THE FUNDAMENTAL GROUP AND COVERING SPACES

Transcription

1 THE FUNDAMENTAL GROUP AND COVERING SPACES JESPER M. MØLLER Astrct. These notes, from first course in lgeric topology, introduce the fundmentl group nd the fundmentl groupoid of topologicl spce nd use them to clssify covering spces. Contents 1. Homotopy theory of pths nd loops Chnge of se point nd unsed homotopies 4 2. Covering spces 5 3. The fundmentl group of the circle, spheres, nd lense spces Applictions of π 1 (S 1 ) 9 4. The vn Kmpen theorem Fundmentl groups of knot nd link complements Ctegories Ctegories of right G-sets Trnsitive right ctions The clssifiction theorem Cyley tles, Cyley grphs, nd Cyley complexes Norml covering mps Sections in covering mps Universl covering spces of topologicl groups 27 References Homotopy theory of pths nd loops Definition 1.1. A pth in topologicl spce X from x 0 X to x 1 X is mp u: I X of the unit intervl into X with u(0) = x 0 nd u(1) = x 1. Two pths, u 0 nd u 1, from x 0 to x 1 re pth homotopic, nd we write simply u 0 u 1, if u 0 u 1 rel I, ie if u 0 nd u 1 re re homotopic reltive to the end-points {0, 1} of the unit intervl I. The constnt pth t x 0 is the pth x 0 (s) = x 0 for ll s I The inverse pth to u is the pth from x 1 to x 0 given y u(s) = u(1 s) If v is pth from v(0) = u(1) then the product pth pth u v given y { u(2s) 0 s 1 (u v)(s) = 2 1 v(2s 1) 2 s 1 where we first run long u with doule speed nd then long v with doule speed is pth from u(0) to v(1). In greter detil, u 0 u 1 if there exists homotopy h: I I X such tht h(s, 0) = u 0 (s), h(s, 1) = u 1 (s) nd h(0, t) = x 0, h(1, t) = x 1 for ll s, t I. All pths in homotopy clss hve the sme strt point nd the sme end point. Note the following rules for products of pths x 0 u u u x 1 Dte: Decemer 10, I would like to thnk Morten Poulsen for supplying the grphics nd Yokun Wu for spotting some errors in n erlier version. The uthor is prtilly supported y the DNRF through the Centre for Symmetry nd Deformtion in Copenhgen. /home/moller/underv/lgtop/notes/chp1/clsscover.tex. 1

2 2 J.M. MØLLER u 1 (s) x 0 x 1 h(s, t) u 0 (s) Figure 1. A pth homototopy etween two pths u u x 0, u u x 1 (u v) w u (v w) u 0 u 1, v 0 v 1 = u 0 v 0 u 1 v 1 These drwings re ment to suggest proofs for the first three sttements: u(s) x 0 u v w x 0 x 1 x 0 x 0 x 0 x 1 x 2 x 3 u(2s) x 1 u(2s) u(1 2s) u v w In the first cse, we first run long u with doule speed nd then stnd still t the end point x 1 for hlf the time. The homotopy consists in slowing down on the pth u nd spending less time just stnding still t x 1. In the second cse, we first run long u ll the wy to x 1 with doule speed nd then ck gin long u lso with doule speed. The homotopy consists in running out long u, stnding still for n incresing length of time (nmely for 1 2 (1 t) s 1 2 (1 + t)), nd running ck long u. In the third cse we first run long u with speed 4, then long v with speed 4, then long w with speed 2, nd we must show tht cn deformed into the cse where we run long u with speed 2, long v with speed 4, long w with speed 4. This cn e chieved y slowing down on u, keeping the sme speed long v, nd speeding up on w. The fourth of the ove rules is proved y this picture, u 1 (2s) v 1 (2s 1) x 0 x 1 x 2 u 0 (2s) v 0 (2s 1) We write π(x)(x 0, x 1 ) for the set of l homotopy clsses of pths in X from x 0 to x 1 nd we write [u] π(x)(x 0, x 1 ) for the homotopy clss contining the pth u. The fourth rule implies tht the product opertion on pths induces product opertion on homotopy clsses of pths (1.2) π(x)(x 0, x 1 ) π(x)(x 1, x 2 ) π(x)(x 0, x 2 ): ([u], [v]) [u] [v] = [u v] nd the other three rules trnslte to similr rules [x 0 ] [u] = [u] = [u] [x 1 ] (neutrl elements) [u] [u] = [x 0 ], [u] [u] = [x 1 ] (inverse elements) ([u] [v]) [w] = [u] ([v] [w]) (ssocitivity)

3 THE FUNDAMENTAL GROUP AND COVERING SPACES 3 for this product opertion. We next look t the functoril properties of this construction. Suppose tht f : X Y is mp of spces. If u is pth in X from x 0 to x 1, then the imge fu is pth in Y from f(x 0 ) to f(x 1 ). Since homotopic pths hve homotopic imges there is n induced mp π(x)(x 0, x 1 ) π(f) π(y )(f(x 0 ), f(x 1 )): [u] [fu] on the set of homotopy clsses of pths. Oserve tht this mp π(f) does not chnge if we chnge f y homotopy reltive to {x 0, x 1 }, π(f) respects the pth product opertion in the sense tht π(f)([u] [v]) = π(f)([u]) π(f)([v]) when u(1) = v(0), π(id X ) = id π(x)(x1,x 2), π(g f) = π(g) π(f) for mps g : Y Z We now summrize our findings. Proposition 1.3. For ny spce X, π(x) is groupoid, nd for ny mp f : X Y etween spces the induced mp π(f): π(x) π(y ) is groupoid homomorphism. In fct, π is functor from the ctegory of topologicl spces to the ctegory of groupoids. Definition 1.4. π(x) is clled the fundmentl groupid of X. The fundmentl group sed t x 0 X is the group π 1 (X, x 0 ) = π(x)(x 0, x 0 ) of homotopy clsses of loops in X sed t x 0. The pth product (1.2) specilizes to product opertion nd to trnsitive free group ctions π 1 (X, x 0 ) π 1 (X, x 0 ) π 1 (X, x 0 ) (1.5) π 1 (X, x 0 ) π(x)(x 0, x 1 ) π(x)(x 0, x 1 ) π(x)(x 0, x 1 ) π 1 (X, x 1 ) so tht π 1 (X, x 0 ) is indeed group nd π(x)(x 0, x 1 ) is n ffine group from the left nd from the right. For fundmentl groups, in prticulr, ny sed mp f : (X, x 0 ) (Y, y 0 ) induces group homomorphism π(f) = f : π 1 (X, x 0 ) π 1 (Y, y 0 ), given y π 1 (f) = f ([u]) = [fu], tht only depends on the sed homotopy clss of the sed mp f. Proposition 1.6. The fundmentl group is functor π 1 : hotop Grp from the homotopy ctegory of sed topologicl spces into the ctegory of groups. This mens tht π 1 (id (X,x0)) = id π1(x,x 0) nd π 1 (g f) = π 1 (g) π 1 (f). It follows immeditely tht if f : X Y is homotopy equivlence of sed spces then the induced mp π 1 (f) = f : π 1 (X, x 0 ) π 1 (Y, y 0 ) is n isomorphism of groups. (See Section 5 for more informtion out ctegories nd functors.) Corollry 1.7. Let X e spce, A suspce, nd i : π 1 (A, 0 ) π 1 (X, 0 ) the group homomorphism induced y the inclusion mp i: A X. (1) If A is retrct of X then i hs left inverse (so it is monomorphism). (2) If A is deformtion retrct of X then i hs n inverse (so it is n isomorphism). Proof. (1) Let r : X A e mp such tht ri = 1 A. Then r i is the identity isomorphism of π 1 (A, 0 ). (2) Let r : X A e mp such tht ri = 1 A nd ir 1 X rel A. Then r i is the identity isomorphism of π 1 (A, 0 ) nd i r is the identity isomorphism of π 1 (X, 0 ) so i nd r re ech others inverses. Corollry 1.8. Let X nd Y e spces. There is n isomorphism (p X ) (p Y ) : π 1 (X Y, x 0 y 0 ) π 1 (X, x 0 ) π 1 (Y, y 0 ) induced y the projection mps p X : X Y X nd p Y : X Y Y. Proof. The loops in X Y hve the form u v where u nd v re loops in X nd Y, respectively (Generl Topology, 2.63). The ove homomorphism hs the form [u v] [u] [v]. The inverse homomorphism is [u] [v] [u v]. Note tht this is well-defined. We cn now compute our first fundmentl group. Exmple 1.9. π 1 (R n, 0) is the trivil group with just one element ecuse R n contins the suspce {0} consisting of one point s deformtion retrct. Any spce tht deformtion retrct onto on of its points hs trivil fundmentl group. Is it true tht ny contrctile spce hs trivil fundmentl group?

