Download Algebraic Topology

1
Introduction
When are two spaces X, Y homeomorphic, that is, when is there a continuous
map f : X → Y with continuous inverse?
Let f, g : X → Y be continuous maps. f is homotopic to g if there exists a
homotopy, H : X × I → Y such that H(x, 0) = f (x) and H(x, 1) = g(x).
Put Ht : X → Y by Ht (x) = H(x, t), then H0 = f, H1 = g.
Notation: f, g, ft with f0 = f, f1 = g from Hatcher. If f is homotopic to g,
then we write f ' g. Similarly for spaces.
Let X, Y be two spaces. They have the same homotopy type if there are
continuous f : X → Y and g : Y → X such that g◦f : X → X and f ◦g : Y → Y
are homotopic to the appropriate identity maps.
Examples: In Rn , we have B n = Dn = {x ∈ Rn : kxk ≤ 1}.
B n has the homotopy type of a point, for example, {0}. Let i : {0} → B n
be the inclusion, and r : B n → {0} by r(x) = 0.
Then r◦i = id{0} and i◦r : B n → B n is homotopic to idB n by H : B n ×I → I
by H : B n × I → B n , H(x, t) = (1 − t)x.
But for n > 1, S n−1 = {x ∈ Rn : kxk = 1} are not homotopy equivalent to
a point.
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Fundamental Group
Definition 2.1 (Path). Let X be a topological space. A path in X is f : I → X
a continuous map.
Two paths from x0 to x1 are homotopic if there exists H : I × I → S such
that H(s, 0) = f (s), H(s, 1) = g(s), H(0, t) = x0 and H(1, t) = x1 .
In Euclidean space (in fact, any convex subspace of Euclidean space) every
path is homotopic to a straight line path, by a straight line homotopy.
Let f be a path from x0 to x1 and g be a path
from x1 to x2 , then f · g is a
f (2t)
0 ≤ t ≤ 1/2
new path from x0 to x2 defined by (f · g)t) =
g(2t − 1) 1/2 ≤ t ≤ 1
If x0 ∈ X, then let cx0 : I → X be the constant path cx0 (t) = x0 .
Easy: If f0 ' f1 from x0 to x1 and g0 ' g1 from x1 to x2 , then f0 ·g0 ' f1 ·g1 .
Definition 2.2 (Loop). Fix x0 ∈ X and call x0 the base point. A loop at x0 is
a path f in X such that f (0) = f (1) = x0 . Let [f ] be all loops at x0 which are
path homotopic to f . We denote this relation by f 'p g.
Definition 2.3. We define π1 (X, x0 ) = {[f ] : f is a loop at x0 }
(We can also define π0 (X, x0 ) = all path components of X with base point
being path component of x0 .)
Example: O(2) has two path components, the base point is x0 = I.
What is π1 (O(2), I)? SO(2) is the set of all rotations of R2 , and it is homeomorphic to the circle.
We claim that π1 (X, x0 ) is a group, and call it the Fundamental Group of
X at x0 .
1
Theorem 2.1. π1 (X, x0 ) with composition as above forms a group.
Proof. [f ][g] = [f · g] is the product. Take the identity to be [cx0 ]. For [f ]−1 ,
we take [f¯] where f¯ = f (1 − t). We must check (f · g) · g 'p f · (g · h),
f · cx0 'p f 'p cx0 · f and f · f¯ 'p f¯ · f 'p cx0 .
We can do these using reparameterizations. If f : I → X is any path, and
γ : I → I is continuous with γ(0) = 0 and γ(1) = 1, then f (γ(t)) is a path
which is homotopic to f . F (s, t) = f ((1 − t)γ(s) + ts).
Now, see page 27.
Lemma 2.2 (Pasting Lemma). Let X be a topological space such that X =
A1 ∪ . . . ∪ A` , each Ai is closed. Let f : X → Y be a function to another
topological space. If ∀i, f |Ai : Ai → Y is continuous, then so is f .
−1
Proof. Let C be a closed set in Y . Then f −1 (C) = f |−1
A1 (C) ∪ . . . ∪ f |A` (C).
Each term in this union is closed in the appropriate Ai , and as Ai is closed in
X, each term is itself closed in X. As the finite union of closed sets is closed,
f −1 (C) is closed in X, and so f is continuous.
Let f, g, h be paths in X such that (f · g) · h is defined. We want to show
that this composition is associative. That is, that (f · g) · h 'p f · (g · h). The
right hand side is also equal to [(f · g) · h] ◦ γ, where
 1
0 ≤ t ≤ 21
 2t
1
t − 2 12 ≤ t ≤ 43
γ=

2t − 1 43 ≤ t ≤ 1

4s
)
0 ≤ s ≤ (t + 1)/4
f ( t+1

g(4s − t − 1) (t + 1)/4 ≤ s ≤ (t + 2)/t
And so, we define H(s, t) =

h( 4s−2−t
(t + 2)/4 ≤ s ≤ 1
2−t )
Change of Base Point
Let X be a topological space, and x0 , x1 ∈ X such that there is a path h
from x0 to x1 . Then ∃βh : π1 (X, x1 ) → π1 (X, x0 ) given by βh ([f ]) = [h · f · h̄]
which is an isomorphism of groups.
And so, if X is path-connected, we can write π1 (X).
Definition 2.4 (Simply-Connected). Call X simply connected if X is path
connected and π1 (X) is trivial.
Quotient Topology
I = [0, 1], and we want to identify 0 ∼ 1.
≈
So I/ ∼ is a space, and we believe it is homeomorphic (→) the circle S 1 ⊂ C.
So we define p : I → S 1 such that p(t) = e2πit . p(0) = p(1) = 1 ∈ S 1 ⊂ C.
Definition 2.5 (Quotient Topology). More generally, let X be a topological
space and ∼ be an equivalence relation. Let g : X → X/ ∼ take x to [x]. Call
V ⊂ X/ ∼ open iff g −1 (V ) is open in X. Then we get a topology on X/ ∼ and
we get that
2
g .
X .. .................................................. X/ ∼
.....
..
..... ∀f
.....
.....
......
.......
........
.......
∃h
Y
Proposition 2.3. π1 (S 1 , 1) ' Z.
Proof. We want to define a homomorphism Φ : Z → π1 (S 1 ) be Φ(n) = [ωn ]
where ωn (s) = e2πins .
We need to show that this is a homomorphism. Let p : R → S 1 by p(s) =
2πis
e
. Note that ωn = p ◦ ω˜n , where ω˜n : I → R is given by ω˜n (s) = ns.
R
.......
......
.....
.....
.
.
.
.....
n ............
..
.....
.
.
.
.
.
.....
....
.....
.....
....
ω˜
ω
...
...
...
...
...
...
...
...
...
..
..........
.
p
...........................n
.............................
S1
Φ(n) = [p ◦ f˜] whenever f˜ is any path in R from 0 to n. Then ω˜n must be
homotopic to f˜.
And so, Φ is a homomorphism.
Let τm : R → R be a translation, τm : x 7→ x + m. Then ω˜m · τm ω˜n is
a path in R from 0 (to m) to n + m. Thus Φ(n + m) = [p ◦ (ω˜m · τm ω˜n ] =
[p ◦ ω˜m ] · [p(τm ω˜n )] = Φ(m) · Φ(n).
We need to show that Φ is surjective and that ker Φ = {0}.
We will need two facts.
I
1. If f : I → S 1 starting at x0 ∈ X and let x˜0 ∈ p−1 (x0 ), then ∃ unique lift
f˜ : I → R starting at x˜0 .
This implies that Φ is surjective, as we let [f ] ∈ π1 (S 1 , 1) and taking
x0 = 1, x˜0 = 0, we apply it and get f˜. Then p(f˜(1)) = f (1) = 1. So
˜
f˜(1) = m ∈ Z, e2πif (1) = 1 so Φ(m) = [p ◦ f˜] = [f ].
2. Let F : I ×I → S 1 be a homotopy of paths, all starting at x0 ∈ S 1 . Choose
some x˜0 ∈ p−1 (x0 ). Then, ∃ a unique lifted homotopy F̃ : I × I → R of
paths starting at x˜0 .
(R, x̃0 )
F̃ ......
..
....
..
.....
..
.....
..
....
.........
.
.....
.
..........
..
....
..
...
...
...
...
...
...
p
F
I × I ........................................ (S 1 , x0 )
We note that p ◦ F̃ = F , and so p ◦ f˜t = ft .
The second fact implies that Φ is injective. Suppose that Φ(n) = e = [c1 ] ∈
(S 1 , 1).
Assume that n 6= 0. Then [ωn ] = e = [c1 ]. Then ωn is path homotopic
to the constant map, and so, by the second fact, we can lift this homotopy
3
to ω˜n 'p ω˜0 = c0 , but these have different endpoints, and so cannot be path
homotopic, which is a contradiction.
All that remains is to prove these two facts.
Observe the property of p : R → S 1 : there exists an open cover {Uα }
of S 1 such that p−1 (Uα ) = ∪i Vαi where the Vαi are disjoint open sets and
p|Vαi : Vαi → Uα is a homeomorphism.
One such cover is the covering of S 1 by two open sets with appropriate
overlap.
We will first prove fact 1:
(R, x¯0 )
....
f˜ ....... ..
..
.....
..
.....
........
......
...
...
...
...
...
...
...
...
...
..
.........
..
p
..
....
f
(I, 0) ........................................ (S 1 , x0 )
The Lebesgue Covering Lemma states that if X is a compact metric space
and {Wα } is an open cover of X, then there exists λ > 0 such that every subset
of X with diameter less than λ lies in some Wα .
So, there exists a partition 0 = t0 < t1 < . . . , t` = 1 such that f ([ti−1 , ti ]) ⊂
some Uα . We construct a lifting f˜ on [0, tk ] by induction on k. We start with
t0 , [0, 0], f˜(0) = x˜0 Assume k < `, have f˜ such that p ◦ f˜ = f . We need
to define f˜ on [tk , tk+1 ]. We know that pf˜(tk ) = f (tk ), f (tk ) ∈ Uα for some
α, so f˜(tk ) ∈ Vαi for some unique i. Put f˜(t) for tk ≤ t ≤ tk+1 , equal to
p|Vαi f (t) ∈ Vα,i ⊂ R, so existence is done.
Now we prove uniqueness. If p : X̃ → X is a covering projection, that is, p
is onto and if given a continuous map
X̃, x˜0
.
.....
h, g ......
.
.....
.
.....
...
...
...
...
...
...
...
...
...
..
.........
.
........
......
p
..
.....
f
Y, y0 ............................................... X, x0
and if g, h : (Y, y0 ) → (X̃, x˜0 ) are two lifting maps of f , then g = h provided
that Y is connected, Y is locally connected. We prove this by putting A = {y ∈
Y : g(y) = h(y)} so that y0 ∈ A. Put B = {y ∈ Y : g(y) 6= h(y)}. Both A and
B are open.
We now must prove fact 2.
R, x˜0
F̃ ......
..
....
..
....
..
.....
