math 757 homology theory · 2011-01-21 · homology theory lecture 8 - 1/18/2011 axiom a3...
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Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Math 757Homology theory
January 21, 2011
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Axiom A3 (Excision)
Goal: Prove Axiom A3 (Excision)
Recall:
Axiom A3. Excision
For any pair (X ,A) if U ⊂ A is open with U ⊂ intA then theinclusion map
i : (X − U,A− U)→ (X ,A)
induces an isomorphism
i∗ : hn(X − U,A− U) ∼= hn(X ,A)
Need to “cut up” singular simplices intersecting U.
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 62
Let U = {Ui} be an open cover of X . The subordinate n-chains arethe free abelian group
CUn (X )
on the set
SUn (X ) ={σ : ∆n → X
∣∣σ(∆n) ⊂ Ui for some Ui ∈ U}
Proposition 63
The inclusion mapι : CU (X )→ C (X )
has a chain homotopy inverse
ρ : C (X )→ CU (X )
(That is ρ ◦ ι is chain homotopic to 1CU (X ) and ι ◦ ρ is chainhomotopic to 1C(X ))
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Recall for v0, · · · , vn ∈ Rm (possibly not all different)
[v0, · · · , vn] : ∆n → Rm
is the function
[v0, · · · , vn](x0, · · · , xn) =n∑
i=0
xivi
Definition 64
The barycenter of λ = [v0, · · · , vn] is the point bλ ∈ Rm
bλ =1
n + 1
n∑i=0
vi .
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 65 (Diameter of a linear simplex)
The diameter of the simplex λ = [v0, · · · , vn] is
diamλ = maxx,y∈Imλ
|x − y |
where | · | is the euclidean norm on Rm.
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 66
diam[v0, · · · , vn] = max0≤i,j≤n
|vi − vj |
Proof.
Let x , y ∈ Im[v0, · · · , vn] where x =∑n
i=0 sivi and y =∑n
j=0 tjvjThen
|x − y | =
∣∣∣∣∣∣x −n∑
j=0
tjvj
∣∣∣∣∣∣=
∣∣∣∣∣∣n∑
j=0
tj(x − vj)
∣∣∣∣∣∣ (since∑
ti = 1)
=
∣∣∣∣∣∣n∑
j=0
tj
((n∑
i=0
sivi
)− vj
)∣∣∣∣∣∣
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Lemma 66 (continued).
=
∣∣∣∣∣∣n∑
j=0
tj
((n∑
i=0
sivi
)− vj
)∣∣∣∣∣∣=
∣∣∣∣∣∣n∑
j=0
tj
(n∑
i=0
si (vi − vj)
)∣∣∣∣∣∣ (since∑
si = 1)
=
∣∣∣∣∣∣∑
0≤i,j≤n
si tj(vi − vj)
∣∣∣∣∣∣≤
∑0≤i,j≤n
si tj |vi − vj |
≤∑
0≤i,j≤n
si tj max0≤i,j≤n
|vi − vj |
= max0≤i,j≤n
|vi − vj |
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 67
LetLSn(Rm) = {[v0, · · · , vn]|v0, · · · , vn ∈ Rm}
The linear chains of Rm are the chain complex
LCn(Rm) =
{Z[LSn(Rm)], n ≥ −10, n ≤ −2
with ∂ inherited from the inclusion LCn(Rm) ⊂ Cn(Rm).
We will define three operators:
1 A “cone λ to the point b” operator (chain homotopy)
b : LCn(Rm)→ LCn+1(Rm)
2 A “barycentrically subdivide λ” operator (chain map)
S : LCn(Rm)→ LCn(Rm)
3 A “chain homotopy from λ to its barycentric subdivision”operator (chain homotopy)
T : LCn(Rm)→ LCn+1(Rm)
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011 Definition 68
b ∈ Rm a point. Let
b : LCn(Rm)→ LCn+1(Rm)
denote the operator
b[v0, · · · , vn] = [b, v0, · · · , vn]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 69
b∂ + ∂b = 1LC(Rm)
so b : LCn(Rm)→ LCn+1(Rm) is a chain homotopy between 1LC(Rm)
and 0
Proof.
