Math 751 Week 6 Notes

Math 751 Week 6 Notes
Joe Timmerman
October 26, 2014
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October 7
Definition 1.1. A map p : E → B is called a covering if
1. P is continuous and onto.
2. For all b ∈ B, there exists an open neighborhood U of b which is evenly covered,
i.e., p−1 (U ) = tα Vα , where the vα are disjoint and open, and p|Vα : Vα → U is a
homeomophism.
1. p : R → S 1 , t 7→ e2πit
Example 1.2.
2. idX : X → X
3. p : X × {1, . . . , n} → X, (x, k) 7→ x
4. p : S 1 → S 1 , z 7→ z n
5. p : S n → RPn , x 7→ [x] = {±x}
6. p : C → C∗ , z 7→ ex
7. Products of covering maps: If pi : Ei → Bi , i = 1, 2 are coverings, then p1 × p2 : E1 ×
E2 → B1 × B2 is a covering.
Remark 1.3.
phism.
1. Being a covering implies the map is open and locally a homeomor-
2. Not any local homeomorphism is a covering.
3. p−1 (b) is discrete, for all b ∈ B (by disjointness.)
Definition 1.4. Let p1 : E1 → B, p2 : E2 → B be two coverings. We say p1 and p2 are
equivalent if there exists a homemorphism f : E1 → E2 such that p2 ◦ f = p1 . Note: This
is an equivalence relation .
Problem: Find all coverings of a space B (up to equivalence.)
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Lemma 1.5. If p : E → B is a covering, B0 ⊂ B, and E0 := p−1 (B0 ), then p|E0 : E0 → B0
is a covering.
Example 1.6. Let p : R2 → T 2 be a covering. Overlay the integer lattice on R2 , and
identity each square with a torus in the usual way. Let p0 = (1, 0) ∈ S 1 , and let B0 =
S 1 × {p0 } ∪ {p0 } × S 1 . Then p−1 (B0 ) = R × Z ∪ ζ × R (so it gets rid of the inside of the
squares.)
Theorem 1.7 (Path lifting property). Let P : E → B be a covering, b0 ∈ B, and e0 ∈
p−1 (b0 ). If γ : I → B is a path in B starting at b0 , then there is a unique lift γ
ee0 : I → E
such that γ
ee0 (0) = e0 .
The proof of this theorem follows from the previous lemma.
Theorem 1.8 (Homotopy lifting property). Let F : I × I → B be a homotopy with b0 :=
F (0, 0). Then there is a unique lift Fe : I × I → E of F such that Fe(0, 0) = e0 .
Corollary 1.9. If γ1 , γ2 are paths in B which are homotopic by some F , then γ1 (0) =
˜
γ2 (0) = b0 , then (γe1 )e0 ∼F (γe2 )e0 . In particular, these lifts have the same endpoints:
(γe1 )e0 (1) = (γe2 )e0 (1).
Definition 1.10. Let b0 ∈ B. For e0 ∈ p−1 (b0 ), define
Φe0 : π1 (B, b0 ) → p−1 (b0 )
[γ] 7→ γ
ee0 (1)
Note that by corollary 1.9, this map is well-defined.
Theorem 1.11. Let Φe0 be defined as above. Then Φe0 is onto if E is path-connected, and
it is injective if E is simply connected.
Proof. First suppose that E is path-connected. Let e1 ∈ p−1 (b0 ), and let δ be a path in E
from e0 to e1 . Then p ◦ δ is a path in B. Further, γ := p ◦ δ : I → B is a loop in B with base
point b0 . Then δ is a lift of γ starting at e0 . Then we have φe0 ([γ]) = γ
ee0 (1) = δ(1) = δ1 , so
φe0 is surjective. Note that the equality γ
ee0 (1) = δ(1) comes from the uniqueness of lifts.
Now suppose E is simply connected. Let γ1 , γ2 be loops in B with base point b0 such
that φe0 ([γ1 ]) = φe0 ([γ2 ]) = e1 . By definition, this means that (e
γ1 )e0 (1) = (e
γ2 )e0 (1). To
show that φe0 is injective, we must show γ1 ∼ γ2 . Since E is simply connected, there is
a unique homotopy class of paths from e0 to e1 , so (e
γ1 )e0 ∼ (e
γ2 )e0 by some homotopy F .
This gives a homotopy p ◦ F : I × I → B from p ◦ (γe1 )e0 = γ1 to p ◦ (γe2 )e0 = γ2 , which shows
that φe0 is injective.