4 4 J.M. MØLLER Our tools to compute π 1 in more interesting cses re covering spce theory nd vn Kmpen s theorem Chnge of se point nd unsed homotopies. Wht hppens if we chnge the se point? In cse, the new se point lies in nother pth-component of X, there is no reltion t ll etween the fundmentl groups. But if the two se points lie in the sme pth-component, the fundmentl groups re isomorphic. Lemm If u is pth from x 0 to x 1 then conjugtion with [u] is group isomorphism. π 1 (X, x 1 ) π 1 (X, x 0 ): [v] [u] [v] [u] Proof. This is immedite from the rules for products of pths nd specil cse of (1.5). We lredy noted tht if two mps re homotopic reltive to the se point then they induce the sme group homomorphism etween the fundmentl groups. We shll now investigte how the fundmentl group ehves with respect to free mps nd free homotopies, ie mps nd homotopies tht do not preserve the se point. Lemm Suppose tht f 0 f 1 : X Y re homotopic mps nd h: X I Y homotopy. For ny point x X, let h(x) π(y )(f 0 (x), f 1 (x)) e the pth homotopy clss of t h(x, t). For ny u π(x)(x 0, x 1 ) there is commuttive digrm f 0 (x 0 ) h(x 0) f 1 (x 0 ) in π(y ). f 0(u) f 1(u) f 0 (x 1 ) f 1 (x 1 ) h(x1) Proof. Let u e ny pth from x 0 to x 1 in X. If we push the left nd upper edge of the homotopy I I Y : (s, t) h(u(s), t) into the lower nd right edge f 1 (u) h(x 0 ) h(x 1 ) f 0 (u) we otin pth homotopy h(x 0 ) f 1 (u) f 0 (u) h(x 1 ). Corollry In the sitution of Lemm 1.12, the digrm commutes. π 1 (Y, f 1 (x 0 )) (f 1) π 1 (X, x 0 ) (f 0) π 1 (Y, f 0 (x 0 )) = [h(x0)] [h(x 0)] Proof. For ny loop u sed t x 0, f 0 (u)h(x 0 ) = h(x 0 )f 1 (u) or f 0 (u) = h(x 0 )f 1 (u)h(x 0 ). Corollry (1) If f : X Y is homotopy equivlence (possily unsed) then the induced homomorphism f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )) is group isomorphism. (2) If f : X Y is nullhomotopic (possily unsed) then f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )) is the trivil homomorphism.

5 THE FUNDAMENTAL GROUP AND COVERING SPACES 5 Proof. (1) Let g e homotopy inverse to f so tht gf 1 X nd fg 1 Y. By Lemm 1.12 there is commuttive digrm π 1 (X, x 0 ) π 1 (Y, f(x 0 )) = f f π 1 (Y, f(x 0 )) g π 1 (X, gf(x 0 )) π 1 (Y, fgf(x 0 )) = π 1 (X, x 0 ) which shows tht g is oth injective nd surjective, ie g is ijective. Then lso f is ijective. (2) If f homotopic to constnt mp c then f followed y n isomorphism equls c which is trivil. Thuse lso f is trivil. We cn now nswer question from Exmple 1.9 nd sy tht ny contrctile spce hs trivil fundmentl group. Definition A spce is simply connected if there is unique pth homotopy clss etween ny two of its points. The spce X is simply connected if π(x)(x 1, x 2 ) = for ll x 1, x 2 X, or, equivlently, X is pth connected nd π 1 (X, x) = t ll points or t one point of X. 2. Covering spces A covering mp over X is mp tht loclly looks like the projection mp X F X for some discrete spce F. Definition 2.1. A covering mp is continuous surjective mp p: Y X with the property tht for ny point x X there is neighorhood U (n evenly covered neighorhood), discrete set F, nd homeomorphism U F p 1 (U) such tht the digrm U F p 1 (U) commutes. pr 1 p p 1 (U) U Some covering spces, ut not ll (7.22), rise from left group ctions. Consider left ction G Y Y of group G on spce Y. Let p G : Y G\Y e the quotient mp of Y onto the orit spce G\Y. The quotient mp p G is open ecuse open susets U Y hve open sturtions GU = g G gu = p 1 G p G(U) (Generl Topology 2.82). The open sets in G\Y correspond ijectively to sturted open sets in Y. We now single out the left ctions G Y Y for which the quotient mp p G : Y G\Y of Y onto its orit spce is covering mp. Definition 2.2. [5, (*) p. 72] A covering spce ction is group ction G Y Y where ny point y Y hs neighorhood U such tht the trnslted neighorhoods gu, g G, re disjoint. (In other words, the ction mp G U GU is homeomorphism.) Exmple 2.3. The ctions Z R R: (n, t) n + t Z/2 S n S n : (±1, x) ±x Z/m S 2n+1 S 2n+1 : (ζ, x) ζx, where ζ C is n mth root of unity, ζ m = 1, {±1, ±i, ±j, ±k} S 3 S 3, quternion multipliction [5, Exmple 1.43], re covering spce ctions nd the orit spces re Z\R = S 1 (the circle), Z/2\S n = RP n (rel projective spce), nd Z/m\S 2n+1 = L 2n+1 (m) (lense spce). The ction Z S 1 S 1 : (n, z) e πi 2n z is not covering spce ction for the orits re dense.

6 6 J.M. MØLLER Exmple 2.4. The mps p n : S 1 S 1, n Z, nd p : R S 1 given y p n (z) = z n, nd p (s) = e 2πs = (cos(2πs), sin(2πs)) re covering mps of the circle with fire p 1 n (1) = Z/nZ nd p 1 (1) = Z. There re mny covering mps of S 1 S 1. The mp S n C 2 \S n = RP n, n 1, is covering mp of rel projective n-spce. The mp S 2n+1 C m \S 2n+1 = L 2n+1 (m) is covering mp of the lens spce. M g N g+1 doule covering mp of the unorientle surfce of genus g + 1 with F = Z/2Z. Cn you find covering mp of M g? Cn you find covering mp of R? Theorem 2.5 (Unique HLP for covering mps). [5, 1.30] Let p: Y X e covering mp, B e ny spce, nd h: B I X homotopy into the se spce. If one end of the homotopy lifts to mp B {0} Y then the whole homotopy dmits unique lift B I Y such tht the digrm commutes. e h0 B {0} Y e h B I X Proof. We consider first the cse where B is point. The sttement is then tht in the sitution h {0} y0 Y eu I X u there is unique mp ũ: I Y such tht pũ = u nd ũ(0) = y 0. For uniqueness of lifts from I see Theorem 2.12.(1). We need to prove existence. The Leesgue lemm (Generl Topology, 2.158) pplied to the compct spce I sys tht there is sudivision 0 = t 0 < t 1 < < t n = 1 of I such tht u mps ech of the closed suintervls [t i 1, t i ] into n evenly covered neighorhood in X. Suppose tht we hve lifted u to ũ defined on [0, t i 1 ]. Let U e n evenly covered neighorhood of u(t i 1 ). Suppose tht the lift ũ(t i 1 ) elongs to U {l} for some l F. Continue the given ũ with (p (U {l})) 1 u [t i 1, t i ]. After finitely mny steps we hve the unique lift on I. We now turn to the generl sitution. Uniqueness is cler for we hve just seen tht lifts re uniquely determined on the verticl slices {} I B I for ny point of B. Existence is lso cler except tht continuity of the lift is not cler. We now prove tht the lift is continuous. Let e ny point of X. By compctness, there is neighorhood N of nd sudivision 0 = t 0 < t 1 < < t n = 1 of I such tht h mps ech of the sets N [t i 1, t i ] into n evenly covered neighorhood of X. Suppose tht h(n [0, t 1 ]) is contined in the evenly covered neighorhood U X nd let Ũ p 1 (U) Y e neighorhood such tht p Ũ : Ũ U is homeomorphism nd h 0 (, 0) Ũ. We cn not e sure tht h 0 (N {0}) Ũ; only if N is connected. Replce N y 1 N h 0 (Ũ). Then h 0 (N {0}) Ũ. Then (p Ũ) 1 h N [0, t 1 ] is lift of h N [0, t 1 ] extending h 0. After finitely mny steps we hve lift defined on N I (where N is possily smller thn the N we strted with). Do this for every point of B. These mps must gree on their overlp y uniqueness. So they define lift B I Y. This lift is continuous since it is continuous on ech of the open tues N I. We emphsize the specil cse where B is point. Let y 0 Y e point in Y nd x 0 = p(y 0 ) X its imge in X. Corollry 2.6 (Unique pth lifting). Let x 0 nd x 1 e two points in X nd let y 0 e point in the fire p 1 (x 0 ) Y over x 0. For ny pth u: I X from x 0 to x 1, the exists unique pth ũ: I Y in Y strting t ũ(0) = y 0. Moreover, homotopic pths hve homotopic lifts: If v : I X is pth in X tht is pth homotopic to u then the lifts ũ nd ṽ re lso pth homotopic. Proof. First, in Theorem 2.5, tke B to e point. Next, tke B to e I nd use the HLP to see tht homotopic pths hve homotopic lifts. Corollry 2.7. Let p: Y X e covering mp nd let y 0, y 1, y 2 Y, x 0 = py 0, x 1 = py 1, x 2 = py 2. p p

