Hatcher §1.3

Ex 1.3.7

The quasi-circle W ⊂ R2 is a compactification of R with remainder W − R = [−1, 1]. There is a quotient map q : W → S1 to the one-point compactification S1 of R obtained by collapsing [−1, 1] to a point. This map is manifestly continuous (but there is also a general reason [2]). We first show that any map from a locally path-connected compact space to W factors through a contractible subspace of W . Let X be any locally path-connected compact space and f : X → W a continuous map. There is an open set V ⊂ W containing [−1, 1] with two path-components, V+ ⊃ [−1, 1] and V−. If x is a point of X such that f(x) ∈ [−1, 1] then there is a path-connected neighborhood of x that is being −1 mapped into V+. Thus f [−1, 1] is contained in an open set U such that f(U) ⊂ V+. Now X − U is a compact subspace of X that is being mapped into the remainder R ⊂ W . By compactness, f(X − U) ⊂ [a, b] for some closed interval [a, b] ⊂ R. Thus the f(X) = f(U) ∪ f(X − U) is contained in U+ ∪ [a, b] which again is contained in a compact subspace of W homeomorphic to [−1, 1] ∨ [0, 1]. In particular, π1(W ) = 0 (and, actually, πn(W ) = 0 for all n ≥ 1). Suppose that f is a lift

R > f | || p || || |  1 W q / S

of the quotient map q. Then f([−1, 1]) ⊂ Z and we may as well assume that f([−1, 1]) = {0}. The restriction of f to R = (0, 1) ⊂ W has the form f(t) = t + n for some n ∈ Z by uniqueness of lifts on connected spaces. But this contradicts continuity of f for there is no neighborhood U of 1 1 0 ∈ [−1, 1] such that f(U) ⊂ (− 2 , 2 ). (Any neighborhood U of 0 ∈ [−1, 1] ⊂ W contains points with f-values near n and also points with f-values near n+1 in that f(U ∩(0, 1)) = (U ∩(0, 1))+n.)

Ex 1.3.9

1 Since π1(X) is finite, the induced homomorphism f∗ : π1(X) → π1(S ) = Z is trivial as there are no nontrivial finite subgroups of Z. By the Lifting criterion [1, 1.33], f : X → S1 factors through the universal covering space R of S1, f = pf˜ where p: R → S1 is the universal covering map. But R is contractible, the identity map of R is homotopic to the constant map, so any map into R is homotopic to constant map. In particular, f˜ ' ∗, so also f = pf˜ ' p∗ = ∗.

1.3.12

1 1 2 2 4 Let N ⊂ G = ha, bi = π1(S ∨S ) be the smallest normal subgroup containing a , b , and (ab) . 1 1 The universal covering space of S ∨ S is the Cayley graph XeG of the right G-set G = {e}\G [1, Example 1.45]. The covering space of S1 ∨ S1 corresponding to the subgroup N < G is the orbit space N\XeG for the N-action on XeG which is equivalent [3] to the Cayley graph

1 1 XeN\G = N\XeG → G\XeG = XeG\G = S ∨ S 1 2 a for the right G-set N\G. This Cayley graph XeN\G is just the set N\G with edges x −→ xa and x −→b xb for all x ∈ N\G.

N(ab)2 7 h a b

a b v ( 6 Nbab Naba h b a

b a Nba u ) Nab i K a b

a b

) Nb Na j I b a

b a Ù * Ne Ex 1.3.14 (The Cayley complex for the infinite dihedral group G = a, b|a2b2 .) Let G be the group G = a, b|a2b2 = Z/2 ∗ Z/2. The subgroup generated by ab is an infinite cyclic normal subgroup with complement hai = Z/2 so that G = Z o Z/2 is a semi-direct product. The group Z/2 = hai acts on Z = habi by taking ab to aaba = ba = (ab)−1. The Cayley complex for the group G,

a 0 a 1 a 2 XeG = D ∪ D ∪ D G G×{a,b} G×{a2,b2} an infinite string (infinite in both directions) of spheres holding hands [1, Example 1.48], is the universal covering space of