..
....
........
.
.....
...
...
...
...
...
...
...
...
...
..
.........
.
p
F
I × I, (0, 0) ............................. S 1 , x0
Proof by picture:
Divide the unit square into a smaller grid. F (lower left square) ⊂ Uα , and so
4
F̃ (segment I) ⊂ Vαi , and you can extend the map by continuing it from square
to adjacent square. Or see Hatcher.
Theorem 2.4 (Brouwer Fixed Point Theorem). Any continuous f : D2 → D2
has a fixed point.
Usually this is proved with the no retraction theorem, though can be proved
the other way. The no retraction theorem states that there is no continuous
r : D2 → S 1 such that r(x) = x for x ∈ S 1 .
Theorem 2.5 (Borsuk-Ulam Theorem). If f : S n → Rn is continuous, then
there exists x ∈ S n such that f (x) = f (−x).
For n = 1, intermediate value theorem proves it. For n = 2, use the fundamental group, see page 32. For n ≥ 2, need homology and transfer or cohomology.
Look at page 38, exercise 9.
π1 as a functor.
(X, x0 ) gets π1 (X, x0 ) the fundamental group.
φ
φ∗
Also, (X, x0 ) → (Y, y0 ) induces a homomorphism of groups π1 (X, x0 ) →
π1 (Y, y0 ) define by φ∗ ([f ]) = [φ ◦ f ]. Note φ = idX ⇒ φ∗ = idπ1 (X,x0 ) and
φ
ψ
(X, x0 ) → (Y, y0 ) → (Z, z0 ) gives ψ∗ ◦ φ∗ : π1 (X, x0 ) → π1 (Z, z0 ) equal to
(ψ ◦ φ)∗ .
Proposition 2.6. π1 (X ×Y ) ' π1 (X)×π1 (Y ), and the isomorphism is obtained
by the induced maps from the projection maps of X × Y to X, Y .
π1 (S 1 × . . . × S 1 ) ' Zn
{z
}
|
n times
S 1 ∨ S 1 is two circles intersecting at a point. Then π1 (S 1 ∨ S 1 ) = F2 . We
will need the Seifert-Van Kampen theorem.
If we take C \ {−1, 1}, we see that it retracts onto S 1 ∨ S 1 .
Proposition 2.7. π1 S n = 0 for n > 1.
Corollary 2.8. π1 (Rn \ {0}) = 0 if n > 2.
S n−1 ⊂ Rn \ {0} is a homotopy equivalence by r(x) =
x
tx + (1 − t) kxk
.
x
kxk ,
and rt (x) =
Definition 2.6 (Retraction). If A ⊆ X, then a retraction of X onto A is
r : X → A such that r(a) = a for all a ∈ A. If i : A → X be the inclusion, then
r ◦ i = idA .
i
∗
Choose x0 ∈ A ⊆ X. Then π1 (A, x0 ) →
π1 (X, x0 ) and r∗ in the other
direction, we now have that r∗ ◦ i∗ = idπ1 (A) . And so, i∗ is injective.
∃ no retraction r : D2 → S 1 . Why? If there were one, then there would be
a homomorphism π1 (S 1 ) → π1 (D2 ) which is injective, that is, an injective map
Z → 0, which is impossible.
Notation: X ≈ Y is used to denote homeomorphic, X ' Y is homotopic.
If f ' g : (X, x0 ) → (Y, y0 ) then f∗ = g∗ : π1 (X, x0 ) → π1 (Y, y0 ).
5
Corollary 2.9. If f : (X, x0 ) → (Y, y0 ) is a homotopy equivalence of spaces
with base points, then f∗ : π1 (X, x0 ) → π1 (Y, y0 ) is an isomorphism.
The reason is that ∃g : (Y, y0 ) → (X, x0 ) such that g ◦ f ' idX , f ◦ g ' idY
as based maps, and so π1 (X, x0 ) → π( Y, y0 ) has an inverse homomorphism, g∗ ,
and so they are inverse isomorphisms.
Definition 2.7 (Deformation Retraction). Suppose A ⊆ X is a subspace. A
deformation retraction of X onto A is a homotopy H : X × I → X such that
for all x ∈ X, H(x, 0) = x, H(x, 1) ∈ A and H(a, t) = a for all t if a ∈ A.
Then i : A → X is a homotopy equivalence, and r = H(x, 1) : X → A is a
homotopy inverse.
x
that S n ' Rn+1 \ {0}
Note S n ⊆ Rn+1 \ {0} shows, by x 7→ tx + (1 − t) kxk
Proposition 2.10 (Prop 1.14). π1 (S n ) = 0 for n ≥ 2.
Proof. S n = U ∪V where U, V are homeomorphic to Rn . U ∩V is path connected
and homotopic to S n−1 × (−, ).
We now show that π1 (S n , x0 ) = {1}.
Claim: If X = U ∪ V , U, V open, x0 ∈ U ∩ V , and U ∩ V path connected,
consider:
π1 (X, x0 )
.........
..
...
..
.
∗ .....
..
...
...
..
.
..
i
......
......
...
...
...
...
...
.
∗ .......
...
..
j
π1 (U, x0 )
π1 (V, x0 )
Then each element of π1 (X, x0 ) is a finite product of elements in Im i∗ or
Im j∗ .
Proof of claim: Let f : I → X be a loop at x0 , f : [0, 1] → X. Use Lebesgue
covering lemma to get partition 0 = t0 < t1 < . . . < t`−1 < t` = 1 such that
f ([ti−1 , ti ]) is contained in U or V .
Assume that f ([tt−1 , ti ]) ⊆ U , then f ([ti , ti+1 ]) ⊆ V . We call the restriction
to each of these fi , and so f 'p f1 · . . . · f` , and it is homotopic to (f1 · g¯1 ) · . . . ·
(g`−1 · f` ).
f 'p f1 · f2 · f3 'p f1 g¯1 · g1 · f2 · g¯2 · g2 · f3 , that is, a loop in U , one in V ,
and then one in U .
Theorem 2.11 (Easy Part of Van Kampen Theorem). Let X be ∪Aα with Aα
path connected and open in X, x0 ∈ Aα for all α. Assume that Aα ∩ Aβ is
aways path connected. Then iα → X the inclusion induces iα∗ : π1 (Aα , x0 ) →
π1 (X, x0 ) so that each Im(iα∗ ) is a subgroup of π1 (X, x0 ).
Then, each element of π1 (X, x0 ) is a product of elements in Im iα∗
Proposition 2.12. Let φ : X → Y be a homotopy equivalence, and x0 ∈ X.
Then φ∗ : π1 (X, x0 ) → π1 (Y, φ(x0 )) is an isomorphism.
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Lemma 2.13. Let φt : X → Y be a homotopy, let x0 ∈ X, and let h(t) =
φt (x0 ), a path in Y from φ0 (x0 ) to φ1 (x0 ). Then here’s a commutative triangle
π1 (Y, φ1 (x0 ))
...........
.....
.....
.....
.
.
.
.
h ............
..
.....
.....
.
.
.
.
...
.
.
.
.
..
.....
....
β
...
...
...
...
...
...
...
...
...
..
.........
.
φ1∗
φ0
π1 (X, x0 ) ..............∗π1 (Y, φ0 (x0 ))
Proof. Put ht (s) = h(ts). Let f be a loop in X at x0 . Inspect ht · (φt f ) · h¯t , a
loop at φ0 (x0 ). If t = 0, get φ0∗ ([f ]). If t = 1, then get βh (φ1∗ ([f ])).
Now we prove the prop:
Proof. φ : X → Y , ψ : Y → X and ψ ◦ φ ' idX , φ ◦ ψ ' idY .
The Lemma gives that ψ∗ ◦ φ∗ is an isomorphism, so φ∗ is injective.
β
α
Also, it says that ψ∗ is an isomorphism, and so A → B → C are groups, with
α, β injective and β ◦α isomorphism, then α is onto, and so an isomorphism.
We now go back to chapter zero to define CW-Complexes.
Definition 2.8 (CW-Complexes). Let X be a Hausdorff Space and assume that
X has an increasing filtration by closed subspaces, that is, there are X 0 ⊂ X 1 ⊂
. . . ⊂ X n ⊂ . . ., such that
1. X = ∪n≥0 X n We call X n the n-skeleton
2. X 0 is discrete called the vertices or 0-cells.
3. X 1 the 1-skeleton adjoins the edges, the 1-cells, gives a graph.
n
n−1
4. X
by adjoining n-cells as follows: X n = (X n−1 t
` is nobtained from Xn
n−1
, then x ∼ φα (x) ∈ X n−1 for φα : ∂Dαn →
=
S
)/
∼
if
x
∈
∂D
D
α
α
α α
X n−1 continuous.
Further, a subset A of X is open (closed) in X iff A ∩ X n is open (closed)
in X n .
Note: When is A ⊂ X n closed? True
∩ X n−1 is closed and under the
` iff A
n
n−1
n
n
characteristic maps Φα : Dα ⊂ X
t β Dβ → X n , Φ−1
α (A) is closed in Dα .
n−1
n
n
n
n
an open n-cell. Φα (Dα , Sα ) →
Also, eα = Φα (int of Dα ) ⊂ X is S
(X n , X n−1 ), Φα |Sαn−1 = φα , X n = X n−1 ∪ α enα .
Third, enα = Φα (Dαn ) may be identified with e˙nα = enα \ enα = φα (Sαn−1 )
Examples
1. Any graph, even infinite.
V
2. α Sα1 for any index set.
7
P
3. If X is a finite CW-complex, then χ(X) = i≥0 (−1)i αi (X) where αi (X)
is the number of i-cells. For T 2 , α0 = 1, α1 = 2, α2 = 1 so χ = 0.
4. S n start with S 1 the circle in C. χ(S 1 ) = 0 because we get n edges and n
vertices.
For S 2 , we see the half spheres are 2-cells, and then notice that we get 2
i-cells for each, which gives χ(S 2 ) = 2.
S n is a CW complex with 2 i-cells for 0 ≤
i ≤ n. More economically,
2 n ∈ 2Z
S n = e0 ∪ en , and so χ(S n ) = 1 + (−1)n =
0 n∈
/ 2Z
5. RPn = S n / ∼ where x ∼ −x. It is also the set of lines through the origin
in Rn+1 .
6. CPn = S 2n+1 ⊂ Cn+1 with x ∼ y if x = λy for λ ∈ S 1 .
RPn = RPn−1 ∪ en ? Well, RPn = S n / ∼ identifying antipodal points, which
is equal to Dn / ∼, identification of antipodal points on the boundary. φ :
Dn → Dn / ∼ gives us the attaching map, because S n−1 → S n−1 / ∼= RPn−1
are contained in Dn .
Example: {S n }, S n−1 ⊂ S n . Then S ∞ = ∪S n is contractible.
RP∞ = ∪RPn and CP∞ = ∪CPn .
Definition 2.9 (Subcomplex). Let X be a CW-complex. Call a subspace A ⊂ X
if A is closed, and if enα ∩ A 6= ∅ then enα ⊆ A.