7 THE FUNDAMENTAL GROUP AND COVERING SPACES 7 (1) By recording end points of lifts we otin mps p 1 (x 1 ) π(x)(x 1, x 2 ) p 1 (x 2 ), p 1 (x 0 ) π 1 (X, x 0 ) p 1 (x 0 ) given y y [u] = ũ y (1) where ũ y is the lift of u strting t y. Multipliction y pth u from x 1 to x 2 slides the fire over x 1 ijectively into the fire over x 2. (2) The covering mp p: X Y induces injective mps π(y )(y 1, y 2 ) p π(x)(x 1, x 2 ), π 1 (Y, y 0 ) p π 1 (X, x 0 ) The suset p π(y )(y 1, y 2 ) π(x)(x 1, x 2 ) consists of ll pths from x 1 to x 2 tht lift to pths from y 1 to y 2. The sugroup p π 1 (Y, y 0 ) π 1 (X, x 0 ) consists of ll loops t x 0 tht lloft to loops t y 0. Definition 2.8. The monodromy functor of the covering mp p: X Y is functor F (p): π(x) Set of the fundmentl groupoid of the se spce into the ctegory Set of sets. This functor tkes point in x X to the fire F (p)(x) = p 1 (x) over tht point nd it tkes pth homotopy clss u π(x)(x 0, x 1 ) to F (p)(x 0 ) = p 1 (x 0 ) p 1 (x 1 ) = F (p)(x 1 ): y y u. (The nottion here is such tht F (p)(uv) = F (p)(v) F (p)(u) for pths u π(x)(x 0, x 1 ), v π(x)(x 1, x 2 ).) In prticulr, the fire F (p)(x) = p 1 (x) over ny point x X is right π 1 (X, x)-set. Corollry 2.9 (The fundmentl groupoid of covering spce). The fundmentl groupoid of Y, π(y ) = π(x) F (p) is the Grothendieck construction of the fier functor (2.8). In other words, the mp π(p): π(y )(y 0, y 1 ) π(x)(x 0, x 1 ) is injective nd the imge is the set of pth homotopy clsses from x 0 to x 1 tht tke y 0 to y 1. In prticulr, the homomorphism p : π 1 (Y, y 0 ) π 1 (X, x 0 ) is injective nd its imge is the set of loops t x 0 tht lift to loops t y 0. Proof. We consider the functor F (p) s tking vlues in discrete ctegories. The ojects of π(x) F (p) re pirs (x, y) where x X nd y F (p)(x) Y. A morphism (x 1, y 1 ) (x 2, y 2 ) is pir (u, v) where u is morphism in π(x) from x 1 to x 2 nd v is morphism in F (p)(x 2 ) from F (p)(u)(x 1 ) = x 1 u to y 2. As F (p)(x 2 ) hve no morphisms ut identities, the set of morphisms (x 1, y 1 ) (x 2, y 2 ) is the set of u π(x)(x 1, x 2 ) such tht y 1 u = y 2. This is precisely π(y )(y 1, y 2 ). Definition For spce X, let π 0 (X) e the set of pth components of X. Lemm Let p: X Y e covering mp. (1) Suppose tht X is pth connected. The inclusion p 1 (x 0 ) Y induces ijection p 1 (x 0 )/π 1 (X, x 0 ) π 0 (Y ). In prticulr, Y is pth connected π 1 (X, x 0 ) cts trnsitively on the fire p 1 (x 0 ) (2) Suppose tht X nd Y re pth connected. The mps π 1 (Y, y 1 )\π(x)(x 1, x 2 ) p 1 (x 2 ) π 1 (Y, y 0 )\π 1 (X, x 0 ) p 1 (x 0 ) π 1 (Y, y 1 )u y 1 u [pu y ] y π 1 (Y, y 0 )u y 0 u [pu y ] y re ijections. Here, u y is ny pth in Y from y 1 or y 0 to y. In prticulr, π 1 (X, x 0 ): π 1 (Y, y 0 ) = p 1 (x 0 ). Proof. The mp p 1 (x 0 ) π 0 (Y ), induced y the inclusion of the fire into the totl spce, is onto ecuse X is pth connected so tht ny point in the totl spce is connected y pth to point in the fire. Two points in the fire re in the sme pth component of Y if nd only if re in the sme π 1 (X, x 0 )-orit. If Y is pth connected, then π 1 (X, x 0 ) cts trnsitively on the fire p 1 (x 0 ) with isotropy sugroup π 1 (Y, y 0 ) t y 0.

8 8 J.M. MØLLER Theorem 2.12 (Lifting Theorem). Let p: Y X e covering mp nd f : B X mp into the se spce. Choose se points such tht f( 0 ) = x 0 = p(y 0 ) nd consider the lifting prolem (B, 0 ) f f (Y, y 0 ) p (X, x 0 ) (1) If B is connected, then there exists t most one lift f : (B, 0 ) (Y, y 0 ) of f over p. (2) If B is pth connected nd loclly pth connected then There is mp f : (B, 0 ) (Y, y 0 ) such tht f = p f f π 1 (B, 0 ) p π 1 (Y, y 0 ) Proof. (1) Suppose tht f 1 nd f 2 re lifts of the sme mp f : B X. We clim tht the sets { B f 1 () = f 2 ()} nd { B f 1 () f 2 ()} re open. Let e ny point of B where the two lifts gree. Let U X e n evenly neighorhood of f(). Choose Ũ p 1 (U) = U F so tht the restriction of p to Ũ is homoemorphism nd f 1 () = f 2 () elongs to Ũ. f 1 1 Then f 1 nd f 1 2 gree on the neighorhood (Ũ) f 2 (Ũ) of. Let e ny point of B where the two lifts do not gree. Let U X e n evenly neighorhood of f(). Choose disjoint open sets Ũ1, Ũ2 p 1 (U) = U F so tht the restrictions of p to Ũ1 nd Ũ2 re homoemorphisms nd f 1 () elongs to Ũ1 nd f 2 () to Ũ2. Then f 1 nd f 2 do not gree on the neighorhood f 1 f 1 2 (Ũ2) of. 1 (Ũ1) (2) It is cler tht if the lift exists, then the condition is stisfied. Conversely, suppose tht the condition holds. For ny point in B, define lift f y f() = y 0 [fu ] where u is ny pth from 0 to. (Here we use tht B is pth connected.) If v is ny other pth from 0 to then y 0 [fu ] = y 0 [fv ] ecuse y 0 [fu fv ] = y 0 s the loop [fu fv ] π 1 (Y, y 0 ) fixes the point y 0 y Lemm We need to see tht f is continuous. Note tht ny point B hs pth connected neighorhood tht is mpped into n evenly covered neighorhood of f() in X. It is evident wht f does on this neighorhood of. A mp f : B S 1 C {0} into the circle hs n nth root if nd only if the induced homomorphism f : π 1 (B) Z is divisile y n. 3. The fundmentl group of the circle, spheres, nd lense spces For ech n Z, let ω n e the loop ω n (s) = (cos(2πns), sin(2πns), s I, on the circle. Theorem 3.1. The mp Φ: Z π 1 (S 1, 1): n [ω n ] is group isomorphism. Proof. Let p: R S 1 e the covering mp p(t) = (cos(2πt), sin(2πt)), t R. Rememer tht the totl spce R is simply connected s we sw in Exmple 1.9. The fire over 1 is p 1 (1) = Z. Let u n (t) = nt e the ovious pth from 0 to n Z. By Lemm 2.11 the mp Z π 1 (S 1, 1): n [pu n ] = [ω n ] is ijective. We need to verify tht Φ is group homomorphism. Let m nd n e integers. Then u m (m + u n ) is pth from 0 to m + n so it cn e used insted of u m+n when computing Φ(m + n). We find tht Φ(m + n) = [p(u m (m + u n ))] = [p(u m ) p(m + u n )] = [p(u m )][p(m + u n )] = [p(u m )][p(u n )] = Φ(m)Φ(n) ecuse p(m + u n ) = pu n s p hs period 1. Theorem 3.2. The n-sphere S n is simply connected when n > 1.

9 THE FUNDAMENTAL GROUP AND COVERING SPACES 9 Proof. Let N e the North nd S the South Pole (or ny other two distinct points on S n ). The prolem is tht there re pths in S n tht visit every point of S n. But, in fct, ny loop sed t N is homotopic to loop tht voids S (Prolem nd Solution). This mens tht π 1 (S n {S}, N) π 1 (S n, N) is surjective. The result follows s S n {S} is homeomorphic to the simply connected spce R n. Corollry 3.3. The fundmentl group of rel projective n-spce RP n is π 1 (RP n ) = C 2 for n > 1. The fundmentl group of the lense spce L 2n+1 (m) is π 1 (L 2n+1 (m)) = C m for n > 0. Proof. We proceed s in Theorem 3.1. Consider the cse of the the covering mp p: S 2n+1 L 2n+1 (m) over the lense spce L 2n+1 (m). Let N = (1, 0,..., 0) S 2n+1 C n+1. The cyclic group C m = ζ of mth roots of unity is generted y ζ = e 2πi/m. The mp ζ j ζ j N, j Z, is ijection C m p 1 pn etween the set C m nd the fire over pn. As S 2n+1 is simply connected there is ijection Φ: p 1 pn = C m π 1 (L 2n+1 (m), pn): ζ j [pω j ] where ω j is the pth in S 2n+1 from N to ζ j N given y ω j (s) = (e 2πisj/m, 0,..., 0). Since ω i+j ω i (ζ i ω j ), it follows just s in Theorem 3.1 tht Φ is group homomorphism. For the projective spces, use the pths ω j (s) = (cos(2πjs), sin(2πjs), 0,..., 0) from N to ( 1) j N, to see tht Φ: p 1 pn = C 2 π 1 (RP n, pn): (±1) j [pω j ] is ijection Applictions of π 1 (S 1 ). Here re some stndrd pplictions of Theorem 3.1. Corollry 3.5. The nth power homomorphism p n : (S 1, 1) (S 1, 1): z z n induces the nth power homomorphism π 1 (S 1, 1) π 1 (S 1, 1): [ω] [ω] n. Proof. (p n ) Φ(1) = (p n ) [ω 1 ] = [p n ω 1 ] = [ω n 1 ] = [ω n ] = Φ(n) = Φ(1) n. Theorem 3.6 (Brouwer s fixed point theorem). (1) The circle S 1 is not retrct of the disc D 2. (2) Any mp self-mp of the disc D 2 hs fixed point. Proof. (1) Let i: S 1 D 2 e the inclusion mp. The induced mp i : Z = π 1 (S 1 ) π 1 (D 2 ) = 0 is not injective so S 1 cn not e retrct y 1.7. (2) With the help of fixed-point free self mp of D 2 one cn construct retrction of D 2 onto S 1. But they don t exist. Theorem 3.7 (The fundmentl theorem of lger). Let p(z) = z n + n 1 z n z + 0 e normed complex polynomil of degree n. If n > 0, then p hs root. Proof. Any normed polynomil p(z) = z n + n 1 z n z + 0 is nonzero when z is lrge: When z > 1 + n , then p(z) 0 ecuse n 1 z n n 1 z n < n 1 z n z n 1 = ( n ) z n 1 < z n Therefore ny normed polynomil p(z) defines mp S 1 (R) C {0} where S 1 (R) is the circle of rdius R nd R > 1 + n In fct, ll the normed polynomils p t (z) = z n + t( n 1 z n z + 0 ), t I, tke S 1 (R) into C {0} so tht we hve homotopy S 1 (R) I C {0}: (z, t) z n + t( n 1 z n z + 0 ) etween p 1 (z) = p(z) S 1 (R) nd p 0 (z) = z n. If p(z) hs no roots t ll, the mp p S 1 (R) fctors through the complex plne C nd is therefore nullhomotopic (s C is contrctile) nd so is the homotopic mp S 1 (R) C {0}: z z n nd the composite mp S 1 z Rz S 1 (R) z zn C {0} z z/ z S 1 But this is simply the mp S 1 S 1 : z z n which we know induces multipliction y n (3.5). However, nullhomotopic mp induces multipliction y 0 (1.14). So n = 0. A mp f : S 1 S 1 is odd if f( x) = f(x) for ll x S 1. Any rottion (or reflection) of the circle is odd (ecuse it is liner).