b ∈ Rm a point. Then
∂b[v0, · · · , vn] = ∂[b, v0, · · · , vn]
= [v0, · · · , vn] +n∑
i=0
(−1)i+1[b, · · · , vi , · · · , vn]
= [v0, · · · , vn] +n∑
i=0
(−1)i+1b[v0, · · · , vi , · · · , vn]
= [v0, · · · , vn]− bn∑
i=0
(−1)i [v0, · · · , vi , · · · , vn]
= [v0, · · · , vn]− b∂[v0, · · · , vn]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Exercise 70 (HW2 - Problem 2)
Show that if M is the cone on a nonempty space X (see Hatcherpage 9) and X ⊂ M is the inclusion X = X × {1} then
Hn(M,X ) ∼= Hn−1(X ) for n ≥ 0
Exercise 71 (HW2 - Problem 3)
Let (X ,A,B) be a triple of spaces (B ⊂ A ⊂ X) show that there is along exact sequence (called the long exact sequence of the triple(X ,A,B))
· · · −→ H2(X ,B) −→ H2(X ,A) −→ H1(A,B) −→ H1(X ,B)
−→ H1(X ,A) −→ H0(A,B) −→ H0(X ,B) −→ H0(X ,A) −→ 0
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Exercise 72 (HW2 - Problem 4)
Given (X ,A) show that if r : X → A is a retraction (see Hatcherpage 3) and i : A→ X is the inclusion map then
i∗ : Hn(A)→ Hn(X )
is an injection.
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 73
The barycentric subdivision operator is
S : LCn(Rm)→ LCn(Rm)
where S : LC−1(Rm)→ LC−1(Rm) is defined on the empty simplex by
S [∅] = [∅]
and inductively assume S : LCn−1(Rm)→ LCn−1(Rm) is defined.
For λ = [v0, · · · , vn] with barycenter bλ set
S(λ) = bλ(S∂λ)
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Example 74
Let’s compute S [v0, v1]
S [v0, v1] = b[v0,v1]
(S∂[v0, v1]
)= b[v0,v1]
(S([v1]− [v0])
)= b[v0,v1]
(b[v1]S∂[v1]− b[v0]S∂[v0]
)= b[v0,v1]
(b[v1]S [∅]− b[v0]S [∅]
)= b[v0,v1]
(b[v1][∅]− b[v0][∅]
)= b[v0,v1]
([v1]− [v0]
)=
[v0 + v1
2, v1
]−[v0 + v1
2, v0
]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 75
If λ = [v0, · · · , vn] is a linear simplex and µ is a simplex of thebarycentric subdivision of λ then
diamµ ≤ n
n + 1· diamλ
Proof.
The inequality holds trivially for 0-simplices
Assume inductively that the inequality holds for n − 1 simplices.
By our inductive definition of barycentric subdivision
µ = [bλ,w0, · · · ,wn−1]
where [w0, · · · ,wn−1] is a simplex in the barycentric subdivision ofsome face [v0, · · · , vi , · · · , vn] of λ.
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Lemma 75 (continued).
Suppose wj ,wk are maximally distant vertices of µ. Then byLemma 66
diamµ = diam[w0, · · · ,wn−1]
≤ n − 1
ndiam[v0, · · · , vi , · · · , vn]
≤ n
n + 1diam[v0, · · · , vi , · · · , vn]
≤ n
n + 1diam[v0, · · · , vn]
Otherwise bλ and wj must be maximally distant vertices of µ where
wj =1
`+ 1
∑k=0
vn(k)
is a barycenter of some `-face of [v0, · · · , vi , · · · , vn]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Lemma 75 (continued).
diamµ =
∣∣∣∣∣bλ − 1
`+ 1
∑k=0
vn(k)
∣∣∣∣∣=
∣∣∣∣∣ 1
`+ 1
∑k=0
bλ − vn(k)
∣∣∣∣∣≤ 1
`+ 1
∑k=0
∣∣bλ − vn(k)∣∣ (ave. of dist. to some verts)
≤ max0≤k≤n
|bλ − vk |
It remains to show that the baryenter of λ is within nn+1 diamλ of
any vertex of λ.
