈I} in the category ℭ, is an object (denoted ∐ C_{i}) together with a family of morphisms

ι_{j} : C_{j} → ∐ C_{i}

satisfying the UMP we saw earlier. (Note: sometimes the coproduct is called a “sum”.)

We’ve seen that the category R-mod has products and coproducts for any family of objects.

Examples: i) If ℭ is the category of commutative rings with identity, then ℤ is an initial object. The zero ring (which has one element 0 = 1) is a terminal object. There is no zero object.

It’s easy to see that R_{1} × R_{2} (with pairwise operations) is a product in the category of RINGS, but it is NOT a coproduct! (There is a coproduct in this category, we’ll see later that it is R_{1} ⊗_{ℤ} R_{2}.) There are no kernels or cokernels in this category since there is no zero object.

ii) Let ℭ be GROUPS. The group {e} is a 0-object in ℭ. The cartesian product G_{1} × G_{2} is easily seen to be a product but not a coproduct in this category. There is a coproduct, but it is not so easy to describe.

Kernels are easy to describe. On the other hand, if f : G_{1} → G_{2} then coker(f) is

π : G_{2} → G_{2}/(normal subgroup generated by Im(f)).

Notice that the inclusion of a subgroup into a group is a mono, but it is not a kernel unless the subgroup is normal in the group. (I suggest you verify that a non-normal subgroup of a group is NOT a kernel in the category GROUPS.)

Sect.4A

Special Kinds of Categories An additive category ℭ is a category such that:

i) There is a zero object in the category; and

ii) For every A, B ∈ Obj(ℭ) we have that ℭ(A, B) is an abelian group and, moreover, the composition

ℭ(A, B) × ℭ(B, C) → ℭ(A, C)

is bilinear, i.e. if f, g ∈ ℭ(A, B) and h ∈ ℭ(B, C) then

h ∘ (f + g) = h ∘ f + h ∘ g

(and the bilinearity on the other side as well!)

iii) If A, B ∈ Obj(ℭ) then A ∏ B exists in ℭ.

It is easy to see that Ab is an additive category, as is R-mod and mod-R. The apparent asymmetry in the definition (which asks only for finite products, but not finite coproducts) is more apparent than real. It’s not hard to prove that in an additive category the two notions agree.

We now come to the most important kind of category that we will be considering this year, the abelian category. We give the complete definition.

Definition: An abelian category is a category ℭ which is an additive category and which satisfies, in addition, the following properties:

i). Every morphism in ℭ has a kernel and a cokernel. ii). Every monic is the kernel of its cokernel. iii). Every epi is the cokernel of its kernel.

Proposition: The category mod-R is an abelian category (and, we’ll see, is the prototype of an abelian category).

Proof: To see i). Let f : A → B be a homomorphism of R-modules and set

K(A) = {a ∈ A | f(a) = 0}

and let ι : K(f) → A the inclusion of K(f) into A.

Claim: The pair (K(f), ι) is a kernel of f.

Proof: Suppose we have a diagram

K(f) ⟶^{ι} A ⟶^{f} B
↗ g
C

and suppose that fg = 0.

If c ∈ C, then g(c) ∈ A and f(g(c)) = 0 and so g(c) ∈ K(f). Let χ : C → K(f) be defined by χ(c) = g(c). Then χ is the required map which makes the diagram commute and is easily verified to be an R-module homomorphism. ⊓ ⊔

Let Im(f) ⊆ B. Then Im(f) is a submodule of B and we can define

π : B → B/Im(f) .

Claim: The pair (π, B/Im(f)) is a cokernel of f.

Proof: Suppose we have a diagram

A ⟶^{f} B ⟶^{π} B/Im(f)
g ↘
C

where gf = 0. We want a unique map B/Im(f) → C making the diagram commute.

Define χ(\overline{b}) = g(b). As soon as one shows this is well-defined it is easily seen to be an R-module homomorphism and is the required map. ⊓ ⊔

As for ii), we must show

a) that every mono is the kernel of its cokernel.

So, let f : A → B be a mono. Then f is 1 − 1 (as we have seen!) We have just seen that the cokernel of f is the natural map π : B → B/Im(f). So, we must show that f is the kernel of π.