10 10 J.M. MØLLER Lemm 3.8. Let f : S 1 S 1 e n odd mp. Compose f with rottion R so tht Rf(1) = 1. The induced mp (Rf) : π 1 (S 1, 1) π 1 (S 1, 1) is multipliction y n odd integer. In prticulr, f is not nullhomotopic. Proof. We must compute (Rf) [ω 1 ]. The HLP gives lift 0 {0} R eω I ω 1 S 1 S 1 nd we hve (Rf) [ω 1 ] = [p ω]. When 0 s 1/2, ω 1 (s+1/2) = ω 1 (s) nd lso Rfω 1 (s+1/2) = Rfω 1 (s) s Rf is odd. The lift, ω of Rfω 1, then stisfies the eqution Rf ω(s + 1/2) = ω(s) + q/2 for some odd integer q. By continuity nd connectedness of the intervl [0, 1/2], q does not depend on s. Now ω(1) = ω(1/2) + q/2 = ω(0) + q/2 + q/2 = q nd therefore (Rf) [ω 1 ] = [p ω] = [ω q ] = [ω 1 ] q. We conclude tht (Rf) is multipliction y the odd integer q. Since nullhomopotic mp induces the trivil group homomorphism (1.14), f is not nullhomotopic. Theorem 3.9 (Borsuk Ulm theorem for n = 2). Let f : S 2 R 2 e ny continuous mp. Then there exists point x S 2 such tht f(x) = f( x). Proof. Suppose tht f : S 2 R 2 is mp such tht f(x) f( x) for ll x S 2. The composite mp f(x) f( x) S 1 x incl f(x) f( x) 2 S is odd so it is not nullhomotopic. But the first mp S 1 S 2 is nullhomotopic ecuse it fctors through the contrctile spce D+ 2 = {(x 1, x 2, x 3 ) S 2 x 3 0}. This is contrdiction. This implies tht ther re no injective mps of S 2 R 2 ; in prticulr S 2 does not emed in R 2. Proposition 3.10 (Borsuk Ulm theorem for n = 1). Let f : S 1 R e ny continuous mp. Then there exists point x S 1 such tht f(x) = f( x). Proof. Look t the mp g(x) = f(x) f( x). If g is identiclly 0, f(x) = f( x) for ll x S 1. Otherwise, g is n odd function, g( x) = g(x), nd g hs oth positive nd negtive vlues. By connectedness, g must ssume the vlue 0 t some point. This implies tht there re no injective mps S 1 R; in prticulr S 1 does not emed in R. 4. The vn Kmpen theorem Let G j, j J, e set of groups indexed y the set J. The coproduct (or free product) of these groups is group i J G j with group homomorphisms ϕ j : G j j J G j such tht (4.1) Hom( j J S 1 G j, H) = j J Hom(G j, H): ϕ (ϕ ϕ j ) j J is ijection for ny group H. The group j J G j contins ech group G j s sugroup nd these sugroups do not commute with ech other. If the groups hve presenttions G j = L j R j then j J L j R j = j J L j j J R j s this group hs the universl property. See [9, 6.2] for the construction of the free product. The chrcteristic property (4.1) pplied to H = j J G j shows tht there is group homomorphism Gj G j from the free product to the direct product whose restriction to ech G j is the inclusion into the product. Exmple 4.2. [9, Exmple II III p 171] Z/2 Z/2 = Z Z/2 nd Z/2 Z/3 = PSL(2, Z). We cn prove the first ssertion: Z/2 Z/2 =, 2, 2 =,, c 2, 2, c = =, 2, cc, c =, 2, c, c = c 1 ut the second one is more difficult.

11 THE FUNDAMENTAL GROUP AND COVERING SPACES 11 Suppose tht the spce X = j J X j is the union of open nd pth connected suspces X j nd tht x 0 is point in j J X j. The inclusion of the suspce X j into X induces group homomorphism ι j : π 1 (X j, x 0 ) π 1 (X, x 0 ). The coproduct j J π 1(X j, x 0 ) is group equipped with group homomorphisms ϕ j : π 1 (X j, x 0 ) j J π 1(X j, x 0 ). Let Φ: j J π 1 (X j, x 0 ) π 1 ( j J X j, x 0 ) = π 1 (X, x 0 ) e the group homomorphism determined y Φ ϕ j = ι j. Is Φ surjective? In generl, no. The circle, for instnec, is the union of two contrctile open suspces, so Φ is not onto in tht cse. But, if ny loop in X is homotopic to product of loops in one of the suspces X j, then Φ is surjective. Is Φ injective? It will, in generl, not e injective, ecuse the individul groups π 1 (X i ) in the free product do not intersect ut the suspces do intersect. Any loop in X tht is loop in X i X j will in the free product count s loop oth in π 1 (X i ) nd in π 1 (X j ). We lwys hve commuttive digrms of the form π 1 (X i, x 0 ) ι ij ι i π 1 (X i X j, x 0 ) π 1 (X, x 0 ) ι ji ι j π 1 (X j, x 0 ) where ι ij re inclusion mps. This mens tht Φ(ι ij g) = Φ(ι ji g) for ny g π 1 (X i X j, x 0 ) so tht (4.3) i, j J g π 1 (X i X j ): ι ij (g)ι ji (g) 1 ker Φ Let N j J π 1(X j, x 0 ) e the smllest norml sugroup contining ll the elements of (4.3). The kernel of Φ must contin N ut, of course, the kernel could e igger. The surprising fct is tht often it isn t. Theorem 4.4 (Vn Kmpen s theorem). Suppose tht X = j J X j is the union of open nd pth connected suspces X j nd tht x 0 is point in j J X j. (1) If the intersection of ny two of the open suspces is pth connected then Φ is surjective. (2) If the intersection of ny three of the open suspces is pth connected then the kernel of Φ is N. Corollry 4.5. If the intersection of ny three of the open suspces is pth connected then Φ determines n isomorphism Φ: j J π 1 (X j, x 0 )/N = π 1 (X, x 0 ) Proof of Theorem 4.4. (1) We need to show tht ny loop u π 1 (X) in X is product u 1 u m of loops u i π 1 (X ji ) in one of the suspces. Let u: I X e loop in X. Thnks to the Leesgue lemm (Generl Topology, 2.158) we cn find sudivision 0 = t 0 < t 1 < t m = 1 of the unit intervl so tht u i = u [t i 1, t i ] is pth in (sy) X i. As u(t i ) X i X i+1, nd lso the se point x 0 X i X i+1, nd X i X i+1 is pth connected, there is pth g i in X i X i+1 from the sepoint x 0 to u(t i 1 ). The sitution looks like this: X 1 X 2 u [0, t 1 ] g 1 u [t 1, t 2 ] g 2 X 3 u [t 2, 1] Now u u [0, t 1 ] u [t 1, t 2 ] u [t m 1, 1] (u [0, t 1 ] g 1 ) (g 1 u [t 1, t 2 ] g 2 ) (g m u [t m 1, 1]) is product of loops where ech fctor is inside one of the suspces.

