TOPOLOGY HW 6
CLAY SHONKWILER
52.4
Let A ⊂ X; suppose r : X → A is a continuous map such that r(a) = a
for each a ∈ A. If a0 ∈ A, show that
r∗ : π1 (X, a0 ) → π1 (A, a0 )
is surjective.
Proof. Let [α] ∈ π1 (A, a0 ). Then α is a loop in A based at a0 ; certainly this
means that α is a loop in X based at a0 , since A ⊂ X, so [α] ∈ π1 (X, a0 ).
Now,
r∗ ([α]) = [r ◦ α] = [α]
since, for any t ∈ I, α(t) ∈ A, so (r ◦α)(t) = r(α(t)) = α(t). Since our choice
of [α] was arbitrary, we see that for any element of π1 (A, a0 ), there is an
element of π1 (X, a0 ) mapped to it by r∗ . In other words, r∗ is surjective. 52.6
Show that if X is path connected, the homomorphism induced by a continuous map is independent of base point, up to isomorphisms of the groups
involved. More precisely, let h : X → Y be continuous, with h(x0 ) = y0 and
h(x1 ) = y1 . Let α be a path in X from x0 to x1 , and let β = h ◦ α. Show
that
β̂ ◦ (hx0 )∗ = (hx1 )∗ ◦ α̂.
This equation expresses the fact that the following diagram of maps “commutes”.
(hx )∗
π1 (X, x0 ) −−−0−→ π1 (Y, yo )




yα̂
yβ̂
(hx )∗
π1 (X, x1 ) −−−1−→ π1 (Y, y1 )
1
2
CLAY SHONKWILER
Proof. Let [f ] ∈ π1 (X, x0 ). Then
(β̂ ◦ (hx0 )∗ )([f ]) = β̂([h ◦ f ])
= [β] ∗ [h ◦ f ] ∗ [β]
= [β ∗ (h ◦ f ) ∗ β]
= [(h ◦ α) ∗ (h ◦ f ) ∗ (h ◦ α)]
= [h ◦ (α ∗ f ∗ α)]
= (hx1 )∗ ([α ∗ f ∗ α])
= (hx1 )∗ ([α] ∗ [f ] ∗ [α])
= (hx1 )∗ (α̂([f ]))
= ((hx1 )∗ ◦ α)([f ])
so β̂ ◦ (hx0 )∗ = (hx1 )∗ ◦ α̂.
53.3
Let p : E → B be a covering map; let B be connected. Show that if
p−1 (b0 ) has k elements for some b0 ∈ B, then p−1 (b) has k elements for
every b ∈ B. In such a case, E is calld a k-fold covering of B.
Proof. We will need the following lemma:
Lemma 0.1. Let b ∈ B. If U is an evenly covered neighborhood of b, then,
for all x ∈ U , |p−1 (x)| = |p−1 (b)|.
Proof. Let U be an evenly covered neighborhood of b. Then
G
p−1 (U ) =
Vα
α∈J
where P |Vα is a homeomorphism of Vα with U . Since p−1 (b) intersects each
Vα in a singleton, we see that |p−1 (b)| = |J|. Hence, for any x ∈ U , p−1 (x)
has exactly |p−1 (b)| elements.
Define Ak := {x ∈ B : |p−1 (x)| = k} and let A∞ := {x ∈ B : p−1 (x) is infinite}.
Then, for α, β ∈ N ∪ {∞} = N∗ , Aα ∩ Aβ = ∅ when α 6= β. Hence,
G
(1)
B=
Aα .
α∈N∗
Furthermore, each Aα is open. To see this, let y ∈ Aα for α ∈ N∗ . Let Uy
be an evenly covered neighborhood of y. Then, by the above lemma, every
element of Uy is contained in Aα , so Uy ⊆ Aα . Since our choice of y was
arbitrary, we conclude that Aα is open.
Suppose b0 ∈ B such that p−1 (b0 ) has k elements. Then b0 ∈ Ak . Since
each Aα is open, Ak is open and
G
A=
Aα
α6=k
is open. Ak ∩ A = ∅ and, by (1), Ak ∪ A = B. Hence, since B is connected,
it must be the case that A = ∅. Therefore, Ak = B. We conclude, then,
that for all b ∈ B, p−1 (b) has exactly k elements.
TOPOLOGY HW 6
3
A
i. Let I/ ∼ be the quotient space of I = [0, 1] obtained by identifying 0
and 1. Show that S 1 is homeomorphic to I/ ∼.