Let bi = 1n (v0 + · · ·+ vi + · · ·+ vn) be the barycenter of
[v0, · · · , vi , · · · , vn]
Then bλ = 1n+1 (nbi + vi ) = n
n+1bi + 1n+1vi is a point on the segment
[vi , bi ] distance nn+1 |vi − bi | from vi .
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 76
S : LC (Rm)→ LC (Rm)
is a chain map.
Proof.
We must show ∂S = S∂.
Let λ = [v0, · · · , vn] then
∂Sλ = ∂bλ(S∂λ)
= (1LC(Rm) − bλ∂)(S∂λ) (Lemma 69)
= S∂λ− bλ∂S∂λ
= S∂λ− bλS∂∂λ (by induction)
= S∂λ
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 77
LetT : LCn(Rm)→ LCn+1(Rm)
T : LC−1(Rm)→ LC−1(Rm) is defined on the empty simplex by
T [∅] = 0 ∈ LC0(Rm)
and inductively assume T : LCn−1(Rm)→ LCn(Rm) is defined.
For λ = [v0, · · · , vn] with barycenter bλ set
T (λ) = bλ(λ− T∂λ)
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Example 78
Let’s compute T [v0, v1]
T [v0, v1] = b[v0,v1][v0, v1]− b[v0,v1]T∂[v0, v1]
=[v0+v1
2 , v0, v1]− b[v0,v1]T ([v1]− [v0])
=[v0+v1
2 , v0, v1]− b[v0,v1](T [v1]− T [v0])
=[v0+v1
2 , v0, v1]− b[v0,v1]
(b[v1]([v1]− T∂[v1])
− b[v0]([v0]− T∂[v0]))
=[v0+v1
2 , v0, v1]− b[v0,v1]
(b[v1][v1]− b[v1]T [∅]
− b[v0][v0]− b[v0]T [∅])
=[v0+v1
2 , v0, v1]− b[v0,v1]
([v1, v1]− b[v1]0
− [v0, v0]− b[v0]0)
=[v0+v1
2 , v0, v1]− b[v0,v1][v1, v1] + b[v0,v1][v0, v0]
=[v0+v1
2 , v0, v1]−[v0+v1
2 , v1, v1]
+[v0+v1
2 , v0, v0]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 79
T : LC (Rm)→ LC (Rm)
is a chain homotopy between 1LC(Rm) and S.
Proof.
We must show ∂T + T∂ = 1LC(Rm) − S .
∂Tλ = ∂bλ(λ− T∂λ)
= (1− bλ∂)(λ− T∂λ)
= λ− T∂λ− bλ∂λ+ bλ∂T∂λ
= λ− T∂λ− bλ(1− ∂T )∂λ
= λ− T∂λ− bλ(T∂ + S)∂λ (induction)
= λ− T∂λ− bλT∂∂λ− bλS∂λ
= λ− T∂λ− bλS∂λ
= λ− T∂λ− Sλ (definition of S)
= (1LC(Rm) − T∂ − S)λ
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
We may now define S and T for singular simplices.
Let {e0, e1, · · · , en} ⊂ Rn be the standard basis.
Note that[e0, · · · , en] : ∆n → ∆n
is the identity map
Definition 80 (Barycentric subdivision of singular simplices)
Let σ : ∆n → X be a singular simplex.
The barycentric subdivision of σ is
Sσ = σ] (S [e0, · · · , en]) .
Extend linearly to get
S : Cn(X )→ Cn(X )
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 81
S : Cn(X )→ Cn(X ) is a chain map
Proof.
We must show ∂S − S∂ = 0. Let σ : ∆n → X be a singular simplex.
