MATH540: Algebraic Topology
PROBLEM SET 3
STUDENT SOLUTIONS
Key Problems
1. Compute π1 of the Mobius strip.
Solution (Spencer Gerhardt):
In other words, M ∼
= I × I/(s, 0) ∼ (1 − s, 1). Let x0 = ( 21 , 0).
Now by horizontal compression (M, x0 ) deformation retracts to the vertical line L = { 12 } × I with
endpoints x0 = ( 21 , 0) ∼ ( 12 , 1) identified. We know an interval with endpoints identified is homotopic
to S 1 , so we have (M, x0 ) ' (L, x0 ) ' (S 1 , x0 ). As π1 is invariant under homotopy equivalence, this
implies π1 (M, x0 ) ∼
= π1 (S 1 , x0 ) ∼
= Z.
2. Compute π1 of the surface of genus two using its presentation by identifying sides of a hexagon using
Seifert van-Kampen theorem as we did in class with the Torus T 2 .
Solution (Hongying Zhao):
Figure 1: X
We can have two open sets U and V that cover the surface of X = genus 2, as shown in the following
figures.
U , V and U ∩ V are path connected. π1 (V ) = Z ∗ Z ∗ Z ∗ Z, π1 (U ) = 1, and π1 (U ∩ V ) = Z. Let
i1 , i2 be inclusion maps from π1 (U ∩ V ) to π1 (U ), π1 (V ) respectively. If γ generates π1 (U ∩ V ), then
i1 (γ) = 1, i2 (γ) = aba−1 b−1 cdc−1 d−1 = [a, b][c, d].
By the Van Kampen Theorem, we have that π1 (X) = π1 (U ∪ V ) =< a, b, c, d|[a, b][c, d] >.
3. (a) Use the cell structure for the orientable surface Mg of genus g given in Hatcher pg 5 to compute
its fundamental group.
1
Figure 3: V
Figure 2: U
(b) Argue that Mg is not homeomorphic, or even homotopy equivalent, to Mh if g 6= h. Hint: Consider
the abelianization of π1 (Mg ).
Solution (Christian Geske): 3. (a) Hatcher instructs us to construct the orientable surface Mg of
W2g
genus g as follows. Let a1 , ..., a2g be 2g copies of S 1 . Take their wedge
sum n=1 an with basepoint x0
Q
g
and attach a single 2-cell via the map ϕ that sends S 1 to the loop n=1 a2n−1 a2n ā2n−1 ā2n to get Mg .
W2g
Qg
1
Let N π1 ( n=1 an , x0 ) be the normal subgroup generated by [ϕ(S )] = [ n=1 a2n−1 a2n ā2n−1 ā2n ] =
Qg
W2g
W2g
−1
[a2n ]−1 . Because n=1 an is path-connected and because π1 ( n=1 an , x0 ) ∼
=
n=1 [a2n−1 ][a2n ][a2n−1 ]
∼
∗2g
Z
(the
free
product
of
2g
copies
of
Z),
by
Hatcher
Proposition
1.26
we
have
that
π
(M
,
x
)
=
1
g
0
n=1
2g
2g
(∗n=1 Z)/N . Because Mg is path-connected, this can be written simply as π1 (Mg ) ∼
= (∗n=1 Z)/N .
2g
(b) Let surface Mg be given. Then π1 (Mg ) ∼
= (∗n=1 Z)/N where N is defined in part (a).
Lemma 1. For all groups G, for all N G, [G, G]/([G, G] ∩ N ) ∼
= [G/N, G/N ] (where [, ] denotes the
commutator subgroup).