Put An = A ∩ X n for all n. Then A is a CW-complex. For example, each
X is a subcomplex of X. We call (X, A) a CW-pair if X is a CW-complex
and A is a subcomplex of X. Then X/A is what you get when you identify all
points of A. X/A can be made naturally into a CW-complex.
FACT: If (X, A) is a CW-pair and A is contractible to a point, then g : X →
X/A is a homotopy equivalence.
Homotopy Extension Property: Given A ⊂ X, A closed and f : A → Y , can
you extend f to f˜ : X → Y ?
Sometimes, given i : A → X and f : A → Y , there exists f ' g such that
g does have an extension g̃ : X → Y . Want to be able to conclude that f has
extension f˜ : X → Y .
n
Definition 2.10 (Homotopy Extension Property: Provisional Def). (X, A), A
closed in X, has the HEP iff whenever f ' g : A → X and g has an extension
f˜ : X → Y so does g.
Definition 2.11 (Homotopy Extension Property). (X, A) has HEP if any map
F : A × I ∪ X × 0 → Y has extension F̃ : X × I → Y .
When does (X, A) have HEP? Answer is iff A × I ∪ X × 0 is a retract of
X × I.
Example: (Dn , S n−1 ) has HEP. Proof: Dn × I is a solid cylinder, S n−1 ×
I ∪ Dn × 0 is the boundary, minus the top. It is a retraction because if we take
8
a point higher up and hold it still, we draw a line through it to any point on
the sides or bottom, and everything on that line gets projected to that point.
This implies that any CW-pair (X, A) has HEP (prop 0.16). Prop 0.17 says
that if (X, A) has HEP and A is contractible, then X → X/A is a homotopy
equivalence.
Topological Properties of CW-complexes: X is path connected iff X 1 is path
connected. X is locally contractible. X is normal. X is compactly generated.
If K is a compact subset of X, then ∃n such that K ⊆ X n , even better, K
is contained in a finite union of closed cells, best, K is contained in a finite
subcomplex.
And now, we return to chapter 1.
We need an algebraic aside so that we can do VK. Problem: Given groups
{Gα }α∈A and homomorphisms φα : Gα → H, H another group, want some
group G together with homomorphisms iα : Gα → G such that ∃Φ : G → H
unique subject to
iα
Gα ........................................................ . G
.
.....
.....
..... φα
.....
..
......
....
.
.
.
.......
...... Φ
G
Then (G, iα ) is the free product or coproduct of the Gα ’s.
Make G = ∗α Gα as all words h1 h2 . . . h` with hi ∈ Gαi \ {1}. We call it a
reduced word if it cannot be simplified further. G is the reduced words under
concatenation.
Theorem 2.14 (Seifert-van Kampen). Let X be a topological space, x0 ∈ X
such that X = ∪α∈I Aα with Aα open, path-connected and x0 ∈ Aα for all
α. If Aα ∩ Aβ ∩ Aγ is always path-connected, then we have a homomorphism
Φ : ∗α π1 (Aα ) → π1 (X) which is surjective, and N = ker Φ is the normal
subgroup of the free product generated by all elements iαβ∗ (ω)i−1
βα∗ (ω) where
iαβ : Aα ∩ Aβ → Aα is the inclusion.
That is, π1 (X) ' ∗α (Aα )/N .
A universal property holds:
G
. ... .
............... ....
. . .
.. . ....
.
.
... . ...
... ... ....
...
...
...
..
..
.
.
...
.
.
... β
α .....
.
...
..
..
...
.
.
..
...
.
...
..
.
...
..
.
...
1
..
.
..
.. .....
.
.
......... ....
.. .........
.
.
.
.
..... ...
..... .
... .........
..
.
∃!
φ
φ
π (X)
π1 (Aα )
........
.........
.....
...
π1 (Aβ )
........
......
.....
.....
π1 (Aα ∩ Aβ )
W
Example: π1 ( α Xα ) with xα ∈ Xα and ∀α, xα ∈ Uα ⊂ Xα , Uα open and
Uα deformation retracts to xα . Then π1 (X) ' ∗π1 (Xα ), where X = ∨α Xα .
Also, assume each Aα is path connected, then so is X.
9
W
W
Proof. Take Aα = Xα ∨( β Uβ ) ' Xα . Then for α 6= β, Aα ∩Aβ = γ Uγ ' x0 .
Now we apply VK Theorem trivially.
Using this, we find that π1 (S 1 ∨S 1 ) = Z∗Z, the free group on two generators.
π1 (S 1 ∨ S 2 ) ' Z.
In fact, if G is any connected graph, then a maximal tree can be contracted
to a point, and so has fundamental group free, and so is a bouquet of circles.
1
π1 (R2 \ N pts) =free group on n generators. This space contracts to ∨N
i=1 S .
2
So π1 (S \ N ) ' free group on n − 1 generators.
Suppose Rn \ K is path connected, K is closed and bounded. π1 (Rn \ K) →
π1 (S n \ K) is an isomorphism when n > 2.
Proof. Take U = Rn \ K and V = S n \ BR (0) where K ⊆ BR (0). Note that
U ∩ V = Rn \ B R (0), which is path connected. (ADD DETAILS LATER)
Let A be a circle in the plane. Then R3 \ A ' S 1 , so π1 (R3 \ A) = Z.
Proposition 2.15. Inspect S p+q−1 ⊂ Pp+q . It has S p−1 , S q−1 as subspheres.
Then there exists a deformation retraction from S p+q−1 \S q−1 onto S p−1 . Thus,
π1 (S 3 \ A) ' π1 (S 1 ).
Proof. Define ft : S p+q−1 \ S q−1 → S p+q−1 \ S q−1 by ft (x, y) = ((1 − t)(x, y) +
x
t( kxk
), 0)/(kT OP k).
R3 \ (A ∪ B) where A, B are unknotted circles contracts to S 1 ∨ S 1 .
R3 \ (A ∪ B) where A, B are linked, then R3 \ (A ∪ B) ' S 1 × S 1 , and so the
fundamental group is Z × Z.
What is π1 (R3 \ (A ∪ B)), the complement of linked circles? It is isomorphic
to π1 (S 3 \ (A ∪ B)). Now, S 3 = {(z, w) ∈ C2 : |z|2 + |w|2 = 1}. Put Tz =
{(z, w) ∈ S 3 : |w|2 ≤ 1/2} and Tw = {(z, w) ∈ S 3 : |z|2 ≤ 1/2}. Note that
{(z, 0) : |z| = 1} ⊂ Tz and {(0, w) : |w| = 1} ⊂ Tw . Now, S 3 = Tz ∪ Tw and
Tz ∩ Tw = {(z, w) ∈ S 3 : |z|2 = |w|2 + 1/2} ' S 1 × S 1 . So Tz ' S 1 × D2 the
solid torus.
Tz ∩ Tw is a deformation retract of S 3 \ (A ∪ B).
Tz = {(z, w) : |z|2 + |w|2 = 1, |w|2 ≤ 1/2}. z = |z|eiθ and |z|2 = 1 − |w|2 so
(z, w) 7→ (eiθ = z/|z|, w) ∈ S 1 × √12 D2 ' S 1 × D2 .
π1 (Km,n ) where (m, n) = 1 is the group ha, b|am b−n i.
Attaching 2-cells creates relations by how they are attached.
Let X be a path connected
x0 ∈ X Attach 2-cells via ϕα : (S 1 , s0 ) →
F space,
2
(X, x0 ) and put Y = (X t α Dα )/ ∼ where ϕα (x) ∼ x for all x ∈ Sα1 .
Proposition 2.16. Then π1 (Y, x0 ) ' π1 (X, x0 )/N , N is the smallest subgroup
containing all [ϕα ] ∈ π1 (X, x0 )
Proof. Case 1: a Single 2-cell. Φ : D2 → e2 ⊂ Y , y = Φ(0). Y = (Y \{y})∪e2 =
e2 \ {y} ' S 1 . VK implies that π1 (Y, Φ(d1 )) ' π1 (X \ {y}, Φ(d1 ))/minimal
subgroup generated by [f ].
π1 (Y, Φ(d1 )) ' π1 (Y, x0 ) ' π1 (X, x0 )/[ϕα ].
10
Case 2: Induction for the case of a finite number of 2-cells.
Case 3: Any number at all: S
Let ϕα : S 1S→ X. Y = X ∪ α∈A e2α , yα = Φα (0).
Uα = X ∪ β (eβ \ yβ ) ∪ Eα = Y = ∪β {yβ } ∪ eα .
Uα will be open and path connected subsets. Thus, α 6= β ⇒ Uα ∩ Uβ =
Y − ∪β∈A {yβ } ' X.
S Let F be all finite subsets of A. Put for F ∈ F , UF = ∪α∈F Uα ' X ∪
αF eα .
We know π1 (UF ) = π1 (X)/normal subgroup generated by [ϕα ] for α ∈ F .
VK Theorem: Note that UF1 ∩UF2 = UF1 ∩F2 , and also for triple intersections.
Thus,
G
. ... .
.........................
.. .. ..
.. . ....
.
... ... ....
...
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...
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...
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.
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...
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1
...
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.
.
.. .......
........ ....
.
..... ...
.. .......
.
.
..... ..
.. ........
.
.
...
.
∃!
π (Y )
π1 (UF1 )
........
.........
.....
...
π1 (VF2 )
........
......
....
.....
π1 (UF1 ∩F2 )
If F is the set of all finite subsets of A and we order F by inclusion, we
obtain an directed ordered set. Given F1 , F2 ∈ F , note that F1 ⊂ F1 ∪ F2 and
F2 ⊂ F1 ∪ F2 .
Thus, π1 (Y ) is the colimit (or direct limit) of {π1 (UF )}F ∈F along with the
homomorphisms from inclusions in F .
An algebraic comment: If D is a directed set α ∈ D with Gα a group and
d ≤ d0 implies there is a homomorphism id,d0 : Gd → Gd0 and d = d0 implies it
is the identity, with d ≤ d0 ≤ d00 gives
Gd ...................................................... Gd00
.....
.....
.....
......
......
.
.....
.....
........
.......
Gd0
Then lim = colim(Gα , iα,α0 ) = ∗d Gα /N where N is the normal subgroup
−→
generated by {gα−1 iα,α0 (gα )} with gα ∈ Gα and α ≤ α0 .
NEXT: Covering Spaces, read the section
Definition 2.12 (Covering Space). A covering map p : X̃ → X is a map such
that X is covered by evenly covered open sets U , that is, p−1 (U ) = ∪α Vα , Vα ’s
open and disjoint in X̃ with p : Vα → U a homeomorphism.
Remarks: p is a local homeomorphism, meaning for all x̃ ∈ X̃, there is an
open nbhd V of x̃ such that p|V : V → p(V ) is a homeomorphism p(V ) open in
X.
p is an open mapping. p is a quotient mapping, provided p is surjective.
11
Examples: ∅ and any X, id : X → X.