∂Sσ = ∂σ]S [e0, · · · , en]
= σ]S∂[e0, · · · , en] (σ] and S chain maps)
= σ]Sn∑
i=0
(−1)i [e0, · · · , ei , · · · , en]
=n∑
i=0
(−1)iσ]S [e0, · · · , ei , · · · , en]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Lemma 81 (continued).
=n∑
i=0
(−1)iσ]S [e0, · · · , ei , · · · , en]
=n∑
i=0
(−1)iσ]S([e0, · · · , ei , · · · , en] ◦ [e0, · · · , en−1])
=n∑
i=0
(−1)iσ][e0, · · · , ei , · · · , en]]S [e0, · · · , en−1]
=n∑
i=0
(−1)iS(σ ◦ [e0, · · · , ei , · · · , en])
= Sn∑
i=0
(−1)i (σ ◦ [e0, · · · , ei , · · · , en])
= S∂σ
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011 Definition 82 (Barycentric subdivision of singular simplices)
Let σ : ∆n → X be a singular simplex.
Tσ = σ]T [e0, · · · , en].
Extend linearly to get
T : Cn(X )→ Cn+1(X )
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 83
T : Cn(X )→ Cn+1(X )
is a chain homotopy between S and 1C(X )
Proof.
∂Tσ = ∂σ]T [e0, · · · , en]
= σ]∂T [e0, · · · , en]
= σ](1− S − T∂)[e0, · · · , en]
= σ][e0, · · · , en]− σ]S [e0, · · · , en]− σ]T∂[e0, · · · , en]
= σ − Sσ − σ]Tn∑
i=0
(−1)i [e0, · · · , ei , · · · , en]
= σ − Sσ −n∑
i=0
(−1)iσ]T [e0, · · · , ei , · · · , en]
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Lemma 83 (continued).
= σ − Sσ −n∑
i=0
(−1)iσ]T [e0, · · · , ei , · · · , en]
= σ − Sσ −n∑
i=0
(−1)iσ]T ([e0, · · · , ei , · · · , en] ◦ [e0, · · · , en−1])
= σ − Sσ −n∑
i=0
(−1)iσ][e0, · · · , ei , · · · , en]]T [e0, · · · , en−1]
= σ − Sσ −n∑
i=0
(−1)iT (σ ◦ [e0, · · · , ei , · · · , en])
= σ − Sσ − Tn∑
i=0
(−1)i (σ ◦ [e0, · · · , ei , · · · , en])
= σ − Sσ − T∂σ
= (1− S − T∂)σ
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
A singular simplex may need to be subdivided many times before itbecomes a subordinate chain so we give a chain homotopy between1C(X ) and Sm
Definition 84
Let Dm : Cn(X )→ Cn+1(X ) be the operator
Dm =m−1∑i=0
TS i
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Lemma 85
Dm is a chain homotopy between 1C(X ) and Sm
Proof.
