Solutions and elaborations for "An introduction to homological algebra" by Charles Weibel (version 26 Oct 2015)

27 Pages·2015·0.331 MB·English

by  Sebastian Bozlee

SOLUTIONS AND ELABORATIONS FOR “AN INTRODUCTION TO
HOMOLOGICAL ALGEBRA” BY CHARLES WEIBEL
SEBASTIAN BOZLEE
Weibel’s homological algebra is a text with a lot of content but also a lot left to the reader.
This document is intended to cover what’s left to the reader: I try to fill in gaps in proofs,
perform checks, make corrections, and do the exercises. It is very much in progress, covering
only chapters 3 and 4 at the moment.
I will assume roughly the same pre-requisites as Weibel: knowledge of some category
theory and graduate courses in algebra covering modules, tensor products, and localizations
of rings. I will assume less mathematical maturity.
1. Missing Details
1.1. Chapter 1.
1.2. Chapter 2.
1.2.1. Section 2.1.
1.2.2. Section 2.2.
1.2.3. Section 2.3.
1.2.4. Section 2.4.
Theorem 2.4.6. It is asserted that the image of a split exact sequence under an additive
functor is exact. See Theorem 4 in “Additional Math” below for a (lengthy, but I think
neat) proof.
1.3. Chapter 3.
1.3.1. Section 3.1.
Proposition 3.1.2. This is the way that a group is a direct limit of its finitely generated
subgroups. Let A be an abelian group.
First, we set up a diagram D whose objects are the finitely generated subgroups of A, and
whose arrows are the inclusion maps ϕ : G → H whenever G ≤ H for finitely generated
subgroups G,H of A. This diagram is indexed by a filtered category, since for any pair G,H
of finitely generated subgroups of A, GH is also a finitely generated group, so G → GH
and H → GH are arrows in the diagram (i.e. we satisfy axiom 1 of a filtered category
in Definition 2.6.13) and there are no parallel arrows (i.e. we satisfy axiom 2 of a filtered
category vacuously.)
The claim is that the colimit of this diagram is A. To see this, first of all, we have a map
ϕ : G → A for each f.g. subgroup G defined by the inclusion of G into A. This commutes
G
with all maps in the diagram since the inclusion of G into H followed by the inclusion of H
into A is just the inclusion of G into A. Therefore A is a cocone of the diagram D.
1
2 SEBASTIAN BOZLEE
It remains to show that A is a universal cocone. Select another cocone D → B. We will
define a map ψ : A → B commuting with everything. Let g ∈ A. The element g is contained
in some f.g. subgroup G of A. We define ψ(g) to be the image of g under the map G → B.
This is the only way to define ψ so that
G
(cid:15)(cid:15) (cid:31)(cid:31)
ψ (cid:47)(cid:47)
A B
commutes, so if ψ defines a map that commutes with the maps D → A and D → B, it is
unique. On the other hand, if H is another f.g. subgroup of A containing g, then since
(cid:47)(cid:47) (cid:111)(cid:111)
G GH H
(cid:33)(cid:33) (cid:15)(cid:15) (cid:125)(cid:125)
B
commutes, the image of g under H → B must be the same as its image under the map
G → B. This shows that ψ is well defined. Finally, ψ is a group homomorphism, since
for any g,h ∈ A, we may select a f.g. subgroup G so that g,h ∈ G. Let ϕ be the map
G → B. Then ψ(g+h) = ϕ(g+h) = ϕ(g)+ϕ(h) = ψ(g)+ψ(h) since ϕ : G → B is a group
homomorphism. So there exists a unique group homomorphism A → B so that everything
commutes, as desired, and A is the colimit.
Intuitively, the direct limit is a fancy way to express a union, and of course A is a union
of its finitely generated subgroups, since each element of A is contained in some finitely
generated subgroup. The argument above shows that the formal definition agrees with this
intuition.
Proposition 3.1.4. The reason that TorZ(Zm,B) = 0 is that Zm is a free Z-module, hence
n
projective, and Tor vanishes on projective modules.