The number of sheets is the index of the subgroup that is the fundamental
group of the covering space. It is normal if you get the same subgroup (rather
than a conjugate one) from a change of basepoint.
Lifting Theorems
Theorem 2.17 (Homotopy Lifting Property). Given
f˜0 .
Y ....................................................... X̃
...
...
...
...
...
...
...
...
...
..
.........
..
inc
...
...
...
...
...
...
...
...
...
..
.........
..
p
{ft }
Y × I .................................................. X
Then there exist a unique {f˜t } : Y × I → X̃.
Theorem 2.18. If p : X̃ → X is any covering map and p(x˜0 ) = x0 then
1. p∗ : π1 (X̃, x̃0 ) → π1 (X, x0 ) is injective
2. Im p∗ = {[f ] : f based at x0 such that the lifting f˜ such that f˜(0) = x̃0 is
a loop}.
Proof.
1. how that ker p∗ = 1. Suppose that [f˜] ∈ π1 (X̃, x̃0 ) such that
[p ◦ f˜] = 1 ∈ π1 (X, x0 ). So p ◦ f˜ = f ' cx0 . The HLT says that we can
lift to get f˜ 'p cx̃0 , and so [f˜] = 1.
2. ⊇ is clear. ⊆ is also clear.
Proposition 2.19. If X̃ and X are path connected, then |p−1 (x)|, x ∈ X is
constant and equal to [π1 (X, x0 ) : p∗ (π1 (X̃, x̃0 ))].
Theorem 2.20 (Lifting Theorem). Assume that Y is path connected and locally
path connected, with f : (Y, y0 ) → (X, x0 ) and p : (X̃, x̃0 ) → (X, x0 ) is a
covering map. A lift f˜ : Y → X̃ exists iff f∗ (π1 (Y, y0 )) ⊆ p∗ (π1 (X̃, x̃0 )).
Proof. It f˜ exists, the inclusion is clear.
Assume that the inclusion holds. Take y ∈ Y run a path γy = γ from y0 to
y. Then f ◦ γy is a path in X from x0 to f (y). Lift to f ◦˜γy which is a path in
X̃ from x̃0 to a point we shall name f˜(y). f˜(y) = f ◦˜γy (1) is set to f (y) by p.
Must check the independence of path from y0 to y, that f˜ : Y → X̃ is
continuous and f˜(y0 ) = x̃0 , though the last part is trivial.
So is f ◦˜γy (1) = f ◦˜ δy (1) for δ, γ paths from y0 to y? (f ◦ γy ) · f ◦ δy '
f ◦ (γy · δ̄y ) 'p p ◦ where is a loop at x̃0 in X̃. Thus f ◦ γy ' (p ◦ ) · (f ◦ δy ),
and so f ◦˜γy 'p · f ◦˜ δy . Thus, the endpoints are the same.
Why must f˜ : Y → X̃ be continuous? Let y ∈ Y , and check continuity at y.
f (y) ∈ U evenly covered, f˜(g) ∈ Vα , p : Vα → U homeo. Show that there is a
12
neighborhood W of y such that f˜(W ) ⊆ Vα . Since U is open and f (y) ∈ U , there
is an open nbhd W of y such that f (W ) ⊂ U hypotheses on Y let us assume that
W is path connected. Recall, f˜(y) = f ◦˜γy (1) ∈ Vα . Let w ∈ W , run a path in
W , ζw from y to w. Then f˜(w) = f ◦ (γ˜y · ζw )(1) = (f ◦ γy ) ˜· (f ◦ ζw )(1). The
first part ends at f (y), and the second part happens in U .
And so, this equals f ◦˜γy ·p−1 (f ·ζw )(1) ∈ Vα . And so, the map is continuous.
Next we construct the universal cover. If H ≤ π1 (X, x0 ) then we construct
X̃H → X such that pα (X̃H ) = H. We will also attempt to classify all covering
spaces of some ”good” X up to isomorphism.
Graphs
For a graph, χ(X) = |V | − |E|, and if X̃ is an n-sheeted cover of X, then
χ(X̃) = nχ(X).
W
take X → X/T ≈ S 1 is a homotopy equivalence, and so χ(X) =
WIf we
χ( S 1 ) = 1−k, where k is the number of circles. Thus, χ(X̃) = n(1−k) = 1−`
and ` = 1 + (n − 1)n.
If X is some nice space, path conn and locally path conn, then an isomorphism of spaces over X is f : X̃1 → X̃2 such that p2 ◦ f = p1 .
We define Cov(X) =all isomorphism classes of coverings of X.
Ỹ = p−1 (X̃)
X̃
...
...
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...
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...
...
...
..
.........
.
p
...
...
...
...
...
...
...
...
...
..
.........
.
p
φ .
Y ....................................................... X
then Ỹ = {(y, x̃) ∈ Y × X̃ : φ(y) = p(x̃)} maps down to Y by πY . Thus, we
get a map φ∗ : Cov(X) → Cov(Y )
Assume that X is as before and also that for each x ∈ X, there is an open
U containing x such that every loop in U is homotopic to the constant loop in
X. Then we call it SLSC (Semi-Locally Simply Connected).
[Losternik-Schrilmann Category of a Space]: RPn = S n /(x ∼ −x). Let
Ui ⊂ RPn be given by xi 6= 0. So RPn is a union of n + 1 contractible sets. In
general, X = V0 ∪ . . . ∪ Vn is an open cover such that ιi : Vi → X is homotopic
to a constant map. This is similar to SLSC.
Theorem 2.21. There exists a simply connected covering space if SLSC holds
for X, and conversely.
Let pu : Xu → X be simply connected covering space of X and let X̃ → X be
any other covering space, X̃ also path connected. Then the diagram commutes:
Xu , xu ......................p.................... X̃, x̃0
.
.........
....
..
....
u .....
.....
.
....
.
pφ
X, x0
13
Note: §1.3 Ex 16 says that φ is also a covering map. Can a s.c. X have
covering space no isom to id : X → X? p : X̃ → X given, then isomorphic as it
must be a one-sheeted covering space.
Easy: If 1-sheet, then p is a homeomorphism. Also if s is a section of the
covering map (p ◦ s = idX ) and s(x) ∈ p−1 (x) for all x ∈ X, then p is a
homeomorphism.
Proposition 2.22. If H is a subgroup of π1 (X, x0 ) then there exists a covering
X̃ → X such that the induced map p∗ (π1 (X̃, x̃0 )) = H.
We define an automorphism of a covering space to be an isomorphism f :
X̃ → X̃. We also call these Deck Transformations. We name the group of these
as G(X̃, p) = G(X̃/X).
If for any two points x̃1 , x̃2 such that p(x̃1 ) = p(x̃2 ) ∈ X, there exists
φ ∈ G(X̃, p) such that φ(x̃1 ) = x̃2 , call p normal.
Proposition 2.23. Let p : (X̃, x̃0 ) → (X, x0 ) be a covering with X̃, X path
conn and X locally path conn. Then
1. p is normal iff p∗ π1 (X̃, x̃0 ) is a normal subgroup of π1 (X, x0 ).
2. Assuming p is normal, there exists an isomorphism G(X̃, p) ' π1 (X, x0 )/p∗ π1 (X̃, x̃0 ).
So for the universal covering, G(X̃, p) ' π1 (X, x0 )
Assume X̃ → X is a universal cover. So G(X̃, p) acts transitively on p−1 (x0 ).
Define φ : π1 (X, x0 ) → G(X̃, p) by if [γ] ∈ π1 (X, x0 ) lift γ to γ̃ a path in X̃
with γ̃(0) = x̃0 and γ̃(1) ∈ p−1 (x0 ). Then there exists g ∈ G(X̃, p) such that
g(x̃0 ) = γ̃(1). Put φ([γ])(x̃0 ) = γ̃(1), and φ([γ]) ∈ G(X̃, p).
This element is unique, as if g1 , g2 ∈ G(X̃, p) and if x̃0 such that g1 x̃0 = g2 x̃0
then g1 = g2 by uniqueness of lifting.
Then we can show easily that φ is a homomorphism of groups and that φ is
surjective. φ is also injective.
Change point of view:
If Y is a given space and G is a group, G × Y → Y an action such that
g : Y → Y is a homeomorphism of Y . Then y ∼ gy for g ∈ G gives an
equivalence relation, and Y → Y /G is the space of orbits.
Favorite Example: Y = S n , G = {±1} ' Z2 by (+1)y = y and (−1)y = −y,
the antipodal map. Y /G = S n /Z2 ' RPn . Also π1 (RPn ) ' G ' Z2 .
Let G act on Y . Call the action even if ∀g ∈ Y , ∃ open neighborhood V of
y such that g 6= h ⇒ gV ∩ hV = ∅.
Lemma 2.24. If G acts evenly on Y , then p : Y → Y /G is a covering map.
Proof. We know p is continuous and open since if W ⊆ Y is open then p−1 (p(W )) =
∪g∈G gW . In fact, this is a disjoin union.
Claim: For V as in definition, then p(V ) is evenly covered. Note that p|gV :
gV → V is bijective. [Clearly surjective, for injective, we say gv1 , gv2 are in the
same orbit, so ∃h ∈ G such that gv1 = hgv2 , disjointness implies that hg = g,
so h = 1, hence v1 = v2 .]
14
We call such a covering space a G-covering.
Isomorphism of G-coverings is φ : Y1 → Y2 over Y1 /G ≈ X ≈ Y2 /G such
that φ is a homeo and φ(gy) = gφ(y).
Example, Y = S 2n−1 ⊂ Cn , G = {z ∈ C : z k = 1}, then G × S 2n−1 → S 2n−1
by v 7→ zv. Then S 2n−1 → S 2n−1 /G gives us a lens space.
Trivial G-covering over a given X is X × G → X by projection where G has
the discrete topology on it. h(x, g) = (x, hg).
Lemma 2.25. Any G-covering Y → X is locally trivial as a G-covering. ie,
X can be covered by open sets U such that gv ∈ p−1 (U ) → U is isomorphic to
U × G → U by projection.
Proposition 2.26. Let p : Y → X be a G-covering. Then we have a homomorphism G → Aut(Y, p) which is injective. The the case where Y is connected,
this is an isomorphism.
Proof. Pick y0 ∈ Y , then for ϕ ∈ Aut(Y, p) = G(Y, p), ϕ(Y0 ) satisfies p(ϕ(y0 )) =
py0 so ∃g ∈ G such that ϕ(y0 ) = gy0 . So ϕ and y 7→ gy0 are 2 elements of
Aut(Y, p) which agree on y0 . Uniqueness props for lifting implies that φ sends
y → gy.
Proposition 2.27. Let p : Y → X be a covering, Y connected, then Aut(Y, p)
acts evenly on Y . If Aut(Y, p) acts transitively on a fiber of p, then the covering
is a G-covering for G = Aut(Y, p).
Chapter 11 and 13 of Fulton.
Proposition 2.28. Let G act evenly on a simply connected and locally path
connected space Y . Then for X = Y /G, π1 X = G.
