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Math 614: Lecture notes
Sep 26, 2007
Left-adjoint functors commute with coproducts
Definition. Given a category C and a set X, we can define a category C X as follows: The objects of C X^ are X-indexed sets of objects (Cx)x∈X of C. A morphism (Cx) → (C x′) is given by an X-indexed set (fx : Cx → C x′)x∈X of C - morphisms. Composition is done coordinatewise: (gx) ◦ (fx) := (gx ◦ fx). For any functor F : C → D, define the functor F X^ : C X^ → DX^ coordinatewise on both objects and morphisms. That is, F X^ ((Cx)x∈X ) := (F (Cx))x∈X and F X^ ((gx)x∈X ) := (F (gx)x∈X ). We define the diagonal functor ∆ := ∆C : C → C X^ by sending C 7 → (C)x∈X and f 7 → (f )x∈X. (i.e. It sends each object and morphism to “X copies of itself”.) If ∆ has a left-adjoint
C :^ C^
X (^) → C , we call it
a coproduct, and we say that C has X-indexed coproducts. (Dually for the right-adjoint to ∆, which is called a product
I said vaguely in a previous lecture that left-adjoint functors commute with coproducts. This is now made precise in the following series of lemmas, the first three of which are given without proof:
Lemma 1. Let C , D, E be categories, and let (F, G, ϕ) : C → D and (F ′, G′, ϕ′) : D → E be adjunctions. Then (F ′^ ◦ F, G ◦ G′, ϕ ◦ ϕ′) : C → E is also an adjunction.
Proof. Purely formal.
Lemma 2. Let C , D be categories, let (F, G, ϕ) : C → D be an adjunction, and let X be any set. Then (F X^ , GX^ , ϕX^ ) : C X^ → DX^ is also an adjunction.
Proof. Formal.
Lemma 3. Let F, F ′^ : C → D and G : D → C be functors such that both F and F ′^ are left-adjoint to G. Then there is an isomorphism of functors F ∼= F ′
Proof. Tricky, but formal. One needs to construct the isomorphism from the available data.
Lemma 4. Let C and D be categories. Suppose that both C and D have
X-indexed coproducts, and let C
F G
D be an adjunction. Then F ◦
C
D ◦F^
X (^) : C X (^) → D. In particular, F (∐ x∈X Cx
x∈X F^ (Cx).
Proof. Consider the following diagram:
C X^ o^ o GX DX
C
∆C
a a CCC CCC CCC CCC CCC CC o o G D
∆D
It’s easy to see that the diagram commutes. Now let’s fill in the diagram with the left-adjoints for each of the maps:
C X^
F X^ //
‘ C
C!!
CCC CC
CCC
CCC
CCC
CC D
X GX
o o ‘ D
A A
AAA
AAA
AAA AA
AAA
C
∆C
a a CC CCC CCC CCC CCC CCC F (^) // o o G D
∆D
Here∐ F X^ is left-adjoint to GX^ by Lemma 2. By Lemma 1, we have that
D ◦F^
X (^) is left-adjoint to GX (^) ◦ ∆D , and also F ◦ ∐ C is left-adjoint to ∆C^ ◦^ G. But GX^ ◦∆D = ∆C ◦G. Hence by Lemma 3, we get the required isomorphism of functors.
Tensor products of A-algebras as coproducts
Let R, S be A-algebras. The structure maps give, in particular, that R, S are A-modules, so we may form the tensor product T := R ⊗A S. In fact, this is a coproduct in the category of A-algebras!
By our work on Proposition 1 of the Sep 24 lecture notes, we have
R ⊗A S ∼=
A[X]
KR
⊗A
A[Y ]
KS
∼= A[X]^ ⊗A^ A[Y^ ]
Im(i ⊗ 1) + Im(1 ⊗ j) ∼= A[X, Y^ ] KRA[X, Y ] + KS A[X, Y ]
Free, projective, and flat
Recall that an R-module P is projective if HomR(P, −) preserves surjections. Proposition 1. The following are equivalent for an R-module P.
a. P is projective.
b. For any surjection π : M N of R-modules and any map g : P → N , there is a “lifting g˜ : P → M of g to M ”. (This just means that π ◦ g˜ = g.)
c. P is a direct summand of a free module.
d. P is a direct summand of any free module that maps onto it.
Proof. (a ⇐⇒ b): This is just by definition. π∗ := Hom(P, π) : Hom(P, M ) → Hom(P, N ) is surjective iff for any g ∈ Hom(P, N ), there is some ˜g ∈ Hom(P, M ) with π ◦ ˜g = π∗(˜g) = g. (c ⇒ b): First we show that free modules are projective. Let F = F (X) = R⊕X^ =
x∈X Rex^ be a free module on^ X^ with basis^ {ex}x∈X^.^ Then let π : M N be a surjection and h : F → N be any homomorphism. For each x ∈ X, pick some mx ∈ M such that π(mx) = h(ex). Then define h˜ : F → M on basis elements by ˜h(ex) := mx. This extends R-linearly to a well-defined homomorphism, and clearly h = π ◦ ˜h. Now let F be a free module that has P as a direct summand. By the exercises, this is equivalent to the existence of a pair F
p j
P of maps such that p ◦ j = 1P. Let M N be a surjection and g : P → N be a homomorphism. Then g ◦ p : F → N has a lifting h : F → M , which means that π ◦ h = g ◦ p. Then g = g ◦ p ◦ j = π ◦ h ◦ j, so that h ◦ j : P → M is a lifting of g. (d ⇒ c) is trivial. (a ⇒ d): Let π : F → P be a surjection, where F is a free module. Since P is projective, the map π∗ : Hom(P, F ) → Hom(P, P ) is a surjection. Hence there is some j ∈ HomR(P, F ) such that π ◦ j = π∗(j) = 1P , which by the exercises is equivalent to P being a direct summand of F.
Proposition 2. If P is a projective R-module, then P is flat.
Proof. First we show that free modules are flat. Let F = R⊕X^ =
x∈X Rex be a free module. Let i : N ↪→ M be an injection of R-modules, and consider
the map F ⊗R N 1 ⊗i −−→ F ⊗R M. A typical element of F ⊗R N looks like z =
∑t j=1 exj ⊗^ nj^ ,^ nj^ ∈^ N^ ,^ xj^ ∈^ X. If^
∑t j=1 exj ⊗^ i(nj^ ) = (1^ ⊗^ i)(z) = 0, then i(nj ) = 0 for each j, so that nj = 0 for all j, which certainly implies that z = 0. Hence 1 ⊗ i is injective and F is flat. By Proposition 1, P is a direct summand of a free module F , so we have
the pair of maps F
p j
P with p ◦ j = 1P. For an injection i : N ↪→ M of
R-modules, we want to show that 1P ⊗R i is injective. We get the following commutative diagram, where (by the exercises) the columns compose to the identity maps:
P ⊗R N i
′ (^) //
j 1
P ⊗R M
j 2 F ⊗R N
i
′′ (^) //
p 1
F ⊗R M
p 2 P ⊗R N i′^ // P ⊗R M
If i′(x) = 0, then j 2 (i′(x)) = i′′(j 1 (x)) = 0, so that i′′^ surjective implies that j 1 (x) = 0, so that we have x = p 1 (j 1 (x)) = p 1 (0) = 0. Hence i′^ is injective, but i′^ = 1P ⊗R i, so P is flat.
Thus, we have shown:
Free =⇒ Projective =⇒ Flat.
Later we will show that for finitely generated modules over a local ring, all three properties are equivalent.