1.3.2. Section 3.2.
Definition 3.2.1. The target category of − ⊗ B is intended to be Ab. If B is an R-S
R
bimodule, wecanalsoview−⊗ B asafunctorfrommod-Rtomod-S. Again, thisisalways
R
rightexact. Itisequivalentfor−⊗B tobeexactasafunctortoAbandasafunctortomod-
S, since in each case exactness is the same as taking every injection A(cid:48) → A of R-modules to
an injective map A(cid:48) ⊗B → A⊗B: we just view the map as a group homomorphism if Ab
is the target and as an S-module homomorphism if mod-S is the target. This means that
flatness does not depend on the target category of −⊗ B, although it does depend on R.
R
Theorem 3.2.2. For an alternate proof without colimits, see [2, Proposition 2.5].
We show that the colimit indicated in the text is the localization, as claimed. Our tech-
nique will be to use an explicit description of filtered colimits. Namely, if I is a filtered
category, and F : I → R-mod is an additive functor, the colimit of F is the module C
whose additive group has underlying set
(cid:32) (cid:33)
(cid:71)
C = F(i) / ∼
i∈I
where a ∈ F(i) and b ∈ F(j) are equivalent a ∼ b if there exist arrows ϕ : i → k and
ψ : j → k so that F(ϕ)(a) = F(ψ)(b). The addition of a ∈ F(i) and b ∈ F(j) is defined
SOLUTIONS AND ELABORATIONS 3
by taking a pair of arrows ϕ : i → k and ψ : j → k (whose existence is assured since I is a
filtered category) and letting a+b be the sum F(ϕ)(a)+F(ψ)(b) in k. For r ∈ R we take
the scalar product of r and a ∈ F(i) to be ra ∈ F(i).
In our particular case, a ∈ F(s ) and b ∈ F(s ) are equivalent if and only if there exists
1 2
a number c ∈ S so that cs a = cs b. Next, note that a ∼ b if and only if a = b in S−1R.
2 1 s1 s2
Therefore the map ϕ : C → S−1R defined by a ∈ F(s ) (cid:55)→ a is a well defined bijection.
1 s1
Next, the sum of a ∈ F(s ) and b ∈ F(s ) is s a+s b ∈ F(s s ), so ϕ preserves addition.
1 2 2 1 1 2
Finally, the scalar product of r ∈ R and a ∈ F(s ) is ra ∈ F(s ), so ϕ preserves scalar
1 1
multiples. Altogether, the colimit C is isomorphic to the localization S−1R, as claimed.
Lemma 3.2.6. When M = R, the map σ is from A∗⊗ R → Hom (R,A)∗, with σ defined
R R
by σ(f ⊗r) : h (cid:55)→ f(h(r)).
There is an isomorphism A∗ → A∗⊗R defined by f (cid:55)→ f ⊗1 and there is an isomorphism
Hom (R,A)∗ = Hom (Hom (R,A),Q/Z) → Hom (A,Q/Z) = A∗
R Ab R Ab
induced by the isomorphism A → Hom (R,A) that takes a to the map ϕ : R → A
R a
determined by ϕ (1) = a. More explicitly, the map Hom (R,A)∗ → A∗ is defined by
a R
h (cid:55)→ h(cid:48) :A → Q/Z
a (cid:55)→ h◦ϕ
a
where ϕ is the element of Hom(R,A) so that ϕ (1) = a. Composing with these isomor-
a a
phisms, we may regard σ as a map σ : A∗ → A∗ defined by σ(f) : m (cid:55)→ f(ϕ (1)) = f(m).
m
That is, σ takes f ∈ A∗ to f ∈ A∗, so σ is an isomorphism, as claimed.
The use of the Five Lemma is with this diagram:
A∗ ⊗Rm (cid:47)(cid:47) A∗ ⊗Rn (cid:47)(cid:47) A∗ ⊗M (cid:47)(cid:47) 0 (cid:47)(cid:47) 0
σ ∼= σ ∼= σ 0 0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
Hom(Rm,A)∗ α∗ (cid:47)(cid:47) Hom(Rn,A)∗ (cid:47)(cid:47) Hom(M,A)∗ (cid:47)(cid:47) 0 (cid:47)(cid:47) 0
Proposition 3.2.9. We will prove that if T is R-flat and P is a projective R-module, then
P ⊗ T is a projective T-module.