Proof. We’ve seen that π1 (X) ' Aut(Y, p) for p : Y → X and prop above
implies that G ' Aut(Y, p), so π1 (X) ' Aut(Y, p) ' G.
Chapter 14 - The Van Kampen Theorem
x0 ∈ X = U ∪ V , x0 ∈ U ∩ V , U, V, U ∩ V path connected, then we can define
the fundamental group and a map π1 (X) → G which is unique to each G that
makes a commutative diagram commute.
What is the meaning of a π1 (X, x0 ) → G?
Answer: G-coverings of X up to isomorphism preserving basepoints.
Suppose given a homomorphism ρ : π1 (X, x0 ) → G. Construct a G-covering
pρ : Yρ → X with yρ ∈ Yρ s.t. pρ (yρ ) = x0 .
Give G the discrete topology. Consider X̃ × G and a left action of π1 (X, x0 )
on it by [σ] · (z, g) = ([σ] · z, g · ρ([σ])−1 ).
Then, put Yρ = (X̃ × G)/π1 (X).
15
3
Homology
Let X be a topological space. We want to define Hi (X) the homology groups
i ≥ 0 with some good properties. We expect it to be functorial and to be useful.
∆-complexes
We look at T 2 .
v
v
a
•.......................................................................................•.......
a
..
.... ...
..
...
.....
.....
..
...
.....
.
.
...
.
...
.
...
.
.
...
.
...
.
...
.
.
.
...
...
.
....
.
.
...
...
.
...
.
.
...
.
...
.
...
.
.
...
.
...
.
...
.
.
...
.
.
...
...
.
.
...
.
.
...
..
... ........
...
...........
..
...............................................................................
c
b
v
•
And calling the inside of the triangles U and L. Let v0 , . . . , vn ∈ Rm , assume
that they are affinely independent. That is, no three are collinear, no four
coplanar, etc.
Equivalently, if c0 v0 + . . . + cn vn = 0 and c0 + . . . + cn = 0 then ci = 0 for
all i.
independent. Then define [v0 , . . . , vn ] = all
PnSo, let v0 , . . . , vn be affinely
P
t
v
where
t
≥
0
and
t
=
1.
i
i
i
i
i=0
P
1
The Barycenter is n+1
vi .
The standard n-simplex is ∆n = [e0 , . . . , en ] ⊆ Rn+1 .
If v0 , . . . , vn are affinely independent points, ∃ a homeomorphism ∆n →
n
[v
P0 , . . . , vn ], as any point in ∆ is of the form (t0 , . . . , tn ), and we can send it to
ti v i .
We can orient the edges by the edge between vi and vj points to the greater
one.
We define simplicial homology for a ∆-complex.
We look at X = ∪enα open n-simplexes and this union is disjoint. We have
for each enα σα : ∆n → [v0 , . . . , vn ], which gives a map from int∆n → enα .
We define ∆n (X) to be the free abelian group generated by all open nsimplexes. We can identify enα ↔ σαn . So the elements look like finite sums of
simplices or of these maps. We call each of these a simplicial chain.
We now attempt to define a boundary operator. ∂([v0 , v1 ]) = v1 − v0 ,
∂([v0 , v1 , v2 ]) = [v0 , v1 ] + [v1 , v2 ]P
− [v0 , v2 ] = [v̂0 , v1 , v2 ] − [v0 , v̂1 , v2 ] + [v0 , v1 , v̂2 ].
In general ∂([v0 , . . . , vn ]) = (−1)i [v0 , . . . , v̂i , . . . , vn ].
v
•
b
∂
And so, we have for ∆-complex X, ∆n (X) → ∆n−1 (X).
Lemma 3.1. ∂∂ = 0.
∂
We define Zn (C) = ker(∆n (X) → ∆n−1 (X)) to be the cycles and Bn (X) =
Im(∆n+1 (X) → ∆n (X)). Bn ⊆ Zn by the lemma.
We define Hn∆ (X) = Zn (X)/Bn (X).
For the torus above, ∆2 (T 2 ) has basis U, L, ∆1 (T 2 ) has basis a, b, c and
∆0 (T 2 ) has basis v.
∂(U ) = ∂(V ) = a + b − c, ∂(a) = ∂(b) = ∂(c) = 0.
H0 = ∆0 /0 ' Z. H1 = ∆1 /(a + b − c) ' Z ⊕ Z. H2 = Z(U − L)/0 ' Z.
16
Singular Homology [Eilenberg]
Let X be any space. A singular n-simplex in X is a continuous map σ : ∆n →
X. Put Cn (X) = theP
free abelian group generated by all singular n-simplexes,
so elements look like
nσ σ where σ is a map ∆n → X where nσ ∈ Z and only
finitely many are nonzero.
Pn
∂ : Cn (X) → Cn−1 (X) is defined by, if σ : ∆n → X, ∂σ = i=0 (−1)i σi
where σi : ∆n−1 → X is the inclusion of the ith face, φi followed by σ.
φi (t0 , . . . , tn−1 ) = (t0 , . . . , ti−1 , 0, ti , . . . , tn−1 ).
Lemma 3.2. ∂ 2 = 0.
If X ∼
= Y , then Hn (X) ∼
= Hn (Y ).
Proposition 3.3. If X = {pt}, then Hn (X) =
Z
0
n=0
else
Proof. For n ≥ 0, σn : ∆n → {pt} is unique. Thus, ∂σn is the alternative sum
of σn−1 ’s, which is 0 if n is odd, 1 if n is even.
Thus, we get the complex Z → Z → Z → . . . → Z. In each, either everything
is a cycle and everything is a boundary, or else none of either is. Except in
dimension zero, as everything is a cycle, but nothing is a boundary, so H1 (X) =
Z.
Proposition 3.4. If X has path components Xα , then Hn (X) ∼
= ⊕α Hn (Xα ).
Proposition 3.5. If X is path connected and nonempty, then H0 (X) ∼
= Z.
P
Proof. Choose x0 ∈ X. Consider C0 (X) = x∈X nx x.
∂
C1P
(X) → C0P
(X) → 0, but we can augment it, so instead we get C0 (X) → Z.
( nx x) = nx .
Note: ∂ = 0 since ∂σ 1 = 1 − 1 = 0. Claim: ker = Im ∂.
We know that Im ∂ ⊂ ker . We need to show the converse.
Let c = n1 x1 + . . . + n` x` ∈ C0 (X) with (c) = 0 so n1 + . . . + n` = 0. Choose
paths fi from x0 to xi (1 ≤ i ≤ `) so fi = σi : ∆1 → X with ∂σi = xi − x0 for
all i.
P
P
P
Then σ( ni σi ) = σni (xi − x0 ) = ni xi − ni x0 = c
Definition 3.1 (Reduced Homology). Let X 6= ∅ be
out space. We have . . . →
Hn (X)
n>0
C2 (X) → C1 (X) → C0 (X) → Z. Define H̃n (X) =
ker / Im ∂ n = 0
Fact: If X is path connected, x0 ∈ X, then there exists a homomorphism
π1 (X, x0 ) → H1 (X) called the Hurewicz homomorphism which is surjective and
has kernel the commutator subgroup of π1 .
Definition 3.2 (Induced Homomorphism). If f : X → Y is a continuous map,
then f] : Cn (X) → Cn (Y ) by composing f with σ : ∆n → X to get σ : ∆n → Y .
17
For c ∈ Cn (X), ∂f] (c) = f] ∂(c).
Note: f] (Zn (X)) ⊆ Zn (Y ) and f] (Bn (X)) ⊂ Bn (Y ) induces f∗ : Hn (X) →
Hn (Y ).
0 → Bn (X) → Zn (X) → Hn (X) → 0, f] gives an homomorphism of short
exact sequences to 0 → Bn (Y ) → Zn (Y ) → Hn (Y ) → 0.
Theorem 3.6. f, g : X → Y and f ' g, then f∗ = g∗ : Hn (X) → Hn (Y ).
Proof. Let F : X × I → Y continuous with F (x, 0) = f (x) and F (x, 1) = g(x)
for all x ∈ X. We want to show that f∗ = g∗ : Hn (X) → Hn (Y ).
Consider ∆n × I. ∆n × {0} = [v0 , . . . , vn ], ∆n × {1} = [w0 , . . . , wn ]. We
have additional n-simplices
[v0 , . . . , vi , wi+1 , . . . , wn ].
Sn
Note: ∆n × I = i=0 [v0 , . . . , vi , wi , . . .P
, wn ] is an n + 1-simplex.
n
Define P : Cn (X) → Cn+1 (Y ) by σ 7→ i=0 (−1)i F ◦(σ×id)|[v0 ,...,vi ,wi ,...,wn ] .
Check: ∂ ◦P = g] −f] −P ◦∂ as hom Cn (X) → Cn (Y ). or ∂P +P ∂ = g] −f] .
(P is a chain homotopy between f] and g] )
Let z ∈ Zn (X). Then z̄ ∈ Hn (X). Then f∗ z̄ = f] z and g∗ z̄ = g] z, so to
show that f∗ = g∗ , we must show that f] z − g] z is a boundary.
f] z − g] z = −(∂P z + P ∂z) = ∂(−P z).
Corollary 3.7. X ' Y implies Hn (X) ' Hn (Y )
What is Hn (S k ) when k, n > 0?
i∗
∆
Want relative homology long exact sequence, → Hn+1 (X, A) → Hn (A) →
j∗
Hn (X) → Hn (X, A) →
Given a space X, A ⊆ X, Z ⊂ A such that Z̄ ⊂int(A).
Have (X \ Z, A \ Z) → (X, A).
Theorem 3.8. i∗ : Hn (X \ Z, A \ Z) → Hn (X, A) is an isomorphism.
Use the above to determine the homology groups of S k by induction on k;
later (or read)
Definition 3.3 (Relative Homology). We define the relative chain complex of a
pair (X, A) where A ⊂ X as Cn (X, A) = Cn (X)/Cn (A). The relative homology
is the homology of this chain complex.
Proof of Theorem:
Proof. We have, easily, i∗ , j∗ , we need ∆ : Hn (X, A) → Hn−1 (A).
Let z ∈ Zn (X, A) with z̄ ∈ Hn (X, A). Pick x ∈ Cn (X) with x 7→ z. Then
there exists a unique a ∈ Cn−1 (A) such that a 7→ ∂x. Note that ∂a = 0 so
a ∈ Zn−1 (A).
Put ∆(z̄) = ā ∈ Hn−1 (A). Must check that ∆ is well-defined.
∆
Theorem 3.9. For (X, A), there exists a long exact sequence Hn+1 X → Hn+1 (X, A) →
δ
Hn (A) → Hn (X) → Hn (X, A) → Hn−1 (A).
18
Proof. Must verify exactness at Hn (A), Hn (X), Hn (X, A).
At Hn (A): i∗ ∆z̄ = 0 is clear from the definition of ∆z̄. Suppose ā0 ∈ Hn (A).
0
a ∈ Zn (A) and i] ā0 = 0 in Hn (X). Then i] a0 = ∂x0 , x0 ∈ Cn+1 (X). Exactness
follows by diagram chasing.
