MATH540: Algebraic TopologyPROBLEM 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 = ( 12 , 0).

Now by horizontal compression (M,x0) deformation retracts to the vertical line L = { 12} × I with

endpoints x0 = ( 12 , 0) ∼ ( 1

2 , 1) identified. We know an interval with endpoints identified is homotopicto S1, so we have (M,x0) ' (L, x0) ' (S1, x0). As π1 is invariant under homotopy equivalence, thisimplies π1(M,x0) ∼= π1(S1, x0) ∼= Z.

2. Compute π1 of the surface of genus two using its presentation by identifying sides of a hexagon usingSeifert 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 followingfigures.

U , V and U ∩ V are path connected. π1(V ) = Z ∗ Z ∗ Z ∗ Z, π1(U) = 1, and π1(U ∩ V ) = Z. Leti1, i2 be inclusion maps from π1(U ∩ V ) to π1(U), π1(V ) respectively. If γ generates π1(U ∩ V ), theni1(γ) = 1, i2(γ) = aba−1b−1cdc−1d−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 computeits fundamental group.

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Figure 2: U Figure 3: V

(b) Argue that Mg is not homeomorphic, or even homotopy equivalent, to Mh if g 6= h. Hint: Considerthe abelianization of π1(Mg).

Solution (Christian Geske): 3. (a) Hatcher instructs us to construct the orientable surface Mg of

genus g as follows. Let a1, ..., a2g be 2g copies of S1. Take their wedge sum∨2gn=1 an with basepoint x0

and attach a single 2-cell via the map ϕ that sends S1 to the loop∏gn=1 a2n−1a2na2n−1a2n to get Mg.

Let N � π1(∨2gn=1 an, x0) be the normal subgroup generated by [ϕ(S1)] = [

∏gn=1 a2n−1a2na2n−1a2n] =∏g

n=1[a2n−1][a2n][a2n−1]−1[a2n]−1. Because∨2gn=1 an is path-connected and because π1(

∨2gn=1 an, x0) ∼=

∗2gn=1Z (the free product of 2g copies of Z), by Hatcher Proposition 1.26 we have that π1(Mg, x0) ∼=(∗2gn=1Z)/N . Because Mg is path-connected, this can be written simply as π1(Mg) ∼= (∗2gn=1Z)/N .

(b) Let surface Mg be given. Then π1(Mg) ∼= (∗2gn=1Z)/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 thecommutator 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 asurjective homomorphism onto [G/N,G/N ] with kernel ([G,G] ∩N). The result follows from the firstisomorphism theorem. �

By Lemma 1 and because N ≤ [∗2gn=1Z, ∗2gn=1Z] , [π1(Mg), π1(Mg)] ∼= [∗2gn=1Z, ∗

2gn=1Z]/N . More-

over, this isomorphism is the identity morphism. Then the abelianization of π1(Mg) is given by

π1(Mg)/[π1(Mg), π1(Mg)] ∼= π1(Mg)/([∗2gn=1Z, ∗2gn=1Z]/N) ∼= ∗2gn=1Z/[∗

2gn=1Z, ∗

2gn=1Z] where the last iso-

morphism follows from the third isomorphism theorem. But this is precisely the direct sum of 2g copiesof Z. If h 6= g, then the direct sum of 2g copies of Z is not isomorphic to the direct sum of 2h copies ofZ. Because the abelianization of a group does not change (up to isomorphism) under an isomorphismof 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 computeits fundamental group.

(b) Argue that Mg is not homeomorphic, or even homotopy equivalent, to Mh if g 6= h. Hint: Considerthe 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 2gcircles. Thus, Π1(X) ∼=< a1, b1, a2, b2, · · · , an, bn >. Let the 2-cell e2 be attached to X viathe map ϕ : S1 → X, then ϕ = a1b1a

−11 b−1

1 a2b2a−12 a−1

2 · · · agbga−1g b−1

g . By Prop 1.26, weconclude that Π1(Mg) ∼= Π1(Mg)/N , where N is the normal subgroup generated by ϕ, i.e.Π1(X) ∼=< a1, b1, · · · , an, bn|a1b1a

−11 b−1

1 · · · agbga−1g b−1

g >.

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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.

4. The space RP 2 can be thought of in several equivalent ways: as the set of lines through the origin inR

3, as the S2 with antipodal points x and −x identified, or as the quotient of D2 with antipodal pointson the boundary ∂D2 identified.

(a) Using the second description and the quotient map S2 → RP 2, use covering space arguments toprove that π1(RP 2, ∗) is isomorphic to Z2.

(b) Using the third description above, thought of a cell complex, compute π1(RP 2, ∗) using resultsproved about cell complexes.

Solution (Kris Joanidis) RPn is the coequaliser of the antipodal map σ and id:

DIAGRAM

let U+k = {x1, x2, ...xn+1) ∈ Sn : xk > 0}

U−k = {x1, x2, ...xn+1) ∈ Sn : xk < 0}

Clearly σ(U+k ) = U−k and σ(U−k ) = U+

k and {U+k , U

−k }k∈{1,2...n+1} is an open cover.

Hence 〈σ〉 acts in a “properly discontinuous” way, so p : Sn → RPn is a covering map.

a;

Using the correspondence between fibres and cosets, we have the following isomorphisms of sets:

π1(RPn)

p∗(π1(Sn))∼= π1(RPn) ∼= p−1(x0) ∼= {•, ◦}

for all n > 1.