R
Recall that an R-module P is projective if and only if P is the direct summand of a free
R-module (Proposition 2.2.1). So let F = P ⊕Q be a free module with direct summand P.
∼ (cid:76)
Since F is free, we also have that F = R for some set S. This gives us the split exact
s∈S
sequence
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
0 P F Q 0.
Since T is R-flat, −⊗ T is exact as a functor from mod-R to mod-T. It therefore takes
R
split exact sequences to split exact sequences (see comment on Theorem 2.4.6), so P ⊗T is
a direct summand of F ⊗ T. Also, − ⊗ T as a functor from mod-R to mod-T has the
R
right adjoint Hom(T,−) (Proposition 2.6.3), so −⊗ T preserves colimits (Theorem 2.6.10).
R
(cid:76) (cid:76) (cid:76)
In particular, F ⊗ T = ( R) ⊗ T = (R ⊗ T) = T, so − ⊗ T takes free
s∈S s∈S s∈S R
R-modules to free T-modules. Together, P ⊗T is a direct summand of a free T-module, so
P ⊗T is a projective T-module.
Lemma 3.2.11. To see why we require that r is central, let µ : P → P be the map p (cid:55)→ rp.
i i i
Then µ is an R-module homomorphism if and only if for all s ∈ R,p ∈ P
i
srp = sµ(p) = µ(sp) = rsp.
4 SEBASTIAN BOZLEE
This holds with no conditions on P if sr = rs for all s ∈ R, that is, if r is central. Had we
i
not required that r were central, we would then have to impose conditions on the projective
resolution P → A to maintain that µ˜ : P → P is a chain map.
Corollary 3.2.13. The equality R ⊗ M = TorRp(A ,B ) follows by first noting
p R n p p
R ⊗ M = R ⊗ TorR(A,B) = TorRp(A⊗ R ,R ⊗ B)
p R p R n n R p p R
from Corollary 3.2.10, and then using that A⊗R and R ⊗B are equal to the localizations
p p
of A and B at p, respectively.
1.3.3. Section 3.3.
Example 3.3.3. Ext0(A,Z) is zero since we have assumed that A is a torsion group: torsion
Z
elements must be sent to torsion elements, so the only homomorphism A → Z is the zero
homomorphism.
The comment that Ext1(Z ,Z) = lim(Z/pn) is proved in Application 3.5.10.
Z p∞ ←−
Example 3.3.5.
(1) This is because A ∼= (cid:76)∞ Z/p, so
i=1
∞ ∞
(cid:89) (cid:89)
Ext1 (A,Z/p) = Ext1 (Z/p,Z/p) = Z/p.
Z/m Z/m
i=1 i=1
where the first equality follows from Proposition 3.3.4 part 1 and the second follows
from Exercise 3.3.2. The last is a vector space of uncountably infinite dimension.
(2) The displayed equation follows from Proposition 3.3.4 part 2 and Calculation 3.3.2.
(Ext1 is intended.) The comment on divisibility follows since A is divisible if and
Z
only if A = pA for all p ≥ 2 if and only if A/pA = 0 for all p ≥ 2.
Heuristically, there seems to be a duality between being torsionfree and divisible.
TorZ(−,Q/Z) tests for torsion and Ext1(−,(cid:81)∞ Z/p) tests for divisibility.
1 Z p=2
Lemma 3.3.6. Weibel asserts that Hom (A,B) is a (left) R-module when R is commuta-
Ab
tive. Hereiswhy. Wedefinerf : A → B forr ∈ Randf ∈ Hom (A,B)by(rf)(b) = f(rb).
Ab
In order for this action to make Hom (A,B) an R-module, we need rf to be an R-module
Ab
homomorphism for all r. Now rf is an R-module homomorphism if and only if for all s ∈ R
and b ∈ B,
f(srb) = sf(rb) = s(rf)(b) = (rf)(sb) = f(rsb).