At Hn (X): Clearly j∗ i∗ = 0. Suppose given z̄ ∈ Hn (X) such that z̄ → 0 in
Hn (X, A), there exists a ∈ Cn (A) and n ∈ Cn+1 X sch that z = ∂n + i∗ a check
that a is a cycle, i∗ n̄ = z̄.
Proposition 3.10. Let f, g : (X, A) → (Y, B) be homotopic as maps of pairs.
Then natural f∗ : Hn (X, A) → Hn (Y, B) agrees with g∗ .
Proof. ∃P : Cn (X) → Cn+1 (Y ) for all n such that P ∂ + ∂P = g] − f] a chain
homotopy. Observe that since f (A) ⊂ B, P : Cn (A) → Cn+1 (B), we pass to
quotients to get P : Cn (X, A) → Cn+1 (Y, B) still a chain homotopy between f]
and g] . Thus, on relative homology, f∗ , g∗ are the same.
Naturality of Long Exact Sequence: Suppose f : (X, A) → (Y, B).
.
.∆.....
.............
........H
...H
............n
...........n (X, A)
Hn+1 (X,.....A)
(A)H
n (X)
...
...
...
...
f∗.............
f∗.............
f∗.............
...∆
...........
..............
.....(B)H
.....H
..........n
Hn+1 (Y, ..B)
n Y, B
n (Y )H
commutes.
Reformulation of Excision Theorem
X ⊃ A, Z ⊂ A put B = X \ X, so A ∩ B = A \ Z. Now, RZ = X \ int(X \ Z).
So Z ⊂ int(A) means that X \ int(B) ⊂ int(A), thus A = (A) ∪ int(B).
f∗.............
Theorem 3.11. If U = {A, B} covers X so that X = int(A) ∪ int(B), then
Hn (B, A ∩ B) → Hn (X, A) is an isomorphism.
Application: X is a space SX is a suspension of X,m that is, X × I/ ∼ with
(x, 1) ∼ (x0 , 1) and (x, 0) ∼ (x0 , 0) for all x, x0 ∈ X.
Then there exists isomorphisms H̃n+1 (SX) → H̃n (X)
Corollary 3.12. H̃k (S n ) is Z for k − n, 0 else.
Proof. We now prove the suspension theorem.
A = SX \ {S}, B = SX \ {N }. A ∩ B ' X, so 0 → H̃n+1 (SX) →
Hn+1 (SX, A) → 0 exact (thus middle is an isomorphism), and also Hn+1 (SX, A)
∆
isomorphic to Hn+1 (B, A∩B), and 0 → Hn+1 (B, A∩B) → H̃n (A∩B) → 0.
Generalized Excision Theorem:
X is given. Let U = {Uα } be a cover of X such that {int(Uα )} is a cover
of X. Let CnU (X) ⊂ Cn (X) be a subgroups generated by all U-small singular
n-complexes σ : ∆n → X meaning σ(∆n ) ⊂ Uα for some α.
Observe that σ U-small implies that ∂σ is U-small.
U
Thus, ∂ maps CnU (X) → Cn−1
(X). So get ι∗ : HnU (X) → Hn (X) is an
isomorphism.
19
More precisely, ι : CnU (X) → Cn (X) is an inclusion which has chain homotopy inverse ρ : Cn (X) → CnU (X), ι and ρ are both chain maps and ρ ◦ ι 'c id
and ι ◦ ρ 'c id where 'c is chain homotopy.
Exercise: From this, you can get the excision theorem.
Proof of the generalized excision theorem
Proof. We will prove that ι∗ : HnU (X) → Hn (X) is an isomorphism. We will
prove it in four parts.
1. Subdivide simplexes, with iterated barycentric subdivision.
If [w0 , . . . , wn ] is an n-simplex in RN and b ∈ RN , then b · [w0 , . . . , wn ] =
[b, w0 , . . . , wn ], an n + 1 simplex.
Aim: Barycentric Subdivision of an n-simplex [v0 , . . . , vn ]. We define the
−1 simplex to be [∅]. We will define this by induction on n.
For n = −1, nothing. For n = 0, also no change.
For n > 0, assume that each (n−1)-face of [v0 , . . . , vn ] has been subdivided
into n! (n − 1)-faces, which we will call [w0 , . . . , wn−1 ]
1
(v0 +. . .+vn ) be the barycenter of [v0 , . . . , vn ] and conThen we let b = n+1
sider all n-simplexes [b, w0 , . . . , wn−1 ] = b · [w0 , . . . , wn−1 ] giving a grand
total of n!(n + 1) = (n + 1)! n-simplexes.
Lemma 3.13. diam[b, w0 , . . . , wn−1 ] ≤
n
n+1 diam[v0 , . . . , vn ].
2. Y ⊆ RN a convex set. Let LCn (Y ) ⊂ Cn (Y ) be the subgroup generated
by all linear maps σ : ∆n → Y .
∂
Note: LCn (Y ) → LCn−1 (Y ), so the linear chains give us a chain complex.
If b ∈ Y , we get b : LCn → LCn+1 by b · [w0 , . . . , wn ] = [b, w0 , . . . , wn ].
Check that ∂b(α) = α−b(∂α) for all α ∈ LCn (Y ). That is, ∂b+b∂ = id −0
on LCn (Y ).
Take α = [w0 , . . . , wn ], then ∂b[w0 , . . . , wn ] = ∂[b, w0 , . . . , wn ] = [w0 , . . . , wn ]−
[b, w1 , . . . , wn ] + . . ., so it works out.
Define S : LCn (Y ) → LCn (Y ) by induction on n, let λ : ∆n → Y , λ =
[w0 , . . . , wn ] and let bλ = the barycenter of [w0 , . . . , wn ]. Put S([∅]) = [∅]
and S([w0 ]) = [w0 ]. In general, S(λ) = bλ · (S∂λ) in LCn (Y ).
Check ∂S = S∂. Next we construct a chain homotopy T : LCn (Y ) →
LCn+1 (Y ) between S and id. We do this inductively by setting T = 0 on
LC−1 (Y ) and T λ = bλ · (λ − T ∂λ) for n ≥ 0.
3. We will now subdivide general chains. S : Cn X → Cn X by Sσ = σ] S ·
[e0 , . . . , en ], where S[e0 , . . . , en ] ∈ LCn .
Check ∂Sσ = S∂σ and putting T σ = σ] T ∆n , then ∂T + T ∂ = id −S.
20
4. Iterated subdivision. Recall that we have S, T such that ∂S = S∂ and
∂T + T ∂ = id −S.
This says that S i is a chain map, and that S i is even chain homotopic to
id.
P
P
Dm = 0≤i≤m T S i = T ( S i ), check that ∂Dm + Dm ∂ = id −S m .
For each σ : ∆n → X, there exists m such that S m (σ) ∈ CnU (X). Put
m(σ) =least m that works. Define D : Cn (X) → Cn+1 (X) by Dσ =
Dm(σ) σ, then ∂Dσ + Dσ∂ = σ − ρ(σ). We discover that ρ(σ) ∈ CnU (X).
Check ρ : Cn (X) → CnU (X) is a chain map, ∂ρ = ρ∂, and moreover
∂D + D∂ = id −ι ◦ ρ where ι is the inclusion of C U into C.
Note that ρι = id on C U , and so we are done.
Theorem 2.13
Definition 3.4 (Good Pair). Hn (X, A) → H̃n (X/A) isomorphism if A =
6 ∅, A
closed, A ⊂ V open, A a deformation retract of V . Then we call (X, A) a good
pair.
Prop 2.21
C∗U (X) → C∗ (X) the inclusion, p in the opposite direction, with pi 'c id
and ip 'c id
Relative version, (X, U ), (A, U |A ).
0 ......................................... C∗U |A (A) .............................. C∗U (X) .......................... C∗U (X, A) ..................................... 0
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
C∗ (A) ....................................... C∗ (X) ................................ C∗ (X, A) ......................................... 0
implies that H∗U (X, A) → H∗ (X, A)Ris an Risomorphism.
Pf. of excision theorem in form X = (A)∪ (B) implies that H∗ (B, A∩B) =
H∗ (X, A).
0
................................................
Proof. Let U = {A, B}
0 .................................................. C∗U |A ....................................... C∗U (X) .......................... C∗U (X, A) ..................................... 0
C∗ (A) C∗ (A) ⊕ C∗ (B)
with vertical columns equivalences.
C∗ B
∗B
Then C∗U (X, A) = C∗ A+C
' C∗ (A)∩C
=
C∗ A
∗ (B)
21
C∗ B
C∗ (A∩B)
= C∗ (B, A∩B).
(X, A) a good pair implies that q : (X, A) → (X/A, A/A) induces an isomorphism
H∗ (X, A) → H∗ (X/A, A/A) → H̃∗ (X/A).
A closed subset of V open, A deformation retract of V implies that A/A
closed subset of V /A, A/A a def retract of V /A.
(X, A, B) triple, X ⊃ A ⊃ B, then 0 → C∗ (A, B) → C∗ (X, B) → C∗ (X, A) →
0 is a short exact sequence of chain complexes, and so we get a long exact sequence on relative homology groups.
'
H∗ (X, A) ........................ H∗ (X, V )
...
...
...
...
...
...
...
...
...
..
.........
.
q∗
...
...
...
...
...
...
...
...
...
..
.........
.
'.....
...........H
H∗ (X/A, A/A)
∗ (X/A, V /A
In fact, we can excise A and get isomorphisms to H∗ (X \ A, V \ A) and to
H∗ (X/A \ A/A, V /A \ A/A).
i : A → V is a homotopy equivalence, so Hn (A) ' Hn (V ) → Hn (V, A) =
0 → Hn−1 (A) ' Hn−1 (V )
Excision holds for CW-pairs. X = A ∪ B, X a CW complex and A, B CWsubcomplexes implies that H∗ (B, A ∩ B) ' H∗ (X, A). induces isomorphisms
....?
W
If X = α Xα where xα ∈ Xα for all α, (X, {xα }) is a good pair for all α,
then ια : Xα → X the inclusions induce ⊕α iα∗ : ⊕H̃∗ (Xα ) → H̃∗ (X).
Theorem 3.14. If U ⊆ Rm , V ⊆ Rn are open and V homeomorphic to U , then
m = n.
Proof. Let x ∈ U . Excision implies that Hm (U, U \ {x}) → Hm (Rm , Rm \ {x}),
is an isomorphism, as are H̃m (S m−1 ) to Hm (Rm , S m−1 ) to it, and so we get
δ`m Z = Hm (S m−1 ). Homeomorphisms induce isomorphisms on homology, so
m = n, as δ`m Z = δ`n Z.
Theorem 3.15. If X is a ∆-complex, then there exists a natural isomorphism
H∗∆ ' H∗ (X).
In fact, if A is a subcomplex of a ∆-complex X, then get isomorphism
H∗∆ (X, A) = H∗ (X, A).
Proof. Take dim X < ∞, A = ∅. Let X k be the union of all simplexes in X of
dim less than or equal to k, call this the k-skeleton.