12 12 J.M. MØLLER (2) Let N π 1 (X i ) e the smllest norml sugroup contining ll the elements (4.3). Let u i π 1 (X ji ). For simplicity, let s cll X ji for X i. Consider the product u 1 }{{} u 2 }{{} π 1(X 1) π 1(X 2) u }{{} m j J π 1 (X j ) π 1(X m) nd suppose tht Φ(u 1 u m ) is the unit element of π 1 (X). We wnt to show tht u 1 u m lies in the norml sugroup N or tht u 1 u m is the identity in the quotient group π 1 (X j )/N. Since u 1 u m is homotopic to the constnt loop in X there is homotopy I I X = X j from the loop u 1 u m in X to the constnt loop. Divide the unit squre I I into smller rectngles such tht ech rectngle is mpped into one of the suspces X j. We my ssume tht the sudivision of I {0} is further sudivision of the sudivision t i/m coming from the product u 1 u m. It could e tht one new vertex is (or more new vertices re) inserted etween (i 1)/m nd i/m. X k X l X i Connect the imge of the new vertex with pth g inside X i X k X l to the se point. Now u i is homotopic in X i to the product (u i [(i 1)/m, ] g) (g u i [, i/m]) of two loops in X i. This mens tht we my s well ssume tht no new sudivision points hve een introduced t the ottom line I {0}. Now pertur slightly the smll rectngles, ut not the ones in the ottom nd top row, so tht lso the corner of ech rectngle lies in t most three rectngles. The lower left corner my look like this: X 5 X 6 u 15 u 16 X 1 u 12 X 2 u 1 The loop u 1 in X 1 is homotopic to the product of pths u 15 u 16 u 12 y homotopy s in the proof of Connect the imge of the point to the se point y pth g 156 inside X 1 X 5 X 6 nd connect the imge of the point to the se point y pth g 126 inside X 1 X 2 X 6. Then u 1 is homotopic in X 1 to the product of loops (u 15 g 156 ) (g 156 u 16 g 126 ) (g 126 u 12 ) in X 1. The first of these loops is loop in X 1 X 5, the second is loop in X 1 X 6, nd the third is loop in X 1 X 2. In π 1 (X j ) nd modulo the norml sugroup N we hve tht u }{{} 1 u 2 = u }{{} 15 g 156 g } {{ } 156u 16g 126 g } {{ } 126u 12 u } {{ } 2 = u }{{} 15 g 156 g } {{ } 156u 16g 126 g } {{ } 126u 12 u 2 } {{ } X 1 X 2 X 1 X 1 X 1 X 2 X 5 X 6 X 2 After finitely mny steps we conclude tht modulo N the product u 1 u m equls product of constnt loops, the identity element. Corollry 4.6. Let X j e set of pth connected spces. Then X j ) j J π 1 (X j ) = π 1 ( j J provided tht ech se point x j X j is the deformtion retrct of n open neighorhood U j X j. Proof. Vn Kmpen s theorem does not pply directly to the suspces X j of X j ecuse they re not open. Insted, let X j = X j i J U i. The suspces X j re open nd pth connected nd the intersection of t lest two of them is the contrctile spce i J U i. Moreover, X j is deformtion retrct of X j. For instnce, punctured compct surfces hve free fundmentl groups.

13 THE FUNDAMENTAL GROUP AND COVERING SPACES 13 Corollry 4.7 (vn Kmpen with two suspces). Suppose tht X = X 1 X 2 where X 1, X 2, nd X 1 X 2 re open nd pth connected. Then for ny sepoint x 0 X 1 X 2. π 1 (X 1 X 2, x 0 ) = π 1 (X 1, x 0 ) π1(x 1 X 2,x 0) π 1 (X 2, x 0 ) = This mens tht when X 1 X 2 is pth connected the fundmentl group functor tkes push out of spces to push out, mlgmted product, of groups X 1 X 2 i 1 i 2 X 2 π 1 π 1 (X 1 X 2 ) (i1) (i 2) π 1 (X 2 ) X 2 X π 1 (X 2 ) π 1 (X) As very specil cse, we see tht spce, tht is the union of two open simply connected suspces with pth connected intersection, is simply connected. This proves, gin (Theorem 3.2), tht S n is simply connected when n > 1. We cn use this simple vrint of vn Kmpen to nlyze the effect on the fundmentl group of ttcing cells. Corollry 4.8 (The fundmentl group of cellulr extension). Let X e pth connected spce. Then { π 1 (X f α D n π 1 (X)/ γ α f α γ α n = 2 α ) = π 1 (X) n > 2 where γ α is pth from the se point of X to the imge of the se point of S 1 α D 2 α. Proof. Let Y e X with the n-cells ttched. Attch strips, fences connecting the se point of X with the se points of the ttched cells, to Y nd cll the results Z. This does not chnge the fundmentl group s Y is deformtion retrct of Z (Corollry 1.7). Let A e Z with the top hlf of ech cell removed nd let B = Z X. Then Z = A B nd A B re pth connected (the fences re there to mke A nd B pth connected) so tht π 1 (Z) = π 1 (A) π1(a B) π 1 (B) y the vn Kmpen theorem in the simple form of Corollry 4.7. Now B is contrctile, hence simply connected (Corollry 1.14), so π 1 (Y ) = π 1 (Z) is the quotient of π 1 (A) y the smllest norml sugroup contining the imge of π 1 (A B) π 1 (A). But A B is homotopy equivlent to wedge α Sn 1 α of (n 1)-spheres. In prticulr, A B is simply connected when n > 2 (Corollry 4.6, Theorem 3.2) so tht π 1 (Y ) = π 1 (Z) = π 1 (A) = π 1 (X). When n = 2, π 1 (A B) is free group nd the imge of it in π 1 (A) = π 1 (X) is generted y the pth homotopy clsses of the loops γ α f α γ α. Corollry 4.9. Let X e CW-complex with skelet X k, k 0. Then π 0 (X 1 ) = π 0 (X), π 1 (X 2 ) = π 1 (X) Corollry The fundmentl groups of the compct surfces of positive genus g re π 1 (M g ) = 1, 1,..., g, g [ i, i ], π 1 (N g ) = 1,..., g 2 i, The compct orientle surfces M g, g 0, re distinct, π 1 (M g ) = Z 2g, nd the compct nonorientle surfces N h, h 1, re distinct, π 1 (N g ) = Z g Z/2. Corollry Let M e connected mnifold of dimension 3. Then π 1 (M {x}) = π 1 (M) for ny point x M. Proof. Apply vn Kmpen to M = M {x} D n, M {x} D n S n 1 nd rememer tht S n 1 is simply connected when n 3. Which groups cn e relized s fundmentl groups of spces? For instnce, C = S 1 nd C m = S 1 m D 2 so tht ny finitely generted elin group cn e relized s the fundmentl group of product of these spces.

14 14 J.M. MØLLER Corollry For ny group G there is 2-dimensionl CW-complex X G such tht π 1 (X G ) = G. Proof. Choose presenttion G = g α r β nd let X G = D 0 {g α} D 1 {r β } e the 2-dimensionl CW-complex whose 1-skeleton is wedge of circles, one for ech genertor, with 2-discs ttched long the reltions. Oserve tht X H G = X H X G. Also, X π1(m g) = M g, X π1(n g) = N g, g Fundmentl groups of knot nd link complements. The complement of pir of unlinked circles in R 3 deformtion retrcts to S 1 S 1 S 2 S 2 nd pir of linked circles to (S 1 S 1 ) S 2. The fundmentl groups re Z Z nd Z Z, respectively. Thus the two complements re not homeomorphic. Let m nd n e reltively prime nturl numers nd K = K mn the (m, n)-torus knot. We wnt to compute the knot group π 1 (R 3 K). According to (4.11), π 1 (R 3 K) = π 1 (S 3 K). Now D 2 S 3 = D 4 = (D 2 D 2 ) = D 2 D 2 D 2 D 2 is the union of two solid tori intersecting in torus S 1 S 1. Let K e emedded in this middle torus. Then S 3 K = ( D 2 D 2 K) (D 2 D 2 K), ( D 2 D 2 K) (D 2 D 2 K) = S 1 S 1 K nd vn Kmpen sys (if we ignore 1 the condition tht the susets should e open) π 1 (S 3 K) = π 1( D 2 D 2 K) π 1 (D 2 D 2 K) π 1 (S 1 S 1 K) Here, D 2 D 2 K deformtion retrcts onto the core circle D 2 {0}, nd S 1 S 1 K (the torus minus the knot) is n nnulus S 1 (0, 1). (Tke n open strip [0, 1] (O, 1) nd wrp it round the torus so tht the end 0 (0, 1) meets the end 1 (0, 1)). The imge of the genertor of this infinite cyclic group is the m power of genertor, respectively the nth power. Hence π 1 (S 3 K) =, m = n = G mn It is now mtter of group theory to tell us tht if G m1n 1 nd G m2n 2 re isomorphic then {m 1, n 1 } = {m 2, n 2 }. In order to nlyze this group, note tht m = n is in the center. Let C e the centrl group generted y this element. The quotient group G mn /C =, m, n = Z/m Z/n hs no center. (In generl the free product G H of two nontrivil groups hs no center ecuse the elements re words in elements from G lternting with elements from H.) Therefore C is precisely the center of G mn. Thus we cn recover mn s the order of the eliniztion of G/Z(G). Also, ny element of finite order in Z/m Z/n is conjugte to n element of Z/m or Z/n. Thus we cn recover the lrgest of m, n s the mximl order of torsion element in G/Z(G). Thus we cn recover the set {m, n}. Corollry There re infinitely mny knots. (Here re some of them.) Another wy of sying this is tht D 2 D 2 K deformtion retrcts onto the mpping cylinder of the degree m, respectively n, mp S 1 S 1. Thus the union of these two spces, S 3 K, deformtion retrcts onto the union of the two mpping cylinders, which is the doule mpping cylinder X mn for the two mps. Thus X mn emeds in S 3 nd R 3 when (m, n) = 1. On the other hnd X 22 is the union of two Möius nds. A Möius nd is RP 2 minus n open 2-disc, so X 22 = RP 2 #RP 2, the Klein ottle, which does not emed in R 3. 1 To fix this, thicken the knot nd enlrge the two solid tori little so tht they overlp.