Proof. Let p : I → S 1 be given by p(θ) = ei2πθ . Then p is certainly continuous and surjective. Furthermore, if C ⊆ I is closed, then, since I is compact,
C is compact. Hence, since p is continuous, p(C) is compact. Now, since S 1
is Hausdorff, p(C) is closed, so we see that p is a closed map. Since p is a
continuous, surjective, closed map, it is a quotient map.
For 0 < θ < 1, p−1 (ei2πθ) ) = θ and
p−1 (ei2π0 ) = 0 = p−1 (ei2π ),
so I/ ∼= {p−1 ({z})|z ∈ S 1 }. By Corollary 22.3, then, since p is a quotient
map, I/ ∼' S 1 .
ii. From above we know that any loop f : I → X based at x0 induces
a map fˆ : S 1 → X such that f = fˆ ◦ p. Show that two loops f, g : I =
[0, 1] → X based at x0 ∈ X are path homotopic if and oly if there exists a
continuous map Ĥ : S 1 × I → X such that
a) Ĥ(z, 0) = fˆ(z), Ĥ(z, 1) = ĝ(z) for all z ∈ S 1 .
b) Ĥ(1, t) = x0 for all t ∈ I.
Proof. (⇒) Suppose two paths f, g : I → X based at x0 are path homotopic.
Then there exists H : I × I → X such that
H(s, 0) = f (s), H(s, 1) = g(s)
H(0, t) = H(1, t) = x0 .
Now, f and g induce maps fˆ : S 1 → X and ĝ : S 1 → X such that f = fˆ ◦ p
and g = ĝ ◦ p. That is to say that fˆ = f ◦ p−1 and ĝ = g ◦ p−1 . Now, define
Ĥ : S 1 × I → X
(z, t) 7→ H(p−1 (z), t).
Then Ĥ is certainly continuous and
Ĥ(z, 0) = H(p−1 (z), 0) = f (p−1 (z)) = (f ◦ p−1 )(z) = fˆ(z),
Ĥ(z, 1) = H(p−1 (z), 1) = g(p−1 (z)) = (g ◦ p−1 )(z) = ĝ(z)
and
Ĥ(1, t) = H(p−1 (1), t) = H(0, t) = x0 .
(⇐) Now, let f, g be loops based at x0 and suppose there is an Ĥ fulfilling
the above requirements. Then define
H : I ×I → X
(s, t) 7→ Ĥ(p(s), t).
Then H is continuous and
H(s, 0) = Ĥ(p(s), 0) = fˆ(p(s)) = (fˆ ◦ p)(s) = f (s),
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CLAY SHONKWILER
H(s, 1) = Ĥ(p(s), 0) = ĝ(p(s)) = (ĝ ◦ p)(s) = g(s),
H(0, t) = Ĥ(p(0), t) = Ĥ(1, t) = x0
and
H(1, t) = Ĥ(p(1), t) = Ĥ(1, t) = x0 ,
so f and g are path homotopic.
B
If A ⊆ X is a subspace and f : A → Y is a continuous map, we say that
F : X → Y is a continuous extension of f or f extends to a continuous map
F : X → Y if
(a) F : X → Y is continuous.
(b) F (x) = f (x) for any x ∈ A.
Now, let f : S 1 → X be a loop in X. Show that [f ] = 1 ∈ π1 (X; x0 ) if and
only if f extends to a continuous map F : D2 → X; that is, there exists a
continuous map F : D2 → X such that F (x) = f (x) for any x ∈ ∂D2 = S 1 .
Proof. (⇐) Suppose f extends to a continuous map F : D2 → X. Define
F̃ : D2 × I → D2
(z, t) 7→ z(1 − t) + 1(t)
where 1 denotes the point (1, 0). Then F̃ is certainly continuous and
F̃ (z, 1) = 1
for all z ∈ S 1 and
F̃ (1, 1) = 1,
so F̃ is a retraction of D2 to {1}.
Define H : S 1 × I → X by H = F ◦ F̃ . Then H is continuous and
H(z, t) = F ◦ F̃ (z, t) = F (z(1 − t) + 1(t)).
Hence
H(z, 0) = F ◦ F̃ (z, 0) = F (z) = f (z),
H(z, 1) = F ◦ F̃ (z, 1) = F (1) = f (1)
and
H(1, t) = F ◦ F̃ (1, t) = F (1) = f (1) = x0 .
Therefore, by the result proved in part A above, f is path homotopic to the
constant map x0 , so [f ] = 1 ∈ π1 (X; x0 ).