If we assume R is commutative, then srb = rsb, so we get the equality above for free.
Otherwise, we would have to impose conditions on A and B to maintain that Hom (A,B)
Ab
is an R-module.
The assertion that multiplication by r gives a chain map when r is central is addressed in
my comment on Lemma 3.2.11.
Lemma 3.3.8. Let’s check the isomorphism for A = R. Then
S−1Hom (A,B) = S−1Hom (R,B) ∼= S−1B.
R R
On the other hand,
Hom (S−1A,S−1B) ∼= S−1B.
S−1R
So both are equal to S−1B as R-modules, as promised.
SOLUTIONS AND ELABORATIONS 5
The implication that the same holds for Rm is due to the fact that additive functors
preserve direct sums. This is a corollary of Theorem 4 and Proposition 3 in “Additional
Math” above.
1.3.4. Section 3.4.
Theorem 3.4.3. Accordingtotheerrata,thetheoremshouldbestatedwithΘ : ξ (cid:55)→ ∂(id ),
B
instead of id . We will see why as we fill in the details of this proof.
A
0
When Weibel writes that X → A is the map induced by the maps B → A and P → A,
he is referring to the universal property of pushouts: since
j (cid:47)(cid:47)
M P
β
(cid:15)(cid:15) (cid:15)(cid:15)
0 (cid:47)(cid:47)
B A
commutes (we get zero when composing M → P → A by exactness of 0 → M → P → A →
0) there is a unique map X → A so that the following diagram commutes:
j (cid:47)(cid:47)
(1) M P
β σ
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47) ϕ
B X
ϕ(cid:48)
(cid:32)(cid:32) (cid:19)(cid:19)
0 (cid:43)(cid:43)
A
That map is the map we are to use in the sequence 0 → B → X → A → 0. More explicitly, if
ϕ is the map P →ϕ A, then the dotted map, ϕ(cid:48) : X → A, is defined by taking the equivalence
class of (p,b) in X to the element ϕ(p) in A. This is well defined since if (p,b) and (p(cid:48),b(cid:48))
are two representatives of the same element of X, then ϕ(p)−ϕ(p(cid:48)) = ϕ(p−p(cid:48)) = ϕ(j(m))
j ϕ
for some m ∈ M. By exactness of 0 → M → P → A → 0, ϕ(j(m)) = 0.
Now we show that the bottom row of the diagram
(cid:47)(cid:47) j (cid:47)(cid:47) ϕ (cid:47)(cid:47) (cid:47)(cid:47)
(2) 0 M P A 0
β σ
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) i (cid:47)(cid:47) ϕ(cid:48) (cid:47)(cid:47) (cid:47)(cid:47)
0 B X A 0
is exact. (The left square commutes by definition of pushout. The right square commutes
since the right triangle of (1) commutes.)
Since ϕ is surjective, and ϕ = ϕ(cid:48)◦σ, it follows ϕ(cid:48) is surjective. This shows exactness at A.
The map i : B → X is defined by b (cid:55)→ [(0,b)]. Suppose that [(0,b)] = [(0,0)]. Then, by
the definition of X, (0,b) = (0,0)+(j(m),−β(m)) for some m ∈ M. Since j is injective, we
must have that m = 0, so [(0,b)] = [(0,0)]. Therefore i is injective. This shows exactness at
B.
Finally, suppose (p,b) is the equivalence class of some element of the kernel of ϕ(cid:48). Then
ϕ(p) = 0, so by exactness of the first row, p = j(m) for some m ∈ M. Therefore, [(p,b)] =
[(j(m),b)] = [(0,b+β(m))] is in the image of i. This shows exactness at X, and we are done.
6 SEBASTIAN BOZLEE
The reference to naturality of ∂ refers to the fact that this diagram, induced from the mor-
phism of short exact sequences (2) by the contravariant cohomological δ-functor Ext∗(−,B),
commutes:
Hom(M,B) ∂ (cid:47)(cid:47) Ext1(A,B)
(cid:79)(cid:79) (cid:79)(cid:79)
β∗ id
Hom(B,B) ∂ (cid:47)(cid:47) Ext1(A,B)
By commutativity of the diagram, ∂(id ) = ∂(β∗(id )) = ∂(β) = x. This gives surjectivity
B B
of Θ.