Example 1.12. Let p : S n → RPn be a covering. For n ≥ 2, S n is path-connected and
simply connected. Then by theorem 1.11, Φe0 : π1 (RPn , b0 ) → (p−1 (b0 ) is a bijection. Since
#p−1 (b0 ) = 2, it must be that π1 (RPn , b0 ) ∼
= Z/2Z.
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Example 1.13. Let p : R → S 1 , t 7→ e2πit . Since R is both simply connected and patheconnected, theorem 1.11 tell us that φe0 : π1 (S 1 , b0 ) → Z is a bijection. To show that
the groups are isomorphic, we now need to show that φe0 is a homomorphism. Let γ, δ ∈
π1 (S 1 , b0 ), and let γ
e0 , δe0 be their lifts in R. Let γ
e0 (1) = n ∈ Z, δe0 (1) = m ∈ Z. By definition,
φe0 ([g]) = n, φe0 ([δ]) = m.
Claim 1.14. φe0 ([g] · [δ]) = n + m (i.e., it is a homomorphism.)
Proof. We have
^
∗ δ)0 (1) = (e
γ0 ∗ δe∗ )(1) = δe∗ (1) = n + m
φe0 ([δ] · [γ]) = φe0 ([δ ∗ γ]) = (γ
Now set δe∗ (t) = n + δe0 (t) so that δe∗ (0) = n, δe∗ (1) = n + m. Thus, φe0 is a homomorphism
and therefore an isomorphism.
Proposition 1.15. If p : E → B is a covering and B is path-connected, and b0 , b1 ∈ B,
there there is a bijection p−1 (b0 ) → p−1 (b1 ).
Proof. Let γ be a path in B from b0 to b1 (which exists because B is path-connected.)
Define the bijection fγ : p−1 (b0 ) → p−1 (b1 ) by e0 7→ γ
ee0 (1). It has inverse (fγ )−1 = fγ .
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October 9
Proposition 2.1. Let E be path connected, p : E → B a covering, and p(e0 ) = b0 . Then
p∗ : π1 (E, e0 ) → π1 (B, b0 ) is injective. Further, if e0 is changed to some other e1 ∈ p−1 (b0 ),
then the images of p∗ are conjugate in π1 (B, b0 ).
Proof. Let p∗ ([γ1 ]) = p∗ ([γ2 ]). Then P ◦ γ1 ∼ p ◦ γ2 by some homotopy F . By homotopy
lifting, we have that (^
p ◦ γ1 )e0 ∼ (^
p ◦ γ2 )e0 , which implies that γ1 ∼ γ2 , by the uniqueness
of lifts. Thus, p∗ is injective.
Now let e1 be a different point in the fiber of p over b0 . Define H1 = P∗ π1 (E, e0 ), H2 =
p∗ π1 (E, e1 ). We want to show these are conjugate. First let δ be a path in E from e0 → e1 .
Then the following diagram commutes:
π1 (E, e0 )
p∗
δ#
π1 (B, b0 )
(p◦δ)#
π1 (E, e1 )
π1 (B, b0 )
Note that δ# is an isomorphism. So we have H1 ∼
= (p ◦ δ)# H2 , by conjugation with [p ◦ δ].
Theorem 2.2. Let E be path-connected, p : E → B a covering map, and e0 ∈ p−1 (b0 ). Let
H := p∗ π1 (E, e0 ) ≤ π1 (B, b0 ). Then:
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a A closed path γ in B based at b0 lifts to a loop in E at e0 iff [γ] ∈ H.
ee0 (1) is a bijection. In particular #p−1 (b0 ) =
b φe0 : H\π1 (B, b0 ) → p−1 (b0 ), [γ] 7→ γ
[π1 (B, b0 ) : p∗ π1 (E, e0 )].
Proof of (b). First show that φe0 is well-defined, i.e., if [δ] ∈ H, then φe0 ([δ[·[γ]) = φe0 ([γ]).
We have
eδ˜e
∗ γ)e0 (1) = (δee0 ∗ γ
φe0 ([γ] · [δ]) = φe0 ([δ ∗ γ]) = (δ]
0 (1)
)(1) = γ
eδ˜e
0 (1)
eδ˜e (1) = φe0 ([γ]), so it’s well
By part (a), since [δ] ∈ H, we have that δee0 (1) = e0 . Thus, γ
0
defined. From last class we know that φe0 is onto, so it remains to show that it’s injective.