(⇒) Now suppose [f ] = 1 ∈ π1 (X; x0 ). Then we want to show that there
exists a continuous extension F : D2 → X of f .
First, we need the following lemma:
Lemma 0.2. S 1 × I is homeomorphic to the annulus A = {z ∈ C|1/2 ≤
|z| ≤ 1}.
TOPOLOGY HW 6
5
Proof. Define
φ : S1 × I → A
(z, t) 7→ z t+1
2
φ is certainly continuous
addition and reciprocals are
since multiplication,
z
1
continuous. If z ∈ A, ||z|| , 2||z|| − 1 ∈ S × I and
z
z (2||z|| − 1) + 1
φ
, 2||z|| − 1 =
= z,
||z||
||z||
2
so φ is surjective.
If (z, t), (z 0 , t0 ) ∈ S 1 × I such that
φ(z, t) = φ(z 0 , t0 )
then
0
(t + 1)
t+1
0 t +1
=z
⇔ z = z0 0
.
z
2
2
(t + 1)
Since ||z|| = 1, it must be the case that tt+1
0 +1 = ±1. However, s + 1 > 0 for
all s ∈ I, so t = t0 . Hence, z = z 0 , so (z, t) = (z 0 , t0 ), so φ is injective.
Furthermore, if C ⊆ S 1 × I is closed, then, since S 1 × I is compact, C
is compact. Hence φ(C) is compact and thus, since A is Hausdorff, closed.
Therefore φ is a closed map. Therefore, since φ is a bijective, closed map, φ
is a homeomorphism.
Now, since [f ] = 1 ∈ π1 (X; x0 ) there exists a homotopy H : S 1 × I → X
such that
H(z, 0) = 1, H(z, 1) = f (z) for all z ∈ S 1 ,
H(1, t) = x0 .
Then H ◦ φ−1 : A → I is continuous since H is continuous and φ−1 is
continuous. Now, if we define B := {z ∈ C|0 ≤ ||z|| ≤ 1/2}, then let
Ĥ : B → X be given by
Ĥ(x) = x0
for all x ∈ B. This Ĥ is certainly continuous. Finally, we define F : D2 → X
by:
Ĥ(z)
0 ≤ ||z|| ≤ 1/2
F (z) =
−1
(H ◦ φ )(z) 1/2 ≤ ||z|| ≤ 1.
If z ∈ C ∩ A, then ||z|| = 1/2, meaning φ−1 (z) = (2z, 0), so
(H ◦ φ−1 )(z) = H((2z, 0) = x0 .
Hence,
(H ◦ φ−1 )(z) = x0 = Ĥ(z).
Therefore, by the pasting lemma, F is continuous. To see that it is, in fact,
our desired map, we simply note that if s ∈ S 1 ,
F (s) = (H ◦ φ−1 )(s) = H(s, 1) = f (s)
by definition of H, so F is an extension of f to D2 .
6
CLAY SHONKWILER
C
Definition 0.1. Let p : E → B be a covering map. A homeomorphism
D : E → E such that p ◦ D = p is called a deck transformation of the
covering p.
Let p : E → B be a covering map and let ∆ = ∆(p) denote the set of
deck transformations of p. Show that ∆ forms a group under composition
of maps.
Proof. First, we show that ∆ is closed under composition of maps. Let φ, ψ ∈
∆. Then, since we know already that function composition is associative,
p ◦ (φ ◦ ψ) = (p ◦ φ) ◦ ψ = p ◦ ψ = p
so φ ◦ ψ ∈ ∆.
Now, let ι denote the identity homeomorphism of E. Then certainly
p ◦ ι = p, so ι ∈ ∆. Now, if φ ∈ ∆ and x ∈ E,
(φ ◦ ι)(x) = φ(ι(x)) = φ(x),
so φ◦ι = φ. Similarly for composition on the left, so ι is the identity element
in ∆.
To show that ∆ contains an inverse for each element, let φ ∈ ∆. Then
φ−1 is a homeomorphism of E and φ ◦ φ−1 = φ−1 ◦ φ = ι, so we need only
assure ourselves that φ−1 ∈ ∆. Then, since p ◦ φ = p,
p ◦ φ−1 = (p ◦ φ) ◦ φ−1 = p ◦ (φ ◦ φ−1 ) = p ◦ ι = p.
Hence, φ−1 ∈ ∆.
Since composition is associative and ∆ is closed under composition and
contains an identity and an inverse for each element, ∆ is a group under
composition.
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]