Now we will check that the maps i : B → X and σ+i◦f : P → X induce an isomorphism
X(cid:48) ∼= X and an equivalence between ξ(cid:48) and ξ. Recall that we have picked two lifts, β and β(cid:48)
of some element x ∈ Ext1(A,B) under the map ∂ in the diagram
Hom(P,B) j∗ (cid:47)(cid:47) Hom(M,B) ∂ (cid:47)(cid:47) Ext1(A,B) (cid:47)(cid:47) 0,
which we arrived at by taking the long exact sequence associated to 0 → M → P → A → 0
under Ext∗(−,B). By exactness, the maps β and β(cid:48) differ by an element of the image of j∗,
so we write β(cid:48) = β +f ◦j, where f : P → B.
First, we check that the diagram
j (cid:47)(cid:47)
(3) M P
β(cid:48) σ+i◦f
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47)
B X
commutes. It does:
i◦β(cid:48) = i◦(β +f ◦j)
= i◦β +i◦f ◦j
= σ ◦j +i◦f ◦j
= (σ +i◦f)◦j.
To get the third equaliity, we used the commutativity of the diagram
j (cid:47)(cid:47)
M P
β σ
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47)
B X
SOLUTIONS AND ELABORATIONS 7
j (cid:47)(cid:47)
Now P is a cocone of the diagram M P so by the universal property
σ+i◦f β(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47)
B X B
of pushouts, there exists a unique map so that the following diagram commutes:
j (cid:47)(cid:47)
(4) M P
β(cid:48) σ(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
B i(cid:48) (cid:47)(cid:47) X(cid:48) σ+i◦f
(cid:32)(cid:32) (cid:20)(cid:20)
i (cid:43)(cid:43)
X
By a symmetric argument, there exists a unique map so that
j (cid:47)(cid:47)
M P
β σ
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47) σ(cid:48)−i(cid:48)◦f
B X
(cid:32)(cid:32) (cid:20)(cid:20)
i(cid:48) (cid:43)(cid:43)
X(cid:48)
commutes. Adding f◦j to β, i◦f to σ, and i(cid:48)◦f to σ(cid:48), we obtain the commutative diagram
j (cid:47)(cid:47)
(5) M P
β(cid:48) σ+i◦f
(cid:15)(cid:15) (cid:15)(cid:15)
i (cid:47)(cid:47) σ(cid:48)
B X
(cid:32)(cid:32) (cid:20)(cid:20)
i(cid:48) (cid:43)(cid:43)
X(cid:48)
with the same arrow from X → X(cid:48).
Now, composing the arrows X(cid:48) → X and X → X(cid:48) in diagrams (4) and (5), we get a
commutative diagram
j (cid:47)(cid:47)
M P
β(cid:48) σ(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
B i(cid:48) (cid:47)(cid:47) X(cid:48) σ(cid:48)
(cid:33)(cid:33) (cid:20)(cid:20)
i(cid:48) (cid:43)(cid:43)
X(cid:48)
By the uniqueness in universal property of pushouts, the dashed arrow must be the identity
on X(cid:48). By a symmetric argument, the other composition X → X(cid:48) → X is the identity on
X. Therefore X and X(cid:48) are isomorphic.
8 SEBASTIAN BOZLEE
Finally, we show that the isomorphism X(cid:48) → X induces an equivalence of extensions.
Consider the diagram
ξ(cid:48) : 0 (cid:47)(cid:47) B i(cid:48) (cid:47)(cid:47) X(cid:48) (cid:47)(cid:47) A (cid:47)(cid:47) 0
(cid:15)(cid:15)
(cid:47)(cid:47) i (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
ξ : 0 B X A 0
where the middle arrow is the isomorphism X(cid:48) → X. The left square commutes since the
the lower triangle of (4) commutes. The right square commutes because the right squares of
(cid:47)(cid:47) j (cid:47)(cid:47) ϕ (cid:47)(cid:47) (cid:47)(cid:47)
0 M P A 0
β σ
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) i (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
0 B X A 0
and
(cid:47)(cid:47) j (cid:47)(cid:47) ϕ (cid:47)(cid:47) (cid:47)(cid:47)
0 M P A 0
β(cid:48) σ(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) i(cid:48) (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
0 B X A 0
commute. Therefore, ξ and ξ(cid:48) are equivalent extensions, as claimed.