Proof. Let π : G → G/N be the quotient map. By the properties of a homomorphism, π([G, G]) ≤
[G/N, G/N ]. Because π is surjective, π([G, G]) = [G/N, G/N ]. Thus the restriction of π to [G, G] is a
surjective homomorphism onto [G/N, G/N ] with kernel ([G, G] ∩ N ). The result follows from the first
isomorphism theorem. 2g
2g
∼ 2g
By Lemma 1 and because N ≤ [∗2g
n=1 Z, ∗n=1 Z] , [π1 (Mg ), π1 (Mg )] = [∗n=1 Z, ∗n=1 Z]/N . Moreover, this isomorphism is the identity morphism. Then the abelianization of π1 (Mg ) is given by
2g
2g
2g
2g
2g
π1 (Mg )/[π1 (Mg ), π1 (Mg )] ∼
= π1 (Mg )/([∗n=1 Z, ∗n=1 Z]/N ) ∼
= ∗n=1 Z/[∗n=1 Z, ∗n=1 Z] where the last isomorphism follows from the third isomorphism theorem. But this is precisely the direct sum of 2g copies
of Z. If h 6= g, then the direct sum of 2g copies of Z is not isomorphic to the direct sum of 2h copies of
Z. Because the abelianization of a group does not change (up to isomorphism) under an isomorphism
of the group, we conclude that π1 (Mg ) is not isomorphic to π1 (Mh ). Solution by Yueyang Ding
(a) Use the cell structure for the orientable surface Mg of genus g given in Hatcher pg 5 to compute
its fundamental group.
(b) Argue that Mg is not homeomorphic, or even homotopy equivalent, to Mh if g 6= h. Hint: Consider
the abelianization of π1 (Mg ).
Proof. (a) Let X be the space of the 0-cell and 1-cells of Mg , then X is the wedge sum of 2g
circles. Thus, Π1 (X) ∼
=< a1 , b1 , a2 , b2 , · · · , an , bn >. Let the 2-cell e2 be attached to X via
−1
−1 −1
1
−1 −1
the map ϕ : S → X, then ϕ = a1 b1 a−1
1 b1 a2 b2 a2 a2 · · · ag bg ag bg . By Prop 1.26, we
∼
conclude that Π1 (Mg ) = Π1 (Mg )/N , where N is the normal subgroup generated by ϕ, i.e.
−1
−1 −1
Π1 (X) ∼
=< a1 , b1 , · · · , an , bn |a1 b1 a−1
1 b1 · · · ag bg ag bg >.
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(b) Since the abelianization of Π1 (Mg ) is the free abelian group of rank 2g, if Π1 (Mg ) ∼
= Π1 (Mh ),
then h = g. Therefore, Mg is not homotopy equivalent to Mh if h 6= g.
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4. The space P 2 can be thought of in several equivalent ways: as the set of lines through the origin in
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, as the S 2 with antipodal points x and −x identified, or as the quotient of D2 with antipodal points
on the boundary ∂D2 identified.
R
(a) Using the second description and the quotient map S 2 →
prove that π1 ( P 2 , ∗) is isomorphic to 2 .
R
Z
RP 2, use covering space arguments to
R
(b) Using the third description above, thought of a cell complex, compute π1 ( P 2 , ∗) using results
proved about cell complexes.
Solution (Kris Joanidis) RP n is the coequaliser of the antipodal map σ and id:
DIAGRAM
let
Uk+ = {x1 , x2 , ...xn+1 ) ∈ S n : xk > 0}
Uk− = {x1 , x2 , ...xn+1 ) ∈ S n : xk < 0}
Clearly σ(Uk+ ) = Uk− and σ(Uk− ) = Uk+ and {Uk+ , Uk− }k∈{1,2...n+1} is an open cover.
Hence hσi acts in a “properly discontinuous” way, so p : S n → RP n is a covering map.
a;
Using the correspondence between fibres and cosets, we have the following isomorphisms of sets:
π1 (RP n ) ∼
= {•, ◦}
= p−1 (x0 ) ∼
= π1 (RP n ) ∼
p∗ (π1 (S n ))
for all n > 1.
Hence |π1 (RP n )| = 2 so π1 (RP n ) ∼
= Z2
b;
We can define the cell structure in the following way:
X 0 = {∗}
X 1 = X 0 ∪ϕ D1
X 2 = X 1 ∪ψ D2
where ϕ is the unique map S 0 → X 0 , and ψ : S 1 → X 1 is given by
ψ(e2πiϑ ) = ϑ ∈ D1 ⊂ X 1 for ϑ ∈ [0, 1)
Using Proposition 1.26 in [?] with ∗ as the basepoint, we obtain:
π1 (RP 2 , ∗) ∼
= ha|a2 i ∼
= Z2
3
R
5. The n-fold dunce cap X is a certain generalization of P 2 . Consider the quotient of D2 by identifying
each point on ∂D2 with its orbit under the map r : S 1 → S 1 sending z 7→ e2πı/n z. Compute π1 (X).
