Solutions to Problem Set 3
December 5, 2015
1. First, let X = (S 1 _ S 1 ) ⇥ [0, 1] and let x0 be the basepoint. The space X can be given
a CW-complex decomposition with two vertices, five edges, and two faces, as shown:
c
a
E
a
d
E
c
b
d
c
c
F
e
F
b
e
c
Let X _ S1 = (a [ b [ c)/ ⇠. Then Tf is X _ S 1 with two added 2–cells:
c
a
a
E
f (a)
c
c
c
b
b
F
f (b)
c
The fundamental group of X _ S 1 is the free group ha, b, ci. The boundaries of the cells
are sent to the paths acf (ā)c̄ and bcf (b̄)c̄, so
⇡1 (Tf ) = ha, b, c | acf⇤ (a) 1 c 1 , bcf⇤ (b) 1 [c] 1 i.
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Now, consider the case that X = S 1 ⇥ S 1 . We give X a CW-complex decomposition
with one vertex, two edges a and b, and one 2–cell E with boundary curve abāb̄. Then
X ⇥ [0, 1] has a decomposition with:
• two vertices, (x0 , 0) and (x0 , 1),
• five edges, (a, 0), (a, 1), (b, 0), (b, 1), and an edge c = x0 ⇥ [0, 1]
• four 2–cells, (E, 0), (E, 1), a ⇥ [0, 1], and b ⇥ [0, 1]
• one 3–cell
Let X _ S 1 = (X, 0) [ c/ ⇠. Then Tf consists of X _ S 1 with two added 2–cells and
one added 3–cell. The 2–cells correspond to a ⇥ [0, 1] (with boundary curve acf (ā)c̄)
and b ⇥ [0, 1] (with boundary curve bcf (b̄)c̄), so
⇡1 (Tf ) = ha, b, c | [a, b], acf⇤ (a) 1 c 1 , bcf⇤ (b) 1 [c] 1 i.
2. (a) If n = 1, then X is a single point, so ⇡1 (X) = 0. More generally, if X is a
connected graph with n vertices and n 1 edges, then every vertex has degree at
least 1, and the total degree of all the vertices is 2n 2, so there is at least one
vertex v with degree 1. Let e be the edge coming out of v and let X 0 be X with
v and e removed. Then ⇡1 (X 0 ) = 0 by induction.
Let A be a neighborhood of e and let B be a neighborhood of X 0 . If A and B
are small enough, then A deformation retracts to e and B deformation retracts to
X 0 and A \ B is path-connected. Then ⇡1 (A) = ⇡1 (B) = 0, so by van Kampen’s
theorem, ⇡1 (A [ B) = 0.
(b) We proceed by induction on (X) = m n + 1. If X is connected and (X) = 0,
then ⇡1 (X) = 0 by the first part of the theorem. If (X) > 0, let T be a spanning
tree for X. There is an edge e that is not in T . Let X 0 be X with e deleted and
let be a path in T between the endpoints of e that has no self-intersections or
backtracking.
Let A be a neighborhood of X 0 and let B be a neighborhood of e [ . Then B
deformation retracts onto the circle e [ , and A deformation retracts onto X 0 .
Furthermore, A \ B deformation retracts onto , which is simply-connected. By
induction, ⇡1 (A) = F (X) 1 , the free group with (X) 1 generators. By van
Kampen’s theorem, ⇡1 (X) = F (X) .
(c) Since X̃ is a cover of X with n sheets, it has n copies of each vertex and edge.
Thus, it has n vertices and nk edges, so (X) = nk n + 1. The rank of ⇡1 (X)
is thus nk n + 1 and the index [H : G] = n, so
rank ⇡1 (x) = [H : G](k
1) + 1.
3. Let (x1 , x2 ) 2 X1 ⇥ X2 and let Ui be an evenly covered neighborhood of xi in Xi . Then
pi 1 (xi ) is a disjoint union of copies of Ui , each taken homeomorphically to Ui by pi .
That is, there are discrete sets Ei such that
pi 1 (xi ) ⇠
= Ui ⇥ Ei ,
2
and pi (u, e) = u.
It follows that
(p1 ⇥ p2 ) 1 (U1 ⇥ U2 ) = p1 1 (U1 ) ⇥ p2 1 (U2 ) = (U1 ⇥ U2 ) ⇥ (E1 ⇥ E2 ),
and (p1 ⇥ p2 )(u1 , u2 , e1 , e2 ) = (u1 , u2 ).
Furthermore,
(p1 ⇥ p2 )⇤ (⇡1 (X̃1 ⇥ X̃2 )) = (p1 )⇤ (⇡1 (X̃)1 ) ⇥ (p2 )⇤ (⇡1 (X̃)2 ),
so if p : X̃ ! X1 ⇥ X2 is such a cover, then p⇤ (⇡1 (X̃)) is a product of a subgroup of
⇡1 (X1 ) and a subgroup of ⇡1 (X2 ).
If X = S 1 ⇥ S 1 , we can construct a cover of X as in the figure below — the indicated
map is a 2–fold cover, where X̃ is a torus, and p⇤ (⇡1 (X̃)) is generated by p⇤ (c) = a 1 b
and p⇤ (d) = ab. The resulting group is
p⇤ (⇡1 (X̃)) = {(x, y) 2 Z2 | x + y is even},
which is not a product of subgroups of Z.
d
a
c
a
c
b
a
b
b
d
b
a
4. Let X̃ be a covering space, let x0 be the basepoint of X, and let x̃0 be a preimage of x0 .
If U is an evenly covered neighborhood of x0 , then U contains infinitely many small
circles through x0 , so X̃ contains infinitely many small circles through x̃0 . The image
of each of these small circles is nontrivial in ⇡1 (X), so they are nontrivial in ⇡1 (X̃).
5. Let q : Y ! X̃ be the covering space illustrated on the next page and let x0 be the
basepoint of X. This is a covering space because if x̃ is a vertex of X̃, then q 1 (x̃)
consists of two vertices of Y . These two vertices are connected by a pair of “crossover”
edges that join one side of Y to the other. A small neighborhood of x̃ that avoids the
crossover edges is evenly covered.
However, if U ⇢ X is a neighborhood of x0 , then U contains a circle C through x. The
preimage of C in Y consists of many edges; most of these form circles homeomorphic
to C, but two of these edges cross from one side of Y to the other. Consequently,
q 1 (p 1 (U )) does not consist of disjoint copies of U taken homeomorphically to U , so
p q is not a cover of X.
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