15 THE FUNDAMENTAL GROUP AND COVERING SPACES 15 A ctegory C consists of [7] 5. Ctegories Ojects,,... For ech pir of ojects nd set of morphisms C(, ) with domin nd codomin A composition function C(, c) C(, ) C(, c) tht to ech pir of morphisms g nd f with dom(g) = cod(f) ssocites morphism g f with dom(g f) = dom(f) nd cod(g f) = cod(g) We require Identity: For ech oject the morphism set C(, ) contins morphism id such tht g id = g nd id f = f whenever these compositions re defined Associtivity : h (g f) = (h g) f whenever these compositions re defined A morphism f C(, ) with domin nd codomin is sometimes written f :. A morphism f : is n isomorphism if there exists morphism g : such tht the two possile compositions re the respective identities. Definition 5.1. A group is ctegory with one oject where ll morphisms re isomorphisms. A groupoid is ctegory where ll morphisms re isomorphisms. Exmple 5.2. In the ctegory Top of topologicl spces, the ojects re topologicl spces, the morphisms re continuous mps, nd composition is the usul composition of mps. In the ctegory hotop, the ojects re topologicl spces, the morphisms re homotopy clsses of continuous mps, nd composition is induced y the usul composition of mps. In the ctegory Grp of groups, the ojects re groups, the morphisms re groups homomorphisms, nd composition is the usul composition of group homomorphisms. In the ctegory Mt R the ojects re the nturl numers Z +, the set of morphisms m n consists of ll n y m mtrices with entries in the commuttive ring R, nd composition is mtrix multipliction. The fundmentl groupoid π(x) of topologicl spce X is groupoid where the ojects re the points of X nd the morphisms x y re the homotopy clsses π(x)(x, y) of pths from x to y, nd composition is composition of pth homotopy clsses. A functor F : C D ssocites to ech oject of C n oject F () of D nd to ech morphism f : in C morphism F (f): F () F () in D such tht F (id ) = id F () nd F (g f) = F (g) F (f). A nturl trnsformtion τ : F = G: C D etween two functors F, G: C D is D-morphism τ() D(F, G) for ech oject of C such tht the digrms F τ() G f F f F τ() G commute for ll morphisms f C(, ) in C. A nturl trnsformtion τ is nturl isomorphism if ll the components τ(), O(C), re D-isomorphisms. Exmple 5.3. The fundmentl group is functor from the ctegory of sed topologicl spces nd sed homotopy clsses of mps to the ctegory of groups. The fundmentl groupoid is functor from the ctegory of topologicl spces to the ctegory of groupoids. Any homotopy h: f 0 f 1 induces nturl isomorphism h: π(f 0 ) = π(f 1 ): π(x) π(y ) etween functors etween fundmentl groupoids (Lemm 1.12). Definition 5.4. Let C nd D e ctegories. The functor ctegory Func(C, D) is the ctegory whose ojects re the functors from C to D nd whose morphisms re the nturl trnsformtions. Gf Definition 5.5. Two ctegories, C nd D, re isomorphic (equivlent) when there re functors C such tht the composite functors re (nturlly isomorphic to) the respective identity functors. Lemm 5.6. A functor F : C D is n equivlence of ctegories if nd only if ny oject of D is isomorphic to n oject of the form F () for some oject of C F G D

16 16 J.M. MØLLER f F (f) F is ijective on morphism sets: The mps C(, ) D(F (), F ()) re ijections for ll ojects nd of C Proof. Suppose tht F : C D is n equivlence of ctegories. Then there is functor G in the other direction nd nturl isomorphisms σ : GF = 1 C nd τ : F G = 1 D. Let d e ny oject of D. The isomorphism τ d : F G(d) = d shows tht d is isomorphic to F for = Gd. Let, e ojects of C. We note first tht C(, ) D(F, F ) C(GF, GF ) is injective for the commuttive digrm GF σ = f GF f GF σ = f shows tht f = σ GF f σ 1 cn e recovered from GF f. Thus C(, ) D(F, F ) is injective. Symmetriclly, lso the functor G is injective on morphism sets. To show tht F is surjective on morphism sets let g e ny D-morphism F F. Put f = σ Gg σ 1. The commuttive digrm F g F GF Gg σ = f σ = σ = GF GF f GF σ = GF shows tht GF f = Gg nd so F f = g since G is injective on morphism sets. Conversely, suppose tht F : C D is functor stisfying the two conditions. We must construct functor G in the other direction nd nturl isomorphisms τ : F G = 1 D nd σ : GF = 1 C. By the first condition, for every oject d D, we cn find n oject Gd C nd n isomorphism τ d : F Gd d. By the second condition, C(Gc, Gd) = D(F Gc, F Gd) for ny two ojects c nd d of D. Here, D(c, d) = D(F Gc, F Gd) ecuse F Gc = c nd F Gd = d. Thus we hve D(c, d) = D(F Gc, F Gd) = C(Gc, Gd). This mens tht for every D-morphism g : c d there is exctly one C-morphism Gg : Gc Gd such tht f F Gc τ c = c F Gg F Gd τ d = d g commutes. Now G is functor nd τ nturl isomorphism F G = 1 D. Wht out GF? Well, for ny oject of C, C(GF, ) = D(F GF, F ) τ F so there is unique isomorphism σ : GF such tht F σ = τ F. This gives the nturl isomorphism σ : GF = 1 C. It follows tht when C of morphism sets. F G D is n equivlence of ctegories then there re ijections C(c, Gd) = D(F c, d) C(Gd, c) = C(d, F c) Lemm 5.7. If C, C nd D, D re equivlent, then the functor ctegories Func(C, D) nd Func(C, D ) re equivlent. The full suctegory generted y some of the ojects of C is the ctegory whose ojects re these ojects nd whose morphisms re ll morphisms in C. Exmple 5.8. The ctegory of finite sets is equivlent to the full suctegory generted y ll sections S <n = {x Z + x < n}, n Z +, of Z +. The ctegory of finite dimensionl rel vector spces is equivlent to the ctegory Mt R. If f : X Y is homeomorphism (homotopy equivlence) then the induced morphism π(f): π(x) π(y ) is n isomorphism (equivlence) of ctegories. The fundmentl groupoid of spce is equivlent to the full suctegory generted y point in ech pth component.

17 THE FUNDAMENTAL GROUP AND COVERING SPACES Ctegories of right G-sets Let G e topologicl group nd F nd Y topologicl spces. Definition 6.1. A right ction of G on F is continuous mp F G F : (x, g) x g, such tht x e = x nd x (gh) = (x g) h for ll g, h G nd ll x F. A topologicl spce equipped with right G-ction is clled right G-spce. A continuous mp f : F 1 F 2 etween two right G-spces is G-mp if f(xg) = f(x)g for ll g G nd x F 1. Definition 6.2. A left ction of G on Y is continuous mp G Y Y : (g, y) g y, such tht e y = y nd (gh) y = g (h y) for ll g, h G nd ll y Y. A topologicl spce equipped with left G-ction is clled left G-spce. A continuous mp f : Y 1 Y 2 etween two left G-spces is G-mp if f(gy) = gf(y) for ll g G nd x Y 1. The orit spces (with the quotient topologies) re denoted F/G = {xg x F } for right ction F G F nd G\Y = {Gy y Y } for left ction G Y Y. The orit through the point x F for the right ction F G F is the su-right G-spce xg = {xg g G} otined y hitting x with ll elements of G; the stilizer t x is the sugroup x G = {g G xg = x} of G. The universl property of quotient spces gives commuttive digrm g xg G xg g xgg xgg xg xg\g of right G-spces nd G-mps (Generl Topology, 2.81). Note tht G-mp x G\G xg: x Gg xg is ijective. (In prticulr, the index of the stilizer sugroup t x equls the crdinlity of the orit through x.) In mny cses it is even homeomorphism so tht the orit xg through x nd the coset spce x G\G of the isotropy sugroup t x re homeomorphic. Proposition 6.3 (G-orits s coset spces). Suppose tht F is right G-spce nd x point of F. Then xg\g xg is homeomorphism G g xg xg is quotient mp Proof. Use tht the ijective quotient mp is homeomorphism, the composition of two quotient mps is quotient, nd if the composition of two mps is quotient thn the lst mp is quotient (Generl Topology, 2.77). By definition, G x G\G is quotient. By right (or left) G-set we just men right (or left) G-spce with the discrete topology. In the following we del with G-sets rther thn G-spces. Definition 6.4. GSet is the ctegory of right G-sets nd G-mps. The ojects re right G-sets F nd the morphisms ϕ: F 1 F 2 re G-mps (mening tht ϕ(xg) = ϕ(x)g for ll x F 1 nd g G) Trnsitive right ctions. The right G-set F is trnsitive if F consists of single orit. If F is trnsitive then F = xg for some (hence ny) point x F so tht F nd H\G re isomorphic G-sets where H is the stilizer sugroup t the point x (Proposition 6.3). Thus ny trnsitive right G-set is isomorphic to the G-set H\G of right H-cosets for some sugroup H of G. Definition 6.6. The orit ctegory of G is the full suctegory O G of GSet generted y ll trnsitive right G-sets. The orit ctegory O G of G is equivlent to the full suctegory of GSet generted y ll G-sets of the form H\G for sugroups H of G. Wht re the morphisms in the orit ctegory O G? Definition 6.7. Let H 1 nd H 2 e sugroups of G. The trnsporter is the set of group elements conjugting H 1 into H 2. N G (H 1, H 2 ) = {n G nh 1 n 1 H 2 } The trnsporter set N G (H 1, H 2 ) is left H 2 -set. Let Let H 2 \N G (H 1, H 2 ) e the set of H 2 -orits.