So, et g : C → B be such that πg = 0. We want to find a unique map χ : C → A making the appropriate diagram commute.

If c ∈ C then g(c) ∈ Ker(π) = Im(f) ≃ A. Thus, χ(c) = f^{−1} ∘ g(a) makes sense and is the only thing that will work.

b) Every epi is the cokernel of its kernel.

I.e. given an epi p : A ↠ B and ι : Ker(p) → A we must show that p is the cokernel of ι.

I.e. given g : A → C, we want to find a unique map χ : B → C making the appropriate diagram commute.

Let b ∈ B, then b = p(a), a ∈ A (I am using the fact that epis are onto in the category R-mod !!!). Define χ(b) = g(a). This is the only map making the diagram commute, so it’s enough to show it is well defined, i.e. if b ∈ B and b = p(a_{1}) = p(a_{2}) then g(a_{1}) = g(a_{2}).

But, p(a_{1}) = p(a_{2}) ⇒ p(a_{1} − a_{2}) = 0 ⇒ a_{1} − a_{2} ∈ Ker(p) ⇒ (gι)(a_{1} − a_{2}) = 0 ⇒ g(ι(a_{1})) = g(ι(a_{2})). Since ιa = a we are done.

Exercise 7A: a) (10) Prove that in an abelian category a morphism which is both mono and epi is an isomorphism.

b) (5) Show, with an example, that this is not necessarily true in an arbitrary category.

Having talked about kernels and cokernels it is possible to speak about short exact sequences in an abelian category.

Exercise 7B: a) (10) If 𝔄 is an abelian category show that Ch(𝔄), the chain complexes over 𝔄 is also an abelian category.

b) (5) Moreover, show that

0 → A_{•} → B_{•} → C_{•} → 0

is a short exact sequence of chain complexes if and only if

0 → A_{n} → B_{n} → C_{n} → 0

is a short exact sequence in 𝔄. (be careful! the objects here may not be sets!)

In the light of this exercise, we can thus talk about chain complexes of chain complexes of chain complexes of .....

The \underline{next} step of this process is interesting.

Definition: Let 𝔄 be an abelian category. A double complex or bicomplex over 𝔄 is a family {C_{p,q}} where C_{p,q} ∈ Obj(𝔄) such that:

d_{p,q} : C_{p,q} → C_{p−1,q} and d′_{p,q} : C_{p,q} → C_{p,q−1}

and d^{2} = 0 and (d′)^{2} = 0.

These double complexes are usually pictured as follows:

↓  ↓  ↓  ↓
⋯ → C_{p+1,q+1} →^{d} C_{p,q+1} →^{d} C_{p−1,q+1} →^{d} ⋯
d′ ↓  d′ ↓  d′ ↓
⋯ → C_{p+1,q}  →^{d} C_{p,q}  →^{d} C_{p−1,q}  →^{d} ⋯
d′ ↓  d′ ↓  d′ ↓
⋯ → C_{p+1,q−1} →^{d} C_{p,q−1} →^{d} C_{p−1,q−1} →^{d} ⋯
d′ ↓  d′ ↓  d′ ↓

where d_{p,q−1} ∘ d′_{p,q} + d′_{p−1,q} ∘ d_{p,q} = 0, i.e. each square is anti-commutative. (Note that this is not quite what the morphisms in Ch(Ch(𝔄)) would be like, but it would be OK if we changed the morphisms in every other column by a minus sign!)

The anticommutativity condition is useful because we can make two new complexes out of the double complex D above, called Total Complexes, as follows:

(T ot∏D)_{n} = ∏_{p+q=n} C_{p,q} , (T ot∐(D))_{n} = ∐_{p+q=n} C_{p,q}

where d̃= d + d′ defines a morphism (in each case) and d̃^{2} = 0 (check this out!).

Exercise 7C: (10) A double complex is bounded if it has only finitely many non-zero terms along each diagonal p + q = n (e.g. if it is concentrated in the first quadrant.)

Prove that T ot(C) is acyclic (i.e. has zero homology modules) when C is a bounded double complex with exact rows (or exact columns). You needn’t do this in an arbitrary abelian category – it is enough for you to do this for R-modules. FUNCTORS

So far, we have only been discussing one single category at a time, but, that is like studying only one group at a time. Just as we have seen that functions are very important in the study of various algebraic and geometric objects, we need to have the appropriate analogue in the study of categories. The appropriate notion appears to be that of a functor.