Induction on k, ∅ ⊂ X 0 ⊂ . . . ⊂ X n−1 ( X n = X, dim X = n.
See diagram on bottom of 128 for (X k , X k−1 ). By the 5-lemma, if the outside
maps are isomorphisms, then the inside one is.
Degree of a map f : S n → S n
Definition 3.5 (Degree). H̃n (S n ) ' Z for all n ≥ 0. f induces f∗ : H̃n (S n ) →
H̃n (S n ), so there exists d ∈ Z such that f∗ (α) = dα for all α ∈ H̃n (S n ). Set
deg f = d.
22
Here are some basic properties:
1. deg(id) = 1
2. f ' g ⇒ f∗ = g∗ ⇒ deg f = deg g.
3. f 'constant⇒ deg f = 0. So if f : S n → S n not surjective, then deg f =
0.
4. deg f g = deg f deg g.
5. R : S n → S n a reflection implies deg R = −1. (Why?)
6. deg A = (−1)n+1 , where Ax = −x for all x.
7. If f : S n → S n has no fixed points, then deg f = (−1)n−1
8. If A ∈ O(n + 1), then deg A = det A.
9. fn : z 7→ z n on S 1 has deg fn = n.
10. f, g : S n → S n and deg f = deg g imply that f ' g.
11. f : S n → S n has suspension Sf : S n+1 → S n+1 indeed, SX for any
f : X → Y induces Sf : SX → SY . Then deg Sf = deg f . [Giver
f : S n+1 → S n+1 , it must be homotopic to a suspension [S n+1 , S n+1 ] ' Z,
h
there exists a map, the Hopf map, S 3 → S 2 such that h 6' constant map.]
12. fP: S n → S n , y ∈ S n , f −1 (y) = {x1 , . . . , xk }, degxi f = ±1, then deg f =
degxi f .
Cellular Homology
S
X a CW-complex, ∅ ⊆ X 0 ⊆ . . . ⊆ X n ⊆ . . ., X = i X i .
Hk (X) ' HkCW (X), the cellular homology given by
d
d
CW
CW
→ Cn+1
(X) → CnCW (X) → Cn−1
(X) →, d2 = 0.
CW
n
n−1
Put Cn (X) = Hn (X , X
) is free abelian, a generator for each n cell
of X.
Why? (X n , X n−1 ) is a good pair, X n \ X n−1 = ∪α enα where enα is homeomorphic to the open ball.
Let xα correspond to the origin in the homeomorphism. Put V = X n \ {xα }
open and there exists a retraction of V onto X n−1 . Hence, Hn (X n , X n−1 ) '
n
H̃n (X n /X n−1 ) = H̃n (∨α Bαn /sn−1
α ) ' ⊕Z with generators eα .
n
n−1
n−1
n−2
We need d : Hn (X , X
) → Hn−1 (X
,X
), we get it by j∗ ◦ ∆
through Hn−1 (X n−1 ) where j : (X n−1 , ∅) → (X n−1 , X n−2 ) is the inclusion but
as a map of pairs.
We must check that d2 = 0, d = ∆◦j∗ : Hn (X n , X n−1 ) → Hn−1 (X n−1 , X n−2 ),
which is true as j∗ ◦ ∆ = 0. Thus H∗ (C∗CW (X)) = H∗CW (X).
Lemma 3.16.
1. Hk (X n , X n−1 ) = 0 if k 6= n
23
2. Hk (X n ) = 0 if k > n and if dim X < ∞, Hk (X) = 0 if k > dim(X)
3. i : X n → X induces isomorphism i∗ : Hk X n → Hk X if k < n.
Proof. Proof of b: Use induction on n. For n = 0, clear. Suppose n > 0,
result is known for smaller skeleton. Then Hk (X n−1 ) = 0 → Hk (X n ) →
Hk (X n , X n−1 ) = 0, so Hk (X n ) = 0.
Proof of c: First, suppose dim X < ∞, X = X N , N = dim X. X n ⊆
n+1
X
⊂ . . . ⊂ X N = X. If n = N , good and simple. Suppose that n <
N for k < n, we get 0 = Hk+1 (X n+1 , X n ) → Hk (X n ) → Hk (X n+1 ) →
Hk (X n+1 , X n ) = 0 so we get isomorphism for all k.
Need the diagram on page 139. Proof by diagram chasing. We will need
the next result to do this. If k < n, Hk (X n ) → Hk (X n+1 ) → . . . → Hk (X N )
isomorphisms, but does not necessarily reach Hk (X).
Lemma 3.17. If K is a compact subset of a CW-complex X, then K ⊂ X n for
some n.
Proof. Suppose not, then for each n, choose xn ∈ K with xn ∈
/ X n . Then
p
K0 = {xn } ⊂ K is an infinite set. Note: K0 ∩ X is finite for all p, so K0 ∩ X
is closed. Thus, K0 is compact. Note that each subset of K0 is closed, and so
K0 has the discrete topology, contradiction.
Easiest Applications
1) Suppose there are no n-cells at all for some n. Then Hn (X) = 0.
2) Suppose there are k n-cells. Then CnCW (X) ' Zk , so ZnCW (X) ' Z` ,
` ≤ k, so Hn (X) has ≤ k generators. In particular, rank Hn (X) ≤ k.
3) Suppose there are n-cells but no n + 1 or n − 1-cells. Then d = 0 in
C∗CW (X), so C∗CW (X) = Hn (X) ' free abelian group on the n-cells.
4) H2k (CPn ) ' Z for 0 ≤ k ≤ n, others are 0.
Application of Degree
Theorem 3.18. Z2 is the only nontrivial group that can map freely on S 2n .
Proof. Any homeomorphism h : S 2n → S 2n has degree ±1, so we get a homomorphism deg : G → {±1}. If g ∈ G, g 6= 1, then gx 6= x for all x. So
we x 7→ gx is homotopic to A : x 7→ −x, so deg g = (−1)2n+1 = −1. Hence,
ker deg = {1}, so deg G → {±1} is injective. Thus |G| ≤ 2.
Let f = fd : S n → S n be a degree d map. X = S n ∪f en+1 , find Hi (X).
Have Φ : (Dn+1 , S n ) → (X, S n ), induces isometried on homology, so
∆
Hn+1 (Dn+1 , S n )......... H̃n (S n )
...
...
...
...
...
...
...
...
...
..
.........
.
Φ∗
...
...
...
...
...
...
...
...
...
..
.........
.
f∗
∆
H̃n+1 (X) .............. Hn+1 (X, S n ) ................... H̃n (S n ) ................................... H̃n (X)
24
With the first map on the bottom injective and the last surjective. So we
conclude that Hn (X) ' Z/dZ, Hn+1 (X) = 0 and Hi (X) = 0 if i 6= n, n + 1.
If f = f2 : S 1 → S 2 is a map of degree two, we can see that RPn has
homology H̃n = Z/2 iff n = 1 and 0 otherwise.
IF X, Y are CW-Complexes and f : X → Y call f cellular if f (X n ) ⊂ Y n
for all n ≥ 0.
FACT: Cellular Approximation Theorem: Any continuous f : X → Y between CW-complexes is homotopic to a cellular map.
Ex: Any f : S m → S n with m < n is homotopic to a constant map.
Then f : (X n , X n−1 ) → (Y n , Y n−1 ) for all n, so get a chain map Hn (X n , X n−1 ) →
Hn (Y n , Y n−1 ), so we get HnCW (X) → HnCW (Y ), thus we get a commutative
square
Hn (Y ) ..................................... Hn (X)
f∗
.
.
........
........
'
....
...
..
...
...
...
...
...
..
'
....
...
..
...
...
...
...
...
..
.
CW
............... H
(X)
HnCW (Y ) ........CW
n
f∗
Cellular Boundary Formula
Prelude: Choose gen of H̃n (S n ) compatibly under isomorphisms H̃n+1 (S n−1 ) →
n
n−1
H̃n (S n ) starting with S 0 . Then also Hn (D
) → H̃n−1 (S n−1 ). Use {enα }
P , S n−1
CW
n
as gens for Cn (X). Then dn (eα ) =
, dαβ ∈ Z the degree of
β dαβ eβ
Sαn−1 → X n−1 →
X n−1
X n−1 \en−1
β
' Sβn−1 .
Homology with Coefficients
If R is a commuative ring with identity (say a field) then we take Cn (X; R)
to be the free R-module generated by the singular simplexes, which is, in fact,
R ⊗Z Cn (X), and every works out with Hn (X; R). So for example, H̃n (X; R) '
R for X = S n .
Let X be aPfinite complex, and αi the number of cells in each dimension.
dim X
Then χ(X) = i=0 (−1)i αi .
Hi (X) = Zβi + Fi where Fi is finite. βi is called the ith Betti number.
[Hi (X) ⊗Z Q has dim βi ].
Fact: For finitely generated abelian groups, if 0 → A → B → C → 0 is
exact, then rank A − rank B + rank C = 0.
Pdim X
Theorem 3.19. χ(X) = i=0 (−1)i βi .
Proof. 0 → BiCW (X) → ZiCW (X) → HiCW (X) → 0 is s.e.s. HiCW (X) '
CW
Hi (X),Pand 0 → ZiCW (X) → P
CiCW (X) → Bi−1
(X) → P
0 s.e.s.
u
i
CW
So (−1) rank
H
(X)
=
(−1)
rank
H
(X)
=
(−1)i [rank ZiCW (X)−
i
i
P
CW
i
CW
CW
CW
rank
(rank CP
(X) − rank Bi−1 (X) −
(X)) =
i
i
P Bii (X)]
P = i(−1)
Prank B
CW
i
CW
i
(−1) αi − (−1) Bi (X) − (−1) rank Bi−1 (X) = (−1) αi = χ(X).
Mayer-Vietoris Sequence
25
(Vietoris) Let A, B ⊂ X, int(A) ∪ int(B) = X. Our aim is a l.e.s. . . . →
Ψ
∆
Φ
Hn (A ∩ B) → Hn (A) ⊕ Hn (B) → Hn (X) → Hn−1 (A ∩ B) → . . .
Recall that Cn (A)+Cn (B) → Cn (X) for all n gives C∗ (A)+C∗ (B) → C∗ (X)
which induces an isomorphism on homology groups.
Next: 0 → C∗ (A ∩ B) → C∗ (A) ⊕ C∗ (B) → C∗ (A) + C∗ (B) ⊂ C∗ (X) where
+ is the join. x
x
x 7→
,
7→ x + y. This is a short exact sequence. This gives:
−x
y
∆
→ Hn (A ∩ B) → Hn (A) ⊕ Hn (B) → Hn (C∗ (A) + C∗ (B)) ' Hn (X) → . . .
Snake Lemma and some applications
4
Cohomology
Why bother with Cohomology? Because it has a ring structure, given by a cup
product H p (X) ⊗ H q (X) → H p+q (X). In fact, it is a graded ring. This is like
differential forms (they are a primary example).