∂Dm + Dm∂ =m−1∑i=0
∂TS i + TS i∂
=m−1∑i=0
∂TS i + T∂S i
=m−1∑i=0
(∂T + T∂)S i
=m−1∑i=0
(1− S)S i
=m−1∑i=0
S i − S i+1
= 1C(X ) − Sm
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
• Let U = {Ui} be an open cover of X
• Given a singular simplex σ : ∆n → X
• Get an open cover σ−1(Ui ) of ∆n
• This open cover of ∆n has a Lebesgue number εσ > 0
• There is some m s.t.(
nn+1
)m< εσ
• Smσ ∈ CU (X )
• Define mσ to be minimum such m
Definition 86
Let D : Cn(X )→ Cn+1(X ) be given by
Dσ = Dmσσ
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Definition 87
Let ρ : Cn(X )→ Cn(X ) be given by
ρσ = Smσσ + Dmσ∂σ − D(∂σ)
• Claim: ρσ ∈ CU (X )
• By definition of mσ we have Smσσ ∈ CU (X )
• Note that m∂ iσ ≤ mσ so
Dmσ∂iσ − D(∂ iσ) =
mσ−1∑k=0
TSk∂ iσ −m∂iσ−1∑k=0
TSk∂ iσ
=mσ−1∑k=m∂iσ
TSk∂ iσ
By definition of m∂ iσ we have Sk∂ iσ ∈ CU (X ) for k ≥ m∂ iσ
• The claim follows
• We may view ρ as a homomorphism ρ : Cn(X )→ CUn (X )
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
ρσ = Smσσ + Dmσ∂σ − D(∂σ)
• By Lemma 85∂Dmσ = 1− Smσ − Dmσ∂
so∂Dσ = σ − Smσσ − Dmσ∂σ
Adding D∂σ to both sides
∂Dσ + D∂σ = σ − Smσσ − Dmσ∂σ + D∂σ
= σ − ρσ
• Claim: ρ : Cn(X )→ CUn (X ) is a chain map
∂ρσ = ∂(1− ∂D − D∂)σ
= ∂σ − ∂D∂σ
ρ∂σ = (1− ∂D − D∂)∂σ
= ∂σ − ∂D∂σ• The claim follows and D is a chain homotopty between ρ and 1
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Five Lemma
A5α5−−−−→ A4
α4−−−−→ A3α3−−−−→ A2
α2−−−−→ A1yγ5 yγ4 yγ3 yγ2 yγ1B5
β5−−−−→ B4β4−−−−→ B3
β3−−−−→ B2β2−−−−→ B1
If
1 If the diagram commutes (e.g. γ4 ◦ α5 = β5 ◦ γ5)
2 and rows are exact
3 If γ2, γ4 are surjective and γ1 is injective then γ3 is surjective.
4 If γ2, γ4 are injective and γ5 is surjective then γ3 is injective.
In the special case that γ1, γ2, γ4, γ5 are isomorphisms it follows thatγ3 is an isomorphism.
Exercise 88 (HW3 - Problem 1)
Prove the Five Lemma
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Theorem 89 (Excision for Singular Simplicial Homology)
For any pair (X ,A) if U ⊂ A is open with U ⊂ intA then theinclusion map
i : (X − U,A− U)→ (X ,A)
induces an isomorphism
i∗ : Hn(X − U,A− U) ∼= Hn(X ,A)
Proof of Excision.
• Let UX = {X − U, intA}• Let UA = {A− U, intA}
We clearly have an injection of chain maps
CUA(A) ↪→ CUX (X )
Or equivalently an exact sequence
0 −→ CUA(A) ↪→ CUX (X ) −→ CUX (X )/CUA(A) −→ 0
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Excision (continued).
We have the commuting diagram of chain complexes
0 −−−−−→ CUA (A) −−−−−→ CUX (X ) −−−−−→ CUX (X )/CUA (A) −−−−−→ 0y y y0 −−−−−→ C(A) −−−−−→ C(X ) −−−−−→ C(X ,A) −−−−−→ 0
Which yeilds commuting diagram of long exact homology sequences
Hn(CUA (A)) −−−−−−−→ Hn(C
UX (X )) −−−−−−−→ Hn(CUX (X )/CUA (A)) −−−−−−−→ Hn−1(C
UA (A)) −−−−−−−→ Hn−1(CUX (X ))y y y y y
Hn(A) −−−−−−−→ Hn(X ) −−−−−−−→ Hn(X, A) −−−−−−−→ Hn−1(A) −−−−−−−→ Hn−1(X )
By the Five Lemma
Hn(CUX (X )/CUA(A)) ∼= Hn(X ,A)
Homology theory
Lecture 8 -1/18/2011
Axiom A3(Excision)
Lecture 9 -1/19/2011
Proof of Excision (continued).
CUX (X ) = C (X − U) + C (intA)
CUA(X ) = C (A− U) + C (intA)
so by the First Isomorphism Theorem from group theory
CUX (X )/CUA(A) ∼= C (X − U)/C (A− U)
= C (X − U,A− U)