18 18 J.M. MØLLER Proposition 6.8. There is ijection τ : H 2 \N G (H 1, H 2 ) O G (H 1 \G, H 2 \G), τ(h 2 n)(h 1 g) = H 2 ng This mp tkes H 2 n to left multipliction H 1 \G H1g H2ng H 2 \G y H 2 n. In cse H 1 = H = H 2, the mp is group isomorphism. τ : H\N G (H) O G (H\G, H\G), τ(hn)(hg) = Hng Proof. The inverse to τ is the mp tht tkes G-mp H 1 \G ϕ H 2 \G to its vlue ϕ(h 1 ) = H 2 n t H 1 H 1 \G. Since H 2 n = ϕ(h 1 ) = ϕ(h 1 H 1 ) = H 2 nh 1, the group element n conjugtes H 1 into H 2. In cse H 1 = H = H 2 nd n 1, n 2 N G (H), we hve so τ is group homomorphism in this cse. τ(hn 1 )τ(hn 2 )(H) = τ(hn 1 )(Hn 2 ) = Hn 1 n 2 = τ(hn 1 n 2 )(H) In prticulr we see tht ll morphisms in O G re epimorphisms ll endomorphisms in O G re utomorphisms every oject H\G of O G is equipped with left nd right ctions (6.9) H\N G (H) H\G G = O G (H\G, H\G) H\G G H\G: Hn Hg m = Hngm where the left ction re the G-utomorphisms of H\G in O G. the mximl G-orit is G = {e}\g nd O G ({e}\g, H\G) = H\G, the miniml G-orit is = G\G nd O G (H\G, G\G) = (G\G = is the finl oject of O G ) Remrk 6.10 (Isomorphism clsses of ojects of O G ). The set of ojects of O G corresponds to the set of sugroups of G. The set of isomorphism clsses of ojects of O G corresponds to the set of conjugcy clsses of sugroups of G: Two ojects H 1 \G nd H 2 \G of the orit ctegory O G re isomorphic if nd only if H 1 nd H 2 re conjugte: If there exist n inner utomorphism tht tkes H 1 into H 2 nd n inner utomorphism tht tkes H 2 into H 1 such tht the composite mps re the respective identity mps of H 1 \G nd H 2 \G, then these inner utomorphisms must in fct give ijections etween H 1 nd H 2 s the Inn(n 1) Inn(n 2) Inn(n 1) fctoriztions H 1 H 2 H 1 H 2 of the respective identity mps imply tht the inner utomorphism Inn(n 1 ) is ijection. 7. The clssifiction theorem In this section we shll see tht covering mps re determined y their monodromy functor. Definition 7.1. Cov(X) is the ctegory of covering spces over the spce X. The ojects re covering mps Y X nd the morphisms Cov(X)(p 1 : Y 1 X, p 2 : Y 2 X) re continuous mps f : Y 1 Y 2 over X (mening tht f preserves fires or p 1 = p 2 f). How cn we descrie the ctegory Cov(X)? We re going to ssume from now on tht X is pth connected nd loclly pth connected. Let Func(π(X), Set) e the ctegory of functors from the fundmentl groupoid π(x) to the ctegory Set of sets. There is functor Cov(X) Func(π(X), Set) which tkes covering mp p: Y X to its monodromy functor F (p): π(x) Set (2.8) nd covering mp morphism to the induced nturl trnsformtion of functors. Conversely, does ny such functor come from covering spce of X? Suppose tht F : π(x) Set is ny functor. Let Y (F ) = x X F (x) e the union of the fires nd let p(f ): Y (F ) X e the ovious mp tking F (x) to x for ny point x X. Definition 7.2. A spce X is semi-loclly simply connected t the point x X if ny neighorhood of x contins neighorhood U of x such tht ny loop t x in U is contrctile in X. The spce X is semi-loclly simply connected if it is semi-loclly simply connected t ll its points. All loclly simply connected spces re semi-loclly simply connected.

19 THE FUNDAMENTAL GROUP AND COVERING SPACES 19 Lemm 7.3. Suppose tht X is loclly pth connected nd semi-loclly simply connected. Then there is topology on Y (F ) such tht p(f ): Y (F ) X is covering mp. The monodromy functor of p(f ): Y (F ) X is F. Proof. Suppose tht x is point in X nd U X n open pth connected neighorhood of x such tht ny loop in U sed t x is nullhomotopic in X. Oserve tht this implies tht there is unique pth homotopy clss u z from x to ny other point z in U so tht U F (x) p 1 (U): (y, z) F (u z )(y) is ijection. For ech y F (x), let (U, y) Y e the imge of U {y} under the ove ijection. By ssumption, the topologicl spce X hs sis of sets U s ove. The sets (U, y) then form sis for topology on Y. The covering mp Y (F ) X determines fire functor (2.8) from the fundmentl groupoid of X to the ctegory of sets. By construction, this fire functor is F. Definition 7.4. A covering mp p: Y X is universl if Y is simply connected. According to the Lifting Theorem 2.12, ny two universl covering spces over X re isomorphic in the ctegory Cov(X) of covering spces over X. We my therefore spek out the universl covering spce of X. Is there lwys universl covering spce of X? By Corollry 2.9 the fundmentl groupoid of Y (F ) hs the set Y (F ) s oject set nd the morphisms re (7.5) π(y (F ))(y 1, y 2 ) = {u π(x)(x 1, x 2 ) F (u)y 1 = y 2 } for ll points x 1, x 2 X nd y 1 F (x 1 ), y 2 F (x 2 ). In prticulr, let x 0 e se point in X. There is right ction F (x 0 ) π 1 (X, x 0 ) F (x 0 ) nd Y (F ) is pth connected The right ction of π 1 (X, x 0 ) on F (x 0 ) is trnsitive Y (F ) is simply connected The right ction of π 1 (X, x 0 ) on F (x 0 ) is simply trnsitive We cn lwys find functor tht stisfies the lst condition in tht F = π(x)(x 0, ): π(x) Set is functor nd the ction of π 1 (X, x 0 ) on F (x 0 ) = π 1 (X, x 0 ) is simply trnsitive. Corollry 7.6. X dmits simply connected covering spce if nd only if X is semi-loclly simply connected. Proof. The covering spce Y (F ) of the functor F = π(x)(x 0, ) is simply connected. Conversely, suppose tht p: Y X is covering mp nd U X nd evenly covered open suspce then U X fctors through Y X. If π 1 (Y ) is trivil then π 1 (U) π 1 (X) is the trivil homomorphism. Exmple 7.7. The Hwiin Erring n Z + C 1/n nd the infinite product S 1 of circles re connected nd loclly pth connected ut not semi-loclly simply connected. Thus they hve no simply connected covering spces. The infinite join S 1 does hve simply connected covering spce since it is CW-complex. Indeed ny CW-complex or mnifold is loclly contrctile [5, Appendix], in prticulr loclly simply connected. Theorem 7.8 (Clssifiction of Covering Mps). Suppose tht X is semi-loclly simply connected. monodoromy functor nd the functor F Y (F ) Cov(X) Func(π(X), Set) re ctegory isomorphisms. Proof. Let p 1 : Y 1 X nd p 2 : E Y X e covering mps over X with ssocited functors F 1 nd F 2. A covering mp Y 1 f p 1 p 2 X induces nturl trnsformtion τ f : F 1 = F 2 of functors given y τ f (x) = f p 1 1 x: p 1 x 1 p 1 x 2. Conversely, ny nturl trnsformtion τ : F 1 = F 2 induces covering mp Y (f): Y (F 1 ) Y (F 2 ) of the ssocited covering spces. Y 2 The

20 20 J.M. MØLLER For exmple, let F : π(x) Set e ny functor, let x 0 X nd y 0 F (x 0 ). There is nturl trnsformtion π(x)(x 0, ) = F whose x-component is π(x)(x 0, x) F (x): u F (u)y y for ny point x of X. This confirms tht the universl covering spce lies ove them ll. Corollry 7.9. The functor is n equivlence of ctegories. Cov(X) π 1 (X, x 0 )Set: (p: Y X) p 1 (x 0 ) Proof. The inclusion π 1 (X, x 0 ) π(x) of the the full suctegory of π(x) generted y x 0 into π(x) is n equivlence of ctegories. The induced functor Func(π(X), Set) Func(π 1 (X, x 0 ), Set) is then lso n equivlence. But Func(π 1 (X, x 0 ), Set) is simply the ctegory of right π 1 (X, x 0 )-sets. In prticulr, the full suctegory Cov 0 (X) of connected covering spces over X is equivlent to the ctegory of trnsitive right π 1 (X, x 0 )-sets which gin is equivlent to the orit ctegory O π1(x,x 0) (6.6). The set of covering spce morphisms from the connected covering spce p 1 : Y 1 X to the connected covering spce p 2 : Y 2 X is nd, in prticulr, Cov(X)(p 1 : Y 1 X, p 2 : Y 2 X) = Func(π(X), Set)(F (p 1 ), F (p 2 )) = π 1 (X)Set(p 1 1 (x 0), p 1 2 (x 0)) = O π1(x)(π 1 (Y 1 )\π 1 (X), π 1 (Y 2 )\π 1 (X)) = π 1 (Y 2 )\N π1(x)(π 1 (Y 1 ), π 2 (Y 2 )) Cov(X)(p: Y X, p: Y X) = π 1 (Y )\N π1(x)(π 1 (Y )) for ny connected covering spce p: Y X over X. If we mp out of the universl covering spce X 1 X this gives Cov(X)(X 1 X, Y X) = π 1 (Y )\π 1 (X) Cov(X)(X 1 X, X 1 X) = π 1 (X) which mens tht the universl covering spce dmits left covering spce π 1 (X)-ction with orit spce π 1 (X)\X 1 X = X. Corollry Let G = π 1 (X) for short. The functor is n equivlence of ctegories. O G Cov 0 (X): H (H\X 1 G\X 1 ) Is this Σ 3 id C 2 \Σ 3 {e}\σ 3 Σ 3 \Σ 3 id C 3 \Σ 3 C 2 picture of the orit ctegory of symmetric group Σ 3 or is it picture of the pth connected covering spces over pth connected, loclly pth connected, nd semi-loclly simply connected spce with fundmentl group Σ 3? Both! The spce could e X Σ3 from Corollry 4.12; see Exmple 7.20 for more informtion. Here re some exmples to illustrte the Clssifiction of Covering Spces. Covering spces of the circle: The ctegory Cov 0 (S 1 ) = O C of pth connected covering spces of the circle S 1 = Z\R consists of the covering spces nz\r Z\R where n = 0, 1, 2,.... There is