Definition: Let ℭ and 𝔇 be two categories. A (covariant) functor from ℭ to 𝔇, written

F : ℭ ⟶ 𝔇

is a rule which assigns to each A ∈ Obj(ℭ) an element F(A) ∈ Obj(𝔇) and to every morphism f ∈ ℭ(A, B) a morphism F(f) ∈ 𝔇(F(A), F(B)) such that

i). F(id_{A}) = id_{F(A)} , and

ii). F(gf) = F(g)F(f) .

Examples: 1). If R is any ring and M is in R-mod, then we have shown that Hom_{R}(M, −) can be viewed as a functor from the category R-mod to the category Ab.

2). If R is a commutative ring and M is a fixed R-module then Hom_{R}(M, −) is a functor from the category R-mod to itself.

3). The functor F : R-mod → Ab which \underline{forgets} that an R-module is an R-module and just remembers that it is an abelian group, is an example of, what is often called, a “forgetful” functor.

4). If A is an abelian group and we let

T (A) = {g ∈ A | ng = 0 for some n ∈ ℤ}

(called the torsion subgroup of A), then the assignment T : A → T (A) describes a functor from the category Ab to itself.

Remark: If, in the definition of a functor, we assume (instead) that

F(f) ∈ 𝔇(F(B), F(A)) and also that F(gf) = F(f)F(g) ,

then F is called a contravariant functor.

The functors Hom_{R}(−, N) (where R is a commutative ring and N is a fixed R-module) are contravariant functors from R-mod to itself.

Special Functors:

In explaining some additional functor possibilities, we shall assume in what follows, that F : ℭ ⟶ 𝔇 is a functor.

i). F is called faithful if, for all A, B, ∈ Obj(ℭ) then the map

ℭ(A, B) ⟶ 𝔇(F(A), F(B))

is a 1 − 1 map of sets.

The forgetful functor above is an example of such a functor.

ii). F is call full if, for all A, B, ∈ Obj(ℭ), the map

ℭ(A, B) ⟶ 𝔇(F(A), F(B))

is an onto mapping of sets.

If 𝔄 is the category of finitely generated abelian groups and T : 𝔄 → 𝔄 is the torsion functor described above, then T is a full (but not faithful) functor.

iii). A functor which is both faithful and full is called fully faithful. The inclusion of a full subcategory of a category into that category is a typical fully faithful functor.

Sect.4B

iv). If ℭ and 𝔇 are both additive categories, then F is called an additive functor if the assignment

F : ℭ(A, B) → 𝔇(F(A), F(B))

is a group homomorphism for every A, B ∈ Obj(ℭ).

v). If now ℭ and 𝔇 are abelian categories we can further speak of left exact, right exact and exact functors. We’ve seen these definitions earlier, done informally, and I leave them to you to formulate exactly!!!

One of the most important theorems of Category Theory talks about (what are called) “small ” categories (i.e. categories for which the collection of objects is a set). We will not give the proof here.

Freyd-Mitchell Embedding Theorem: If 𝔄 is a small abelian category, then there is a ring R and an exact, fully faithful functor from 𝔄 into R-mod which embeds 𝔄 as a full subcategory of R-mod.

(This means that a great many arguments in a general abelian category can be made by using the same argument in the category R-mod).

Natural Transformations of Functors

Definitions: Let F and G be two functors from ℭ to 𝔇.

A natural transformation η : F → G is a rule which associates to every object A ∈ Obj(ℭ), a fixed morphism η_{A} : F(A) → G(A) in 𝔇 such that, for every morphism f ∈ ℭ(A, B) in ℭ, we have a commutative diagram

A  F(A) ⟶^{η_{A}} G(A)
f ↓  F(f) ↓  ↓ G(f)
B  F(B) ⟶^{η_{B}} G(B)

If each η_{A} is an isomorphism we say that η is a natural isomorphism and write F ≃ G.

A functor F : ℭ → 𝔇 is called an equivalence of categories if there is a functor G : 𝔇 → ℭ such that

G ∘ F ≃ id_{ℭ} and F ∘ G ≃ id_{𝔇} .

Example: A simple example of an equivalence of categories is the following: let Ṽ_{k} = category of \underline{based} vector spaces, i.e. the objects are vector spaces over k, each with a fixed basis. Then the morphisms are appropriately sized matrices.

Let V_{k} be the category of all vector spaces and vector space linear transformations. The claim is that these categories are equivalent. To see this, let F be the forgetful functor which “forgets” the chosen basis. Clearly,

many objects get sent to the same thing by this functor. Let G be the functor which picks a basis for each vector space. Then it is easy to see that F ∘ G = id and clearly G ∘ F ≠ id.

Claim: G ∘ F ≃ id_{V_{k}}

Proof: We want to find, for each (V, β) ∈ Obj(Ṽ_{k}), where β is a basis for the vector space V, an isomorphism between

GF(V, β) and (V, β)

making the appropriate diagram commute. i.e. we want:

(V, β′) = GF(V, β) ⟶^{[id]_{β′}^{β}} (V, β)
[f]_{β′}^{γ′} ↓  [f]_{β}^{γ}
GF(W, γ) = (W, γ′) ⟶^{[id]_{γ′}^{γ}} (W, γ)

But, this is standard from linear algebra.

Tensor Products Let R be a commutative ring and let E_{1}, …, E_{n}, F be R-modules. A function

f : E_{1} × … × E_{n} ⟶ F

is called multilinear if:

given i, 1 ≤ i ≤ n and, for each j ≠ i any elements x_{j} ∈ E_{j}, then the function

f(x_{1}, …, x_{i−1}, −, x_{i+1}, …, x_{n}) : E_{i} ⟶ F

is an R-module homomorphism.

Example: If V is a vector space over the field k (with a fixed basis) and dim(V) = n, then if we let V_{i} = V for i = 1, …, n, then

det : V_{1} × ⋯ × V_{n} ⟶ k

defined by det(v_{1}, …, v_{n}) = the determinant of the n × n matrix whose columns are the coordinates of the v_{i} with respect to the chosen basis, is a well-known multilinear form.

We denote by M(E_{1}, …, E_{n}; F) the set of all such multilinear maps. This set can be made into an R-module by using the R-module structure on F, via

(rf)(x_{1}, …, x_{n}) = r(f(x_{1}, …, x_{n}))

and

(f + g)(x_{1}, …, x_{n}) = f(x_{1}, …, x_{n}) + g(x_{1}, …, x_{n}) .

( Note: As earlier, this may not work if R is not commutative.)

We define the tensor product of E_{1}, E_{2}, ⋯, E_{n} over R as follows:

let M be the free R-module on the set E_{1}×⋯×E_{n}. We will denote the basis element in M associated to the element (e_{1}, …, e_{n}) by 1_{(e_{1},...,e_{n})} and let

ϕ : (e_{1}, …, e_{n}) → 1_{(e_{1},...,e_{n})}

be a function from E_{1} × ⋯ × E_{n} to M. Now let N be the submodule of M generated by all the elements of the form:

1_{(x_{1},...,x_{i−1},x_{i}+x′_{i},x_{i+1},...,x_{n})} − 1_{(x_{1},...,x_{i−1},x_{i},x_{i+1},...,x_{n})} − 1_{(x_{1},...,x_{i−1},x′_{i},x_{i+1},...,x_{n})} ,

and

1_{(x_{1},...,x_{i−1},ax_{i},x_{i+1},...,x_{n})} − a1_{(x_{1},...,x_{i−1},x_{i},x_{i+1},...,x_{n})} a ∈ R .

If we now consider η ∘ ϕ : E_{1} × ⋯ × E_{n} → M→^{η}M/N, then we see that we have factored out just enough to make η ∘ ϕ a multilinear map.

Definition: We define M/N = E_{1} ⊗_{R} E_{2} ⊗_{R} ⋯ ⊗_{R} E_{n} and call this the tensor product of E_{1}, …, E_{n} over R. It is an R-module.

If we let χ = η ∘ ϕ, then χ(x_{1}, …, x_{n}) will be denoted by x_{1} ⊗ x_{2} ⊗ ⋯ ⊗ x_{n}. Note that not every element of the tensor product is of this form. Rather it is a sum of elements of this form!