Suppose that φe0 ([γ1 ]) = φe0 ([γ2 ]). By definition, this means that (γe1 )e0 (1) = (γe2 )e0 (1).
Thus, (γe1 )e0 ∗ (e
γ2 )e0 is a loop in E based at e0 , which is in turn a lift of γ1 ∗ γ2 . By (a),
[γ1 ∗ γ2 ] ∈ H. Finally, [γ1 ] = [γ1 ∗ γ2 ∗ γ2 ] = [γ1 ∗ γ2 ] · [γ2 ]. Note that [γ1 ∗ γ2 ] ∈ H, so γ1 , γ2
are equivalent in the set of cosets. Thus, the function is injective.
Theorem 2.3 (Lifting Lemma). Let E, B, Y be path-connected and locally path-connected
(i.e., for all x ∈ X and for all neighborhoods Ux of x, there exists a Vk which is path
connected-connected, contains x, and is contained in Ux .) Let p : E → B be a cover, b0 ∈ B,
e0 ∈ p−1 (b0 ), and f : Y → B such that f (y0 ) = b0 . Then there exists a lift fe: Y → E such
that fe(y0 ) = e0 , p ◦ fe = f iff f∗ (π1 (Y, y0 ) ⊂ p∗ π1 (E, e0 ).
E, e0
∃fe
Y, y0
f
p
B, b0
Proof. The ⇒ direction is clear. Let y ∈ Y . How should we define fe(y)? Let α be a path
inY from y0 to y1 . Then f ◦ γ is a path in B starting at b0 . Define fe(y) := (f]
◦ α)(1). We
e
e
]
have (p ◦ f )(y) = p ◦ (f ◦ α)e0 (1) = f ◦ α(1) = f (y). Thus, f is actually a lift.
Now we need to show fe is well-defined (i.e., independent of α). If β is another path inY
from y0 to y, then α ∗ β ∈ π1 (Y, y0 ), so f ◦ (α ∗ β) ∈ f∗ π1 (Y, y0 ). Now by assumption, we
have f π (Y, y ) ⊂ p π (E, e ). This means that (f ◦^
(α ∗ β)) is a loop at e . Then we
∗ 1
0
∗ 1
0
e0
0
have
(f]
◦ α)e0 ∗ (f]
◦ β)(]
f ◦α)
e0
^
^
^
= (f]
◦ α) ∗ (f
◦ β) = (f
◦ α) ∗ (f
◦ β)
This means that (f]
◦ α)e0 = (f]
◦ β)e0 , which is what we wanted to show.
Now we need to show that fe is continuous. Let y ∈ Y , and let U be a path connected
neighborhood of f −1 (y1 ) ∈ B (which exists by the locally-connected assumption). Let V
be the slice in p−1 (U ) which contains fe(y1 ). By the continuity of f , there is some pathconnected neighborhood of y, say W , in Y such that f (W ) ⊂ U . Then fe = (p|V )−1 ◦ f |W
is continuous.
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Corollary 2.4. If Y is simply connected, then such a lift always exists.
Proposition 2.5. (Lift uniqueness) If Y is connected and fe1 , fe2 : Y → E are two lifts as
in the previous theorem, then fe1 = fe2 .
Proof. Let A = {y : fe1 (y) = fe2 (y)} 6= ∅. We will show A = Y by showing that A is both
open and closed. Let y ∈ Y , and let U be an evenly covered neighborhood of f (y) in B.
eα such that p| e : U
eα → U is a homeomorphism. Let U
e1 , U
e2 be
Then we have p−1 (u) = tα U
Uα
eα containing fe1 (y) and fe2 (y), respectively. Note that the fei are continuous, so there
the U
e1 and fe2 (N ) ⊂ U
e2 . If fe1 (y) 6= fe2 (y), then
is a neighborhood N of y such that fe1 (N ) ⊂ U
e
e
e
e
e
e
U1 6= U2 , so U1 ∩ U2 = ∅. This means that f1 6= f2 on N , so A is closed (as the complement
e1 = U
e2 , which implies that
of an open set). On the other hand, if fe1 (y) = fe2 (y), then U
e
e
e
e
e
e
f1 = f2 on N (since pf1 = pf2 = f , and p is injective on U1 = U2 ). Thus, A is open, which
proves the proposition.
Exercise 1. Any continuous map RP2 → S 1 is null-homotopic.
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