21 THE FUNDAMENTAL GROUP AND COVERING SPACES 21 covering mp nz\r mz\r if nd only if m n nd in tht cse there re m such covering mps, nmely the mps S 1 ζz m/n z n S 1 z m where ζ is ny mth root of unity. Covering spces of projective spces: The ctegory Cov 0 (RP n ) = O C2 of connected covering spces of rel projective n-spce RP n, n 2, hs 2 ojects, nmely the trivil covering mp RP n RP n nd the universl covering mp S n RP n. Covering spces of lense spces: The universl covering spce of the lense spce L 2n+1 (m) = C m \S 2n+1, n 1, is S 2n+1. The other covering spces re the lense spces L 2n+1 (r) = C r \S 2n+1 for eh divisor r of m. The ctegory of connected covering spces of L 2n+1 (m) is equivlent to the orit ctegory O Cm. Covering spces of surfces: The ctegory Cov 0 (M g ) = O π1(m g) is hrder to descrie explicitly. Any finite sheeted covering spce of compct surfce is gin compct surfce. The pper [8] contins informtion out covering spces of closed surfces. Exmple 7.11 (Covering spces of the Möius nd). The cylinder S 1 [ 1, 1] = Z\(R [ 1, 1]) where the ction is given y n (x, t) (x + n, t) The Möius nd MB = Z\(R [ 1, 1]) where the ction is given y n (x, t) (x + n, ( 1) n t) S 1 Every even-sheeted covering spce of the Möuis nd is cylinder, every odd-sheeted covering spce is Möuis nd. Exmple 7.12 (Covering spces of S 1 m (S 1 I)). Let X m = S 1 m (S 1 I) e the mpping cylinder of the degree m mp of the circle. We cn construct X m in the following wy: Tke (codomin) circle of circumference 1/m nd squre [0, 1] [0, 1]. Wrp the ottom edge [0, 1] {0} of the squre m times round the circle in screw motion so tht ech time the squre trvels once round the circle it is lso eing rotted n ngle of 2π/m. Finlly, glue the two ends, {0} [0, 1] nd {1} [0, 1], of the squre together. There is picture of X m in [5, Exmple 1.29]. The codomin circle is the core circle nd the domin circle is the oundry circle. The fundmentl group π 1 (X m ) is Z since X m deformtion retrcts onto the codomin (core) circle so tht the inclusion S 1 i1 X m S 1 is homotopy equivlence. The inclusion S 1 i0 X m S 1 of the domin (oundry) circle induces multipliction y m on the fundmentl groups; this is simply ecuse of the generl mpping cylinder digrm which ecomes (S 1 I) m S 1 = X m S 1 i0 m i 1 S 1 in this specil cse. It my help to envision the oundry circle in X m sliding towrds the core circle. The universl covering spce of X m is X m 1 = CZ/m R where CZ/m = (Z/m I)/(Z/m {1}) is the cone on the set Z/m with m points. (CZ/m is strfish with m rms). We my relize CZ/m R in DR

22 22 J.M. MØLLER Figure 2. The universl covering spce of X m R 3 with CZ/m plced horizontlly in the XY -plne nd R s the verticl Z-xis. The covering spce ction of the unit 1 Z on CZ/m R is then the screw motion ([[] m, t], x) ([[ + 1] m, t], x + 1/m) with mtrix cos(2π/m) sin(2π/m) 0 sin(2π/m) cos(2π/m) /m tht rottes CZ/m counterclockwise 1/mth of full rottion nd moves up long the Z-xis 1/mth of unit. (In Figure 7.12 the R-xis isn t exctly verticl since tht would tke up too much spce. The covering spce ction tkes the indicted lines, situted t distnce 1/m, to ech other.) Wht is the lift of the domin nd the codomin circles of X m to the universl covering spce X m 1? (One of them will lift to loop.) Since m Z cts trivilly on CZ/m there is n m-sheeted covering mp CZ/m S 1 = CZ/m mz\r = mz\(cz/m R) Z\(CZ/m R) = X m with mz\z s deck trnsformtion group. Wht is the lift of the domin nd the codomin circles to this m-fold covering spce? Let X = X 1 X 2 e CW-complex tht is the union of two connected sucomplexes X 1 nd X 2 with connected intersection X 1 X 2. According to vn Kmpen, the fundmentl group G = π 1 (X) = G 1 A G 2 is the free product of G 1 = π 1 (X 1 ) nd G 2 = π 1 (X 2 ) with A = π 1 (X 1 X 2 ) mlgmted. We will ssume tht the homomorphisms G 1 A G 2 re injective. Then lso the homomorphisms G 1 G G 2 of the push-out digrm A G 2 G 1 re injective ccording to the Norml Form Theorem for Free Products with Amlgmtion [6, Thm 2.6]. Let X 1 e the universl covering spce of X = G\X 1 nd let p: X 1 X e the covering projection mp. The spces p 1 (X 1 ) nd p 1 (X 2 ) re left G-spces with intersection p 1 (X 1 ) p 1 (X 1 ) = p 1 (X 1 X 2 ). Let y 0 p 1 (X 1 X 2 ) e se point. The commuttive digrm [3, II.7.5] G π 1 (p 1 X 1, y 0 ) π 1 (X 1, p(y 0 )) π 1 (p 1 X, y 0 ) = {1} π 1 (X, p(y 0 )) tells us tht the component of p 1 (X 1 ) contining y 0 is simply connected so it is the universl covering spce X 1 1 of X 1 = G 1 \X 1 1. We see from this tht there is homeomorphism of left G-spces G G1 X 1 1 p 1 (X 1 ) induced y the mp G X 1 1 p 1 (X 1 ) sending (g, y) to gy. Similr rguments pply to p 1 (X 2 ) nd p 1 (X 1 X 2 ), of course, nd hence X 1 = G G1 X 1 1 G A (X 1 X 2) 1 G G2 X 2 1 is the union of the two G-spces G Gi X i 1, i = 1, 2. This mens tht the universl covering spce of X is the union of the G-trnsltes of the universl covering spces of X 1 nd X 2 joined long G-trnsltes of the universl covering spce of X 1 X 2. The next exmple demonstrtes this principle.

23 THE FUNDAMENTAL GROUP AND COVERING SPACES 23 Exmple [5, 1.24, 1.29, 1.35, 1.44, 3.45] Let X mn = X m S 1 X n e the doule mpping cylinder for the degree m mp nd the degree n mp on the circle. X mn is the union of the two mpping cylinders with their domin (oundry) circles identified, X m X n = S 1. By vn Kmpen, the fundmentl group hs presenttion π 1 (X mn ) = π 1 (X m ) π1(s 1 ) π 1 (X n ) =, m = n = G mn with two genertors nd one reltion. We shll now try to uild its universl covering spce. We my equip X mn with the structure of 2-dimensionl CW-complex. The 1-skeleton of X mn consists of two circles, nd, joined y n intervl, c, nd X mn = Xmn 1 m c n c D2 is otined y ttching 2-cell c Figure 3. 1-skeleton X 1 mn of X mn long the loop m c n c. (If we use the corollry to vn Kmpen [5, 1.26] insted of the vn Kmpen theorem itself we get tht π 1 (X mn ) =, cc m (cc) n.) The universl covering spce X mn 1 is lso 2-dimensionl CW-complex. The inverse imge in X m 1 of the left hlf of the 1-skeleton is the verticl line R with spirling rungs ttched 1/mth of unit prt. Rungs with verticl distnce 1 point in the sme direction so they cn e joined up with the inverse imge in X n 1 of the right hlf of the 1-skeleton. Now fill in 2-cells in ech of the rectngles with sides m, c, m n Figure 4. Prt of 1-skeleton of X mn 1 n nd c. Continue this process. There will e similr rectngles shifted up 1/mth unit long the left xis nd rotted 2π/m or up 1/nth unit long the right xis nd rotted 2π/n. The 2-dimensionl CW-complex X mn 1 uilt in this wy is the universl covering spce; it is the product T mn R of tree T mn nd the rel line, hence contrctile [10, Chp 3, Sec 7, Lemm 1]. The element G mn cts y skew motion round one of the verticl lines in X m 1 nd G mn cts y skew motion round one of the verticl lines in X n 1. Note tht m = n cts y trnslting one unit up. Wht is the lift of X m X n (the circle prllel to circle ut pssing through the point of the 1-skeleton) to the universl covering spce? Wht is the universl elin covering spce G mn\x mn 1 of X mn? Its deck trnsformtion group is G mn\g mn = (G mn ) =, m = n, = = Z Z/d where d = (m, n) is the gretest common divisor. Wht is the mn fold covering spce with fundmentl group equl to the norml closure N of m, 1 1 nd deck trnsformtion group N\G =, m = n, m, = =, m, n, = = Z/m Z/n? Wht is the lift of X m X n to this covering spce? Cyley tles, Cyley grphs, nd Cyley complexes. [6, III.4] [3] For ny group presenttion G = g α r β there exists (Corollry 4.12) