Homework 2 Solution

MATH 5713
Section: 001
Topology II
MWF 11:50 – 12:40
KIMP 308
Spring 2015
Prof. Matthew Clay
Homework 2 – Solution
1. Let X be the quotient space of S 2 obtained by identifying the north and south poles to a single
point. Put a cell structure on X and use this to compute π1 (X).
Proof. We put a cell structure on X as follows. Let X (0) consist of a single point x. Attach a single
1–cell e1 by identifying both endpoints with the single vertex x to obtain X (1) . Finally, attach a
single 2–cell e2 via the map ∂D2 → X (1) which maps ∂D2 to the loop e1 · e1 to obtain X (2) = X.
Specifically, identifying the 1–skeleton X (1) and ∂D2 with {z ∈ C | |z| = 1} this map is:
(
z 2 , if Im(z) ≥ 0
z 7→
z¯2
if Im(z) ≤ 0.
As ∂D2 → X (1) represents the trivial element, by van Kampen’s theorem we find: π1 (X) ∼
=
π1 (X (1) ) ∼
= Z.
2. Compute the fundamental group of the space obtained from two tori S 1 × S 1 by identifying a
circle S 1 × {x0 } in one torus with the corresponding circle S 1 × {x0 } in the other torus.
Proof. We put a cell structure on the this union, X, as follows. For X (1) we take a wedge of three
circles, labeled a, b and c. We consider a, b and c as generators for the free group π1 (X (1) ). Then we
attach two 2–cells e21 and e22 where ∂D12 maps to the loop given by the commutator [a, b] and ∂D22
maps to the loop given by the commutator [b, c] to get X = X (2) . Observe that both e21 ∪ a ∪ b and
e22 ∪ b ∪ c are homeomorphic to a torus and that the two subcomplexes intersection along the single
circle b. By van Kampen’s theorem, we obtain the presentation π1 (X) = ha, b, c | [a, b], [b, c]i.
3. Suppose X = S 1 ∨ S 1 and f : X → X is a basepoint-preserving map. Compute a presentation
for fundamental group of the mapping torus:
Tf = X × I/(x, 0) ∼ (f (x), 1)
using the induced map f∗ : π1 (X) → π1 (X).
(1)
Proof. There is a cell structure on Tf described as follows. The 1–skeleton, Tf is the wedge sum
of three circles, labeled a, b and t. We consider a and b as the circles summands in X and t as
the subset {x0 } × I/(x0 , 0) ∼ (x0 , 1). As before, we consider a, b and t as generators for the free
(1)
group π1 (Tf ). Then Tf is obtained by attaching two 2–cells e21 and e22 where ∂D12 maps to the
loop tat−1 f∗ (a)−1 and ∂D22 maps to the loop tbt−1 f∗ (b)−1 . By van Kampen’s theorem, we obtain
the presentation π1 (Tf ) = ha, b, t | tat−1 f∗ (a)−1 , tbt−1 f∗ (b)−1 i.
e → X be a covering space and A ⊆ X a subspace. Show that the restriction of p to
4. Let p : X
−1
e = p (A) is a covering space.
A
1
e is a disjoint union of
Proof. LetG
{Uα } be an open covering of X such that for each Uα , p−1 (Uα ) ⊆ X
eα,β , each of which maps by p homeomorphically to Uα . Let pA = p e. We have that
open sets
U
A
β
−1
−1
−1
{A
G ∩ Uα } is an open covering of A. For each A ∩ Uα , we have pA (A ∩ Uα ) = p (A) ∩ p (Uα ) =
e∩U
eα,β , a disjoint union of open sets. The map pA is a homeomorphism between A
e∩U
eα,β and
A
β
eα,β and Uα . Thus pA : A
e → A is a
A ∩ Uα as it is the restriction of the homeomorphism between U
covering space.
5. Construct a simply-connected covering space for S 2 ∨ S 1 .
Proof. Fix a point z ∈ S 2 . Without loss of generality, we can assume the wedge sum X = S 2 ∨ S 1
is obtained by identifying z ∈ S 2 with 1 ∈ S 1 .
Let p0 : R → S 1 be the covering map p0 (t) = e2πit . For each integer n ∈ Z, let hn : (Sn2 , xn ) →
e by:
(S 2 , z) be a homeomorphism. We define X
G
e =Rt
Sn2 /n ∼ xn .
X
n∈Z
e is
In other words, we attach the sphere Sn2 to R by identifying xn ∈ Sn2 with 1 ∈ S 1 . The space X
e
e is a contractible subcomplex. The space X/R
e
homotopy equivalent to the space X/R
as R ⊂ X
is
2
homeomorphic to a wedge of countably many S ’s and hence simply-connected by van Kampen’s
e is simply-connected.
theorem. Therefore X
e → X by:
We define p : X
(
p0 (t)
if t ∈ R
p(x) =
hn (x) if x ∈ Sn2
As p0 (n) = 1 = z = hn (xn ), this map is well-defined and continuous. Consider the open cover of X
by U1 = S 2 ∪ {eiθ ∈ S 1 | −π < θ < π} and U2 = {eiθ ∈ S 1 | 0 < θ < 2π}. Then we have:
G
p−1 (U1 ) =
(n − 1/2, n + 1/2) ∪ Sn2 , and
n∈Z
−1
p (U2 ) =
G
(n, n + 1).
n∈Z
For each open set in the decomposition of p−1 (U1 ), p restricts to a homeomorphism onto U1 . Likewise
e → X is a covering space.
for p−1 (U2 ). Thus p : X
2