0 a 1 a 2 2 2 XG = G\XeG = D ∪ D ∪ D = RP ∨ RP , {a,b} {a2,b2} the wedge of two copies of the real projective plane. The universal covering space action G×XeG → XeG is such that • ab translates the infinite string two places to the right, • a flips the whole string around the S2 with hands e and a and acts as the antipodal map on this S2. Since a(ab)a = ba = (ab)−1, the subgroup habi generated by ab is normal. The quotient group has order two and there is a short split exact sequence of the form 0 → Z → G → Z/2 −→` 0 where the homomorphism ` records word length parity. Thus the subgroups of G = Z o Z/2 are n n Hn = h(ab) i and Kn = h(ab) , ai for some n ≥ 0. Let us imagine that the sphere S2 has two hands placed at opposite points. The quotient space RP 2 then has only one hand. Thus RP 2 ∨ RP 2 is two RP 2s in love and holding hands. The covering spaces of RP 2 ∨ RP 2 consist of S2s and RP 2s holding hands such that there are no free hands. The functor

OG → Cov(G\XeG): H\G → H\XeG = XeH\G is an equivalence of categories. We must figure out what these quotients, the Cayley complexes for the orbit spaces H\G, are. 3

The group Hn has index 2n and the Cayley complex for Hn\G [3] is the space Xn consisting of 2n S2s holding hands in a circle, a necklace with 2n pearls. The group Kn has index n and the Cayley complex for Kn\G is the space Yn consisting of a string of n − 1 spheres holding hands and holding a P 2 at each end. 2 2 Let Y∞ consist of a string of S holding hands, infinite in one direction, and holding a RP at one end. The image of π1(Y∞) = Z/2 is the subgroup hai. What is the deck transformation group? The space Xn admits a Z/2 covering space action reflecting the whole arrangement across a line 2 through two S s opposite each other. The quotient space is Yn, 1 ≤ n ≤ ∞ [3, 2.6]. 2 2 Since there are no more subgroups of π1(RP ∨ RP ), we have described all covering spaces of RP 2 ∨ RP 2. Ex 1.3.18

The space X has a universal covering space action G × Y → Y where G = π1(X, x) is the fundamental group and Y is the universal covering space. Any covering map over X = G\Y is isomorphic to a covering map of the form pGH : H\Y → G\Y for some subgroup H < G. The Noether isomorphism theorems [4, 1.4] in group theory tell us that 0 pGH : H\Y → G\Y is abelian ⇔ H is normal in G and G/H is abelian ⇔ G < H where G0 = [G, G] is the derived subgroup generated by all commutators. This means that the 0 covering space pGG0 : G\Y → G \Y corresponding to the derived subgroup is an abelian covering map which is universal in the sense that it lies above any other abelian covering space of X because H\Y = (G0\H)\(G0\Y ). Its deck transformation group is (G0\H)\(G0\G). 0 Write Gab = G/G for the largest abelian quotient of G. We may characterize the universal abelian covering map by either of these two equivalent properties:

• The right G-set corresponding to the universal abelian covering map is isomorphic to Gab (which is a right G-set by the standard action Gab × G → Gab). • The universal abelian covering map is the normal covering map with Gab as deck trans- formation group. Suppose that X = S1 ∨ S1 is a wedge of two circles. Let X0 = Z × R ∪ R × Z ⊂ R2 be the space of all integer grid lines in R2. There is an obvious covering map p: X0 → X. The fibre, −1 the integer grid points p (∗) = Z × Z, is isomorphic to the right Z ∗ Z-set (Z ∗ Z)ab = Z × Z. Therefore p: X0 → X is the universal abelian covering map over X = S1 ∨ S1. The covering space (mZ × nZ)\X0 (a graph that can be drawn on a torus (mZ × nZ)\(R × R) is an abelian covering space of S1 ∨ S1 with Z/m × Z/n as its deck transformation group. It is easy to generalize this example to X = S1 ∨ ... ∨ S1. Ex 1.3.20 (The Cayley complex for G = a, b|abab−1 ) −1 1 1 2 Consider the group G = a, b|abab . Then XG = (S ∨ S ) ∪abab−1 D = N2 is the nonori- −1 entable compact surface N2 of genus 2, the Klein Bottle. From the relation ab = ba we see that ab2 = aab = aba−1 = ba−1a−1 = b2a, i.e. that a and b2 commute. The Cayley complex for 2 G, the universal covering space of N2, is the plane R and the universal covering space action [1, Example 1.42] G × R2 → R2, a(x, y) = (x + 1, y) is horizontal translation by one unit and b(x, y) = (0, 1)+(−x, y) is the glide-reflection consisting of vertical translation by one unit followed by reflection across the y-axis. (These two isometries of the plane do indeed satisfy the relation aba = b as they should.) By covering space theory, the (nonnormal) covering spaces of N2 correspond to conjugacy classes of (nonnormal) subgroups of G = π1(N2). We are looking for nonnormal subgroups. All subgroups of index 2 are normal, so they will not do. The subgroup a, b2 =∼ Z2, generated by the two translations a and b2, is normal because bab−1 = a−1 and a and b2 commute. The quotient group is a, b|abab−1, a, b2 = b|b2 = Z/2. There is thus an extension 0 → Z × Z → G → Z/2 → 0 −1 0 Conjugation by b induces an automorphism of a, b2 =∼ Z × Z with matrix B = with 0 1 respect to the basis {a, b2}. Now note that: 4 • Any subgroup of a, b2 that is not invariant under B is a nonnormal subgroup of G. 2 • Any finite index subgroup H1 of a, b will determine a 2|G: H1|-sheeted covering space 2 2 2 H1\R → G\R of the Klein Bottle by a torus; the total space H1\R is a torus because 2 ∼ 2 it is a compact surface with fundamental group π1(H1\R ) = H1 = Z . 3 2 2 The subgroup H1 = a , ab is a concrete example of an index 3 subgroup of a, b that is not invariant under matrix B. The corresponding covering map is a 6-sheeted nonnormal covering of the Klein Bottle by a torus. 3 The subgroup H2 = a , b is not normal for it is not invariant under conjugation with a as −1 −1 2 −1 2 aba = abbab = ab ab = a b does not belong to H2. (An element of H2 translates a multiple 2 2 of 3 units in the horizontal direction.) Thus H2\R → G\R = N2 is a [G: H2] = 3 sheeted 2 nonnormal covering space of the Klein Bottle N2. The total space H2\R is a Klein Bottle for it is 2 a compact surface with π1(H2\R )ab = (H2)ab = Z × Z/2 = π1(N2)ab. (Indeed, H2 is isomorphic 3 to G by the isomorphism a 7→ a, b 7→ b.) The deck transformation group H2\NG(H2) is trivial because H2 turns out to be self-normalizing, NG(H2) = H2. Try to picture what these covering maps look like geometrically by using the grid shown in [1, Example 1.42]. What are the universal covering spaces of the other compact surfaces (i.e. the Cayley complexes of their fundamental groups)? Using the double covers Mg → Ng+1 it will be enough to find the universal covering spaces of the orientable compact surfaces. See [1, Examples 1B.2, 1B.14] 1.3.21 1 1 1 1 Let X = (S × S ) ∪S1 MB be the union of torus S × S and a M¨obius band MB with the 1 1 1 boundary circle of the M¨obius band identified to S × {s0} ⊂ S × S . According to the van Kampen theorem 1 1 1 2 2 2 1 π1(X) = π1(S × S ) ∗π1(S ) π1(MB) = a, b, c | ab = ba, a = c = b, c | bc = c b (Strictly speaking, the three subspaces here are not open but they have open neighborhoods of which they are deformation retracts.) The universal covering space Xe consists of a plane R2 with strips attached along the horizontal lines R × {n} for all n ∈ Z. On top of each strip is attached a plane with similar strips attached. The element b of the fundamental group translates one unit in vertical direction, orthogonally to the strips; the element c of the fundamental group translates one half unit in the horizontal direction and moves the plane to the plane above it. Why? X is connected, locally path-connected and semi-locally simply connected for it is a connected CW-complex. Thus all connected covering spaces over X are of the form H\Xh1i and they all have Xh1i as their universal covering space. The annulus is a double covering space of the M¨obius

'$- - k

Figure 1. The double&% covering of the M¨obius band band. The covering space action by Z/2 rotates the annulus through an angle of π and reflects across a circle midway between the boundary circles. The two boundary circles are interchanged. Let X1 be the space that is the union of a small torus inside a larger torus and joined by an annulus such that small boundary circle of the annulus is a meridian circle on the small torus and the large boundary circle a meridian circle on the large torus. The Z/2-action on the annulus extends to an action on X1 with X as orbit space. (In fact, X1 = N\Xh1i where N < G is the 2 normal subgroup containing b and c .) X1 is not the universal covering space for it is not simply connected (you may use van Kampen to compute the fundamental group). 1 Let X2 be the space that is the union of two infinite cylinders S ×R inside each other and with 1 the circles S × {n}, n ∈ Z, on the two cylinders joined by annuli. There is a Z-action on X2 with