Solution by Nicolle Sandoval Gonzalez
Solution. As defined we have X = D2 with each point of the boundary identified with its orbit under
z → e2πi/n z.
So then let V = X − a, for a ∈ interior of X, and U be some open neighborhood of a not intersecting
the boundary.
Z
Then π1 (U ) = 0 and π1 (U ∩ V ) = . Now V can be deformation retracted to the boundary where the
path γ such that γ(0) = z and γ(1) = e2πi/n z generates π1 (V ). In particular, we can represent V as
an n-sided polygon with all the edges identified.
Z
Z
Thus, V ' S 1 so π1 (V ) = . By Van Kampen’s theorem we know π1 (X) = /N where N is the least
normal subgroup contaning i1 (1)[i2 (1)]−1 = γ −n . Thus modding by N induces the relation γ n = e.
Since we said hγi = π1 (V ) we obtain that π1 (X) = hγ|γ n = ei = /n .
Z Z
R3. Compute π1(R3 − A).
(b) Let A and B be disjoint circles in R3 . Compute π1 (R3 − (A ∪ B).
(c) How does π1 (R3 − (A ∪ B) change if the circles are linked?
6. (a) Let A be a single circle in
Solution (by Michael Hankin): Main source: Hatcher, page 47
R3 - a circle Taking A to be the embedding of S 1 in R3 and X = R3 −A, X is deformation retractable
to the image below,
which is homeomorphic to S 2 ∧ S 1 as in the image below.
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We have π1 (S 1 , x0 ) ∼
= Z, π1 (S 2 , x0 ) ∼
= 1, and π1 (S 1 ∩ S 2 , x0 ) ∼
= 1 Thus, by SVK fundamental group of
X is π1 (X, x0 ) ∼
=Z
R3 - 2 disjoint circles Taking A and B to be the embeddings of two unlinked copies of S 1 in R3
and X = R3 − (A ∪ B), X is deformation retractable to S 2 ∧ S 1 ∧ S 1 ∧ S 2 as in the image below.
Using the same logic as in the previous problem, the fundamental group of X is π1 (X, x0 ) ∼
=Z?Z
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R3 - 2 linked circles Taking A and B to be the embeddings of two linked copies of S 1 in R3 and
X = R3 − (A ∪ B), X is deformation retractable to S 2 ∧ T 2 as in the image below.
Using the following calculation:
π1 (S 2 ∧ T 2 , x0 ) ∼
= π1 (S 2 , x0 ) ?π1 ({x0 },x0 ) π1 (T 2 , x0 ) ∼
= 1 ?1 π1 (T 2 , x0 ) ∼
= π1 (T 2 , x0 )
We can calculate the fundamental group of X:
π1 (X, x0 ) ∼
= π1 (S 2 ∧ T 2 , x0 ) ∼
= π1 (T 2 , x0 ) ∼
= π1 (S 1 × S 1 , x0 ) ∼
=Z×Z
7. (a) Define the nth dihedral group Dn .
(b) Define the infinite dihedral group D∞ .
(c) Let X be the Klein bottle, see Hatcher pg 51 for a cell structure of the Klein bottle. Find a
presentation of π1 (X, ∗) and construct a homomorphism π1 (X, ∗) → D∞ .
8. Let X ⊂
R3 be the union of n lines through the origin. Compute
π1 (R3 − X).
Solution (Giuseppe Martone) The sphere S 2 \ {2n points} is homotopically equivalent to R3 \ X
since the lines are all different and all passing through the origin. So we can define a retraction
r : R3 \ X → S 2 \ {2n points}
x 7→ `0x ∩ S 2
where `0x is the half line starting in 0 and passing through x. But S 2 \{2n points} ∼
= R2 \{2n−1 points}.
So we only need a formula that tell us the fundamental group of R2 \ {m points} for all m ∈ N. We
want to prove that π1 (R2 \ {m points}) is the free group on m generators.