Definition 3.4.4. I think it is helpful to compute the Baer sum of a simple example. Let
us take the Baer sum of the extensions
ξ : 0 (cid:47)(cid:47) Z/p p (cid:47)(cid:47) Z/p2 1 (cid:47)(cid:47) Z/p (cid:47)(cid:47) 0,
ξ(cid:48) : 0 (cid:47)(cid:47) Z/p p (cid:47)(cid:47) Z/p2 2 (cid:47)(cid:47) Z/p (cid:47)(cid:47) 0.
Now the pullback X(cid:48)(cid:48) of Z/p2 →1 Z/p and Z/p2 →2 Z/p is the submodule {(x,y) ∈ Z/p2 ×
Z/p2 | x ≡ 2y (mod p)} of Z/p2 ×Z/p2. The pullback diagram is
X(cid:48)(cid:48) π2 (cid:47)(cid:47) Z/p2
π1 2
(cid:15)(cid:15) (cid:15)(cid:15)
Z/p2 1 (cid:47)(cid:47) Z/p
We let Y be X(cid:48)(cid:48)/{(px,−px) | x ∈ Z/p}. Then the Baer sum of ξ and ξ(cid:48)is the sequence
0 (cid:47)(cid:47) Z/p f (cid:47)(cid:47) Y g (cid:47)(cid:47) Z/p (cid:47)(cid:47) 0
where f takes x to [(px,0)] = [(0,px)] and g takes [(x,y)] to x (mod p) or equivalently to
2y (mod p).
SOLUTIONS AND ELABORATIONS 9
Ifp (cid:54)= 3, thisisequivalentwiththeextensionassociatedtoi = 2 (mod p), viathefollowing
3
equivalence:
0 (cid:47)(cid:47) Z/p f (cid:47)(cid:47) Y g (cid:47)(cid:47) Z/p (cid:47)(cid:47) 0
σ
(cid:15)(cid:15)
2
0 (cid:47)(cid:47) Z/p p (cid:47)(cid:47) Z/p2 3 (cid:47)(cid:47) Z/p (cid:47)(cid:47) 0
where σ : [(x,y)] (cid:55)→ x+y (mod p2). This is well defined, since any other representative of
(x,y) is of the form (x−pz,y+pz), and (x−pz,y+pz) (cid:55)→ (x−pz)+y+pz = x+y. To see
that the bottom right map is 2, i.e. multiplication by 2·3−1 (mod p), note that (2,1) (cid:55)→g 2,
3
while (2,1) (cid:55)→σ 3 (mod p2), and a morphism between cyclic groups must be multiplication
by some number.
Corollary 3.4.5. When Weibel refers to using the notation of the previous theorem, he
means that we have constructed the following diagrams from the extensions ξ and ξ(cid:48):
(cid:47)(cid:47) j (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
(6) 0 M P A 0
γ τ
(cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) i (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
ξ : 0 B X A 0
and
(cid:47)(cid:47) j (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
(7) 0 M P A 0
γ(cid:48) τ(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
ξ(cid:48) : 0 (cid:47)(cid:47) B i(cid:48) (cid:47)(cid:47) X(cid:48) (cid:47)(cid:47) A (cid:47)(cid:47) 0.
We construct them by first using the lifting property of the projective module P to obtain
a map τ : P → X and a map τ(cid:48) : P → X(cid:48). Then by commutativity of the right squares,
τ(j(M)) ⊆ ker{X → A} and τ(cid:48)(j(M)) ⊆ ker{X(cid:48) → A}. The exactness of the bottom rows
allows us to lift τ and τ(cid:48) to maps γ = i−1 ◦τ ◦j and γ(cid:48) = (i(cid:48))−1 ◦τ ◦j respectively.