1Suppose that X = A ∪ B where A, B, and A ∩ B are open, nonempty and path-connected. Let AAo ∩ B /B be the inclusion maps. If iA iB

π1(A) = ha1,... | r1,...i , π1(B) = hb1,... | s1,...i , π1(A ∩ B) = hc1,... | t1,...i , then π1(X) = π1(A) ∗π1(A∩B) π1(B) = ha1, . . . , b1, . . . , c1,... | r1, . . . , s1,..., (iA)∗(c1) = (iB ) ∗ (c1),...i. 5

Figure 2. A double covering space of T ∪S1 MB (Thanks to Morten Poulsen

Figure 3. An infinite covering space of T ∪S1 MB (Thanks to Morten Poulsen

2 X1 as orbit space. (In fact, X2 = c \Xh1i is the normal covering space with fundamental group 2 2 π1 = c = Z and b, c | c = Z ∗ Z/2 as group of deck translations.) X2 is not the universal covering space either for it is not simply connected. 3 2 2 Let X3 ⊂ R be the space that is the union of the two horizontal planes R × {0} and R × {1} together with the strips R × Z × [0, 1]. The abelian group Z × Z acts on X3 via the commuting 1 maps b(x, y, z) = (x, y + 1, z) and c(x, y, z) = (x + 2 , y, 1 − z) and the orbit space is X (so we could have skipped X1 and X2 and gone directly to X3). (In fact, X3 = [G, G]\Xh1i is the universal abelian covering space of X, the group of covering translations is Gab = Z × Z generated by b and c.) 6

Figure 4. The infinite covering space X3 (Thanks to Morten Poulsen)

2 Now, X3 is the product space L × R where L ⊂ R is an infinite ladder. Thus the universal

Figure 5. The infinite ladder L covering space of X3 and of X is Xh1i = Lh1i × R where Lh1i, the universal covering space of L, is the infinite graph suggested by this figure

Figure 6. An approximation to the universal covering space of L

2 Let Y = RP ∪S1 MB be the union of the real projective plane and a M¨obius band with S1 = RP 1 ⊂ RP 2 identified to the boundary circle of the M¨obius band. According to van Kampen 2 2 2 4 1 π1(Y ) = π1(RP ) ∗π1(S ) π1(MB) = a, b | a , b = a = b | b = C4 2 is cyclic of order four. The universal covering space Y{e} is the union of a small sphere S , a big sphere S2, and an annulus S1 × I such that the small sphere fits in the center hole of the annulus which fits inside the big sphere. The annulus here is the four sheeted covering space of the M¨obius band so it has an action of the cyclic group C4. The small (inner) and the big (outer) sphere meet the annulus at their equators. The generating element b of the fundamental group acts on the outer sphere by rotating through an angle of π/2 around the vertical axis followed by a projection to the inner sphere; it acts on the inner sphere by rotating, followed by projection to the outer sphere, followed by reflection in the horizontal plane. Thus b2 is multiplication by −1.

The space Y has a double covering space YC2 → Y corresponding to the order two subgroup of 2 1 2 2 1 1 the fundamental group. YC2 = RP ∪S ×0 (S × I) ∪S ×1 RP is the union of two RP s and an annulus (the double covering space of the Møbius band) such that the inner boundary circles of the annulus has been identified to the generating circle S1 = RP 1 in one RP 2 and the outer boundary 2 circle to the generating circle of the other RP . By van Kampen, π1(YC2 ) = Z/2 ∗Z Z/2 = 2 2 2 a, b | a , b , a = b = a | a = C2. 7

2 Figure 7. The universal covering space of RP ∪S1 MB (Thanks to Morten Poulsen)

1.3.30 1 2 1 2 The Cayley complex for Z ∗ Z/2 = π1(S ∨ RP ) is the universal covering space of S ∨ RP .

Figure 8. Cayley complex for Z ∗ Z/2

References [1] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002. MR 2002k:55001 [2] Jesper M. Møller, General topology, http://www.math.ku.dk/ moller/e03/3gt/notes/gtnotes.dvi. [3] Jesper M Møller, Classification of covering spaces, http://www.ku.dk/~moller/f03/algtop/covering.pdf, 2003. [4] Derek J. S. Robinson, A course in the theory of groups, second ed., Springer-Verlag, New York, 1996. MR 96f:20001