A cellular chain, if X is a CW-complex, looks like 0 → Cn → Cn−1 → . . . →
C1 → C1 → 0.
f
Cochains: If A is an abelian group, so is hom(A, Z). A → B, then we get
f∗
hom(A, Z) ← hom(B, Z), so this is a contravariant functor. Similarly, if we have
f
g
A → B → C, we get a similar sequence of homs.
If A = Zk , B = Z` , then f : A → B a function. What is f ∗ : hom(B, Z) →
hom(A, Z)? Well hom(Zk , Z) ' Zk , but this is not a natural isomorphism.
So a cochain is 0 ← hom(Cn , Z) ← hom(Cn−1 , Z) ← . . . ← hom(C1 , Z) ←
hom(C0 , Z) ← 0. Traditionally, we write the map as δ = ∂ ∗ and call it the
coboundary. We have δδ = 0.
And so, we define H k (X) = ker δ/ Im δ. We often denote hom(Cn , Z) as C n .
Why did we do this? This comes from differential forms on manifolds. A
p-form isR useful, because it can be integrated over a p-simplex to obtain a real
number ∆p α. That is, α ∈ C p (M, R). For differential forms, there is a map d
from p-forms
to (p
R
R + 1)-forms called the exterior differential. The big property
is that ∆ dα = ∂∆ α, called Stokes’ Theorem. If α ∈ C p is not necessarily a
∂
α
p-form, then
P what is δα? We have Cp+1 → Cp → R. So δα = α ◦ ∂. That is,
δα(∆) = (−1)i α(∆i ).
We hope that H k (X) = hom(Hk (X), Z).
x2
0
Example: X = RP2 . We have 0 → Z → Z → Z → 0, the chain complex
C∗ (X).
x2
0
C ∗ (X) is 0 ← Z ← Z ← Z ← 0, so H 0 (X) = Z. H 1 (X) = 0 and H 2 (X) =
Z2 . So our hopes fail...but we did get the same groups, just in the wrong places.
For S 2 , we have C∗ is 0 → Z → 0 → Z → 0, so C ∗ is 0 ← Z ← 0 ← Z ← 0,
so the cohomology groups are the same as the homology groups.
In fact, this is a general phenomenon, for finitely generated abelian groups,
the free part stays when switching to cohomology, but the torsion part switches
26
dimensions.
Let f∗ : A∗ → B∗ be a chain map between chain complexes. We want to
define the algebraic mapping cylinder.
Definition 4.1 (Mapping Cylinder). f : X → Y is a continuous map of topological spaces. Take (X × I ∪ Y )/(x, 1) ∼ f (x). We call this M (f ).
The mapping cylinder has nice properties, like M (f ) is homotopy equivalent
to Y an that X ,→ M (f ), and the diagram
M (f )
X
..........
..
.....
.....
.....
.
.
.
.
...
.
.
.
.
....
.....
.....
.....
.....
.
.
.
.
..
.....
.....
........................................................
...
...
...
...
...
...
...
...
...
..
.........
.
'
Y
It gives us a pair (M (f ), X) for the map between X and Y , which we can
use to obtain long exact sequences on homology.
Definition 4.2 (Algebraic Mapping Cylinder). We define M (f )k = Ak ⊕ Bk ⊕
Ak−1 . Then the map M (f )k → M (f )k−1 is given by ∂ on Ak , Bk and to ∂ on
Ak−1 to Ak−2 , but also f∗ from Ak−1 to Bk−1 and the identity from Ak−1 to
Ak−1 . That is, with rows summed across:
Ak−1
Ak
Bk
...
...
...
...
...
...
...
...
...
..
..........
.
∂
...
...
...
...
...
...
...
...
...
..
..........
.
-∂
...
...
...
...
...
...
...
...
...
..
..........
.
∂
a................
.
Ak−1
Ak−2
Bk−1
with the sign change made to make sure that this is a complex. We call this
0
0
complex
the algebraic
mapping cylinder,
so ∂(a, alabel
, b) =specification
(∂a+a0 , −∂a0 ,Error:
∂b+f aIncorrect
).
label specification Er
Error:
Incorrect
label specification
Error: Incorrect
Need to show that M (f )∗ is chain homotopy equivalent to B∗ .
We are aiming at the following:
Proposition 4.1. Suppose that C∗ and D∗ are chain complexes of free abelian
groups and f∗ : C∗ → D∗ is a chain map such that f∗ induces an isomorphism
on homology. Then f∗ is a chain homotopy equivalence. That is, there exist
g∗ : D∗ → C∗ , S : C∗ → C∗+1 and T : D∗ → D∗+1 so that ∂S + S∂ = 1 − g∗ f∗
and ∂T + T ∂ = 1 − f∗ g∗ .
This implies that f ∗ : H ∗ (D∗ ) → H ∗ (D∗ ) is an isomorphism.
Proof. S ∗ δ + δS ∗ = 1 − f ∗ g ∗ and T ∗ δ + δT ∗ = 1 − g ∗ f ∗ . Check this!
Proposition 4.2. Suppose that C∗ is a chain complex of free abelian groups
and Hk (C∗ ) is finitely gneerated for each k. Hk (C∗ ) ' Fk ⊕ Tk where Fk is free
and Tk is torsion. Then H k (C∗ ) ' Fk ⊕ Tk−1
27
So we were at f∗ : A∗ → B∗ , M (f )k = Ak ⊕ Ak−1 ⊕ Bk with ∂(a, a0 , b) =
(∂a + a0 , −∂a0 , ∂b − f a0 ).
s : M (f )∗ → M (f )∗+1 , s(a, a0 , b) = (0, a, 0).
∂S + S∂(a, a0 , b) = ∂(0, a, 0) + s(∂a + a0 , −∂a0 , ∂b − f a0 ) = (a, −∂a, −f a) +
(−, ∂a, a0 , 0) = (a, a0 , −f a) = 1 − r. r(a, a0 , b) = (0, 0, b + f a).
And so we get
B∗
. ......
.......... .........
.....
...
.....
..
.....
..... ∗
...
.....
...
.....
.....
∗ ....
.....
...
.....
..
.....
.......
...
.......
..
f
0
r
A∗ ............................................... M (f )∗ ......................................... C∗ (f ) .................................................. 0
Get that f∗ induces isos on homology iff H∗ (C(f )) = 0.
C(f )∗ acyclic implies that f∗ is a chain equiv (later)
........................................................
Definition 4.3 (Free Resolution). Let H be an abelian group. A resolution is
a long exact sequence . . . → F2 → F1 → F0 → H → 0.
Proof. Resolutions are unique up to chain homotopy.
........................................................
........................................................ F
........................................................ F
........................................................ F
H
2
1
0
...
...
...
...
...
...
...
...
...
..
.........
.
φ0
........................................................
0
...
...
...
...
...
...
...
...
...
..
.........
.
φ
G2 ........................................................ G1 ........................................................ G0 ........................................................ K ........................................................ 0
Extends to a chain map which is unique up to chain homotopy, as we can
suppose that there are φ̄’s such that the diagram also commutes with them.
We want S : F∗ → G∗+1 with ∂S + S∂ = φ − φ̄. We want s : F0 → G1 with
∂s = φ − φ̄. Want ∂sz = (φ − φ̄)z, this gets hit by ∂ iff ∂(φ − φ̄)z = 0, and that
is (φ∂ − φ∂)z = 0.
Want s : F1 → G2 such that ∂s + s∂ = φ − φ̄, z a generator in F1 . Want
∂sz = (φ−φ̄−s∂)z. There is an element sz with the right ∂ iff ∂(φ−φ̄−s∂)z = 0,
and this is (∂φ − ∂ φ̄ − ∂s∂)z = (φ0 ∂ − φ̄0 ∂ − ∂s∂)z = (φ0 − φ̄0 − ∂s)∂z.
........................................................
So we can take H and a free resolution of H, and take hom(−, G) of it, it
28
may develop homology, and we thus obtain derived functors.
MISSED A LECTURE
f∗
Suppose that A∗ → B∗ is such that f∗ induces an isomorphism on Homology.
Then H∗ (C(f )) = 0 because 0 → A∗ → M (f )∗ → C(f )∗ → 0 is exact.
H∗ (C(f )) = 0 ⇒ C(f )∗ is chain contractible, as C(f )∗ is a resolution of 0,
so chain equivalent to the zero complex by uniqueness of resolution.
C(f )∗ is chain contractible implies that f has a chain homotopy inverse,
as
S
: C(f
)∗ → C(F )∗+1 is a chain contraction, ∂
: C(f )∗→ C(f )∗01 , then
−∂ 0
k `
: A∗−1 ⊕ B∗ → A∗−2 ⊕ B∗−1 . S =
: A∗−1 ⊕ B∗ →
−f ∂
m n
A∗ ⊕ B∗+1 . ∂S + S∂ = I, and can be worked out via matrix multiplication.
There is a splitting H n (X, G) ' hom(Hn (X), G) ⊕ Ext(Hn (X), G), but it is
not natural, so it must be used carefully, if at all.
Cup Product
(Cohomology is into a commutative ring Λ with identity)
Let ϕ ∈ C k (X, Λ) and ψ ∈ C ` (X, Λ), then ϕ∪ψ ∈ C k+` (X, Λ) by (ϕ∪ψ)σ =
ϕ(σ[v0 , . . . , vk ])ψ(σ[vk , . . . , vk+` ])
δ(ϕ ∪ ψ) = δϕ ∪ ψ + (−1)k ϕ ∪ δψ.
Why is this good?
If ϕ, ψ are cocycles then ϕ ∪ ψ is a cocycle, ie, if δϕ = δψ = 0 then δ(ϕ ∪
ψ) = 0. Also (ϕ + δα) ∪ (ψ + δβ) = ϕ ∪ ψ + ϕ ∪ δβ + δα ∪ ψ + δα ∪ δβ =
ϕ ∪ ψ + δ(α ∪ ψ) + δ(α ∪ δβ) + (−1)k (ϕ ∪ β), so it is ϕ ∪ ψ + δγ for some γ.
Thus, this defines H k (X) × H ` (X) → H k+` (X).
We should check the formula for
Pδ(ϕ ∪ ψ). We take σ = hv0 , . . . , vk+`+1 i.
δ(ϕ ∪ ψ)(σ) = (ϕ ∪ ψ)δσ = (ϕ ∪ ψ) (−1)i hv0 , . . . , vˆi P
, . . . , vk+`+1 i.
δϕ∪ψ(σ) = δϕ[v0 , . . . , vk+1
]ψ[v
,
.
.
.
,
v
]
=
(−1)i ϕ[v0 , . . . , vˆi , . . . , vk+1 ]ψ[vk+1 , . . . , vk+`+1 ]
k+1
k+`+1
P
i
ϕ ∪ δψ(σ) = ϕ[v0 , . . . , vk ] (−1) ψ[vk , . . . , vˆi , . . . , vk+`+1 ]
Sign counting to check that it’s right, see various sources for details.
Poincare Duality: For a closed orientable manifold, there is an isomrophism
H n−k (M ) → Hk (M n ) by a ”cap product”
29