Hence |π1(RPn)| = 2 so π1(RPn) ∼= Z2

b;

We can define the cell structure in the following way:

X0 = {∗}X1 = X0 ∪ϕ D1

X2 = X1 ∪ψ D2

where ϕ is the unique map S0 → X0, and ψ : S1 → X1 is given by

ψ(e2πiϑ) = ϑ ∈ D1 ⊂ X1 for ϑ ∈ [0, 1)

Using Proposition 1.26 in [?] with ∗ as the basepoint, we obtain:

π1(RP 2, ∗) ∼= 〈a|a2〉 ∼= Z2

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5. The n-fold dunce cap X is a certain generalization of RP 2. Consider the quotient of D2 by identifyingeach point on ∂D2 with its orbit under the map r : S1 → S1 sending z 7→ e2πı/nz. 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 underz → e2πi/nz.

So then let V = X − a, for a ∈ interior of X, and U be some open neighborhood of a not intersectingthe boundary.

Then π1(U) = 0 and π1(U ∩ V ) = Z. Now V can be deformation retracted to the boundary where thepath γ such that γ(0) = z and γ(1) = e2πi/nz generates π1(V ). In particular, we can represent V asan n-sided polygon with all the edges identified.

Thus, V ' S1 so π1(V ) = Z. By Van Kampen’s theorem we know π1(X) = Z/N where N is the leastnormal subgroup contaning i1(1)[i2(1)]−1 = γ−n. Thus modding by N induces the relation γn = e.Since we said 〈γ〉 = π1(V ) we obtain that π1(X) = 〈γ|γn = e〉 = Z/nZ.

6. (a) Let A be a single circle in 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?

Solution (by Michael Hankin): Main source: Hatcher, page 47

R3 - a circle Taking A to be the embedding of S1 in R3 and X = R3−A, X is deformation retractableto the image below,

which is homeomorphic to S2 ∧ S1 as in the image below.

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We have π1(S1, x0) ∼= Z, π1(S2, x0) ∼= 1, and π1(S1 ∩ S2, x0) ∼= 1 Thus, by SVK fundamental group ofX is π1(X,x0) ∼= Z

R3 - 2 disjoint circles Taking A and B to be the embeddings of two unlinked copies of S1 in R3

and X = R3 − (A ∪B), X is deformation retractable to S2 ∧ S1 ∧ S1 ∧ S2 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 S1 in R3 andX = R3 − (A ∪B), X is deformation retractable to S2 ∧ T 2 as in the image below.

Using the following calculation:π1(S2 ∧ T 2, x0) ∼= π1(S2, 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(S2 ∧ T 2, x0) ∼= π1(T 2, x0) ∼= π1(S1 × S1, 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 apresentation 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 S2 \ {2n points} is homotopically equivalent to R3 \Xsince the lines are all different and all passing through the origin. So we can define a retraction

r : R3 \X → S2 \ {2n points}x 7→ `0x ∩ S2

where `0x is the half line starting in 0 and passing through x. But S2\{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. Wewant 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 S1 (straightline homotopy). We can procede by induction on m. If we have m different points in the plane, thereexists a line ` that divides the plane in two halves with k points on one side and m − k on the otherwith 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 setof points is finite (and so it is the set of the lines passing through p and at least one of the other mpoints), we can find a line ` with the required property. Again, since our set of points is finite, thereexists an open interval I such that ` × I doesn’t contain any of the m points. (The measure of I hasto be less than the smallest distance from one of the points to `). Now take open sets U and V thatcontain `× I and one of the two (path) connected components of R \ (X ∪ `× I). Now, we can applySeifert-van Kampen theorem and thanks to our induction hypotesis and the fact that U ∩ V = ` × Iis contractible, we have that

π1(R2 \ {m points}) = π1(U) ∗π1(U∩V ) π1(V ) = (∗ki=1Zi) ∗ (∗m−ki=1 Zi) = ∗mi=1Zi.

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 Proposition1.26).

(b) Use this fact to show that the complement of a discrete subspace of Rn is simply-connected ifn ≥ 3.

Solution (Matthew Donner):

Proposition 0.1. Suppose that Y is obtained from a path-connected subspace X by attaching n-cellsfor 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 . Letλα be a path in X from x0 to the point ϕ(s0), where s0 is a basepoint of Sn−1. Now, we beginby expanding Y to a space Z that deformation retracts onto Y . We obtain Z from Y by attachingrectangular strips Sα with lower edge attached along λα and right edge attached along an arc in enαand 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 theare along which Sα is attached. Let A = Z − ∪α{yα} and let B = Z − X. Then A deformationretracts 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 ∩ Bis covered by the open sets Aα = (A ∩ B) − ∪β 6=αenβ . Since Aα deformation retracts onto Sn−1,Π1(Aα) ∼= 1. Thus, by Van Kampen’s Theorem, Π1(A ∩ B) ∼= 1. Now, since A, B, and A ∩ B arepath connected and A and B are open sets, we can apply Van Kampen’s Theorem: since Π1(Aα) ∼= 1,Π1(Y ) ∼= Π1(Z) ∼= Π1(A) ?Π1(B) ∼= Π1(X) ? 1 ∼= Π1(X). Thus, Π1(Y ) ∼= Π1(X).

Proposition 0.2. Use this fact to show that the complement of a discrete subspace of Rn is simply-connected 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 away that none of the disks intersect. Let T denote the union of all these n-disks. We can deformationretract 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-disksis Rn, which has a trivial fundamental group. Thus Rn − Q deformation retracts to a space whosefundamental group is trivial. Thus, the fundamental group of Rn−Q is trivial. Thus, Rn−Q is simplyconnected.

10. (a) Define the topological space RPn.

(b) Write RPn as a cell complex.

(c) Use the results of (??) and (9) to compute π1(RPn).

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