Recall that X(cid:48)(cid:48) is the pullback of the diagram
X(cid:48)(cid:48) (cid:47)(cid:47) X
(cid:15)(cid:15) (cid:15)(cid:15)
X(cid:48) (cid:47)(cid:47) A.
Now
τ (cid:47)(cid:47)
P X
τ(cid:48)
(cid:15)(cid:15) (cid:15)(cid:15)
X(cid:48) (cid:47)(cid:47) A
10 SEBASTIAN BOZLEE
commutes, since each composition is the map P → A, by the diagrams (6) and (7) above.
Accordingly, by the universal property of pullbacks, we get a unique map P → X(cid:48)(cid:48) so that
P
τ
(cid:32)(cid:32) (cid:36)(cid:36)
X(cid:48)(cid:48) (cid:47)(cid:47) X
τ(cid:48)
(cid:28)(cid:28) (cid:15)(cid:15) (cid:15)(cid:15)
X(cid:48) (cid:47)(cid:47) A
commutes. This is the induced map τ(cid:48)(cid:48) : P → X(cid:48)(cid:48). More explicitly, τ(cid:48)(cid:48) is the map P →
X(cid:48)(cid:48) ⊆ X ×X(cid:48) defined by τ(cid:48)(cid:48)(p) = (τ(p),τ(cid:48)(p)). The map τ : P → Y is obtained from τ(cid:48)(cid:48) by
composition with the quotient map X(cid:48)(cid:48) → Y.
Next, Weibel claims that
(cid:47)(cid:47) j (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
0 M P A 0
γ+γ(cid:48) τ
(cid:15)(cid:15) (cid:15)(cid:15)
ξ +ξ(cid:48) : 0 (cid:47)(cid:47) B i(cid:48)(cid:48) (cid:47)(cid:47) Y (cid:47)(cid:47) A (cid:47)(cid:47) 0.
commutes. To see this, we first check that the left square commutes. Recall that the map
i(cid:48)(cid:48) : B → Y is defined by b (cid:55)→ [(i(b),0)] = [(0,i(cid:48)(b))]. Then composing i(cid:48)(cid:48) ◦(γ +γ(cid:48)), we get
(i(cid:48)(cid:48) ◦(γ +γ(cid:48)))(m) = i(cid:48)(cid:48)(γ(m)+γ(cid:48)(m))
= [(i(γ(m)),0)]+[(0,i(cid:48)(γ(cid:48)(m)))]
= [(τ(j(m)),0)]+[(0,τ(cid:48)(j(m)))]
= [(τ(j(m)),τ(cid:48)(j(m)))]
= τ(j(m)),
so the square commutes. To do: the right square commutes because...
1.3.5. Section 3.5.
Definition 3.5.1. We check that the kernel of ∆ is limA . To do
←− i
Lemma 3.5.2. The snake lemma mentioned only gives the map limC →∂ lim1A .
←− i ←− i
To see why we may take the higher derived functors of lim to vanish, note that the
←−
induced map on cokernels lim1B → (cid:81)lim1C is onto, since the map (cid:81)B → (cid:81)C is onto.
←− i ←− i i i
Accordingly, the sequence
0 → limA → limB → limC →∂ lim1A → lim1B → lim1C → 0
←− i ←− i ←− i ←− i ←− i ←− i
is exact.
Example 3.5.5. Applying the δ-functor lim∗− to the short exact sequence 0 → {piZ} →
←−
{Z} → {Z/pi} → 0, we get the long exact sequence
limpiZ (cid:47)(cid:47) limZ (cid:47)(cid:47) limZ/piZ ∂ (cid:47)(cid:47) lim1piZ (cid:47)(cid:47) lim1Z.
←− ←− ←− ←− ←−
This simplifies to
0 (cid:47)(cid:47) Z (cid:47)(cid:47) Zˆ ∂ (cid:47)(cid:47) lim1piZ (cid:47)(cid:47) 0.
p ←−