Theorem: The multilinear map

χ : E_{1} × ⋯ E_{n} ⟶ E_{1} ⊗_{R} E_{2} ⊗_{R} ⋯ ⊗_{R} E_{n}

satisfies the following UMP:

If f is any multilinear map out of E_{1} × ⋯ E_{n} then there is a unique linear map f_{∗} making the following diagram commute:

E_{1} × ⋯ E_{n}  ⟶^{f} F
χ ↓  ↗ f_{∗}
E_{1} ⊗_{R} E_{2} ⊗_{R} ⋯ ⊗_{R} E_{n}

Proof: We have a diagram

E_{1} × ⋯ E_{n}  ⟶^{f} F
ϕ ↓  ↗ g
M
η ↓  ↗ g
M/N = E_{1} ⊗_{R} E_{2} ⊗_{R} ⋯ ⊗_{R} E_{n}

Let g(1_{(x_{1},...,x_{n})}) = f(x_{1}, …, x_{n}). Then g ∘ ϕ = f as set functions. We can extend that to a unique homomorphism of R-modules, since M is free, still called g.

Since f is multilinear, this implies that g vanishes on N, so g passes to an R-module homomorphism, also denoted by g, which makes the diagram commute. The function g is unique, as constructed. ⊓ ⊔

It is not too surprising that the UMP for the tensor product is one of the fundamental features of the object. It “converts” multilinear maps into linear ones!

Simple Properties of the Tensor Product

Proposition: Let R be a commutative ring. There is a natural isomorphism of R-modules,

f : R ⊗_{R} M → M given by f(r ⊗ m) = rm.

Proof: Consider g : R × M → M by g(r, m) = rm. Since M is an R-module, g is multilinear.

By the UMP we have a map g_{∗} making the following diagram commute:

R × M ⟶^{g} M
↓ χ  ↗ g_{∗}
R ⊗_{R} M

where g_{∗}(r ⊗ m) = rm.

Clearly g_{∗} is onto, so it suffices to show that g_{∗} is 1 − 1.

A typical element x of R ⊗_{R} M is x = ∑_{i=1}^{n} a_{i}(r_{i} ⊗ m_{i}) where a_{i} ∈ R, r_{i} ∈ R and m_{i} ∈ M. By the standard properties of the tensor product we have:

a_{i}(r_{i} ⊗ m_{i}) = (a_{i}r_{i}) ⊗ m_{i} = 1 ⊗ (a_{i}r_{i}m_{i}) = 1 ⊗ m′_{i} .

So,

x = ∑_{i=1}^{n} 1 ⊗ m′_{i} = 1 ⊗ (∑_{i=1}^{m} m′_{i}) = 1 ⊗ m

for a suitable m ∈ M.

Thus, g_{∗}(x) = g_{∗}(1 ⊗ m) = 1m = m, so g_{∗}(x) = 0 implies that m = 0, i.e. x = 1 ⊗ 0. But then x = 0 and so g_{∗} is 1 − 1 as we wanted to show. ⊓ ⊔

We now want to see how the tensor product behaves with respect to homomorphisms. Let f : M_{1} → M_{2} and let g : N_{1} → N_{2} be R-module homomorphisms.

Claim: There is an R-module homomorphism denoted f ⊗ g defined as follows

f ⊗ g : M_{1} ⊗ N_{1} → M_{2} ⊗ N_{2} by (f ⊗ g)(m_{1} ⊗ n_{1}) = f(m_{1}) ⊗ g(n_{1}) .

Proof: The method should now be becoming clear! Consider the map (f, g) : M_{1} × N_{1} → M_{2} ⊗ N_{2} given by (f, g)(m_{1}, n_{1}) = f(m_{1}) ⊗ g(n_{1}).

If we prove that (f, g) is a multilinear map, then the induced map on the tensor product is exactly what we are looking for. ⊓ ⊔

Note: If f and g are onto, so is f ⊗ g.

Exercise 8A: i) (10) Show that E_{1} ⊗_{R} E_{2} ≃ E_{2} ⊗_{R} E_{1}. ii) (10) Show that

(E_{1} ⊗ E_{2}) ⊗ E_{3} ≃ E_{1} ⊗ E_{2} ⊗ E_{3} ≃ E_{1} ⊗ (E_{2} ⊗ E_{3}) .