If m = 1 then we alrealdy know that R2 minus a point is homotopically equivalent to S 1 (straight
line homotopy). We can procede by induction on m. If we have m different points in the plane, there
exists a line ` that divides the plane in two halves with k points on one side and m − k on the other
with 0 < k < m. To see that, choose another point p in the plane such that the m points don’t lie
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on the same line passing through p. Since the lines passing through p are uncountable and our set
of points is finite (and so it is the set of the lines passing through p and at least one of the other m
points), we can find a line ` with the required property. Again, since our set of points is finite, there
exists an open interval I such that ` × I doesn’t contain any of the m points. (The measure of I has
to be less than the smallest distance from one of the points to `). Now take open sets U and V that
contain ` × I and one of the two (path) connected components of R \ (X ∪ ` × I). Now, we can apply
Seifert-van Kampen theorem and thanks to our induction hypotesis and the fact that U ∩ V = ` × I
is contractible, we have that
π1 (R2 \ {m points}) = π1 (U ) ∗π1 (U ∩V ) π1 (V ) = (∗ki=1
m
Zi) ∗ (∗m−k
i=1 Zi ) = ∗i=1 Zi .
9. (a) Suppose that Y is obtained from a path-connected subspace X by attaching n-cells for a fixed n ≥ 3.
Show that the inclusion X → Y induces an isomorphism on π1 . (Hint: see Hatcher Proposition
1.26).
(b) Use this fact to show that the complement of a discrete subspace of
n ≥ 3.
Rn
is simply-connected if
Solution (Matthew Donner):
Proposition 0.1. Suppose that Y is obtained from a path-connected subspace X by attaching n-cells
for a fixed n ≥ 3. Show that the inclusion X 7→ Y induces a isomorphism on Π1 .
Proof. We pick a basepoint x0 in X. Let {enα }α∈I denote the n-cells attached to X to form Y . Let
λα be a path in X from x0 to the point ϕ(s0 ), where s0 is a basepoint of S n−1 . Now, we begin
by expanding Y to a space Z that deformation retracts onto Y . We obtain Z from Y by attaching
rectangular strips Sα with lower edge attached along λα and right edge attached along an arc in enα
and all the left edges of the strips attached to each other. The top edges are not attached to anything,
so Z deformation retracts onto Y as desired. Now, in each cell enα we choose a point yα not in the
are along which Sα is attached. Let A = Z − ∪α {yα } and let B = Z − X. Then A deformation
retracts onto X. We show that B and A ∩ B have trivial fundamental groups. Since B is contractible,
Π1 (B) ∼
= 1. Next, we apply Van Kampen’s Theorem to show that Π1 (A ∩ B) ∼
= 1 Now, A ∩ B
is covered by the open sets Aα = (A ∩ B) − ∪β6=α enβ . Since Aα deformation retracts onto S n−1 ,
Π1 (Aα ) ∼
= 1. Thus, by Van Kampen’s Theorem, Π1 (A ∩ B) ∼
= 1. Now, since A, B, and A ∩ B are
path connected and A and B are open sets, we can apply Van Kampen’s Theorem: since Π1 (Aα ) ∼
= 1,
Π1 (Y ) ∼
= Π1 (X).
= Π1 (X). Thus, Π1 (Y ) ∼
= Π1 (X) ? 1 ∼
= Π1 (A) ? Π1 (B) ∼
= Π1 (Z) ∼
Proposition 0.2. Use this fact to show that the complement of a discrete subspace of Rn is simplyconnected if n ≥ 3.
Proof. Let Q be a discrete subset of Rn . Then for each q ∈ Q we can find an n disk containing q in a
way that none of the disks intersect. Let T denote the union of all these n-disks. We can deformation
retract Rn − Q onto Rn − T . Moreover, we can attach n-disks to Rn − T , ”filling in” all of the space,
without chaning the fundamental group (by part a). The space that results from attaching n-disks
is Rn , which has a trivial fundamental group. Thus Rn − Q deformation retracts to a space whose
fundamental group is trivial. Thus, the fundamental group of Rn − Q is trivial. Thus, Rn − Q is simply
connected.
10. (a) Define the topological space
(b) Write
RP n.
RP n as a cell complex.
R
(c) Use the results of (??) and (9) to compute π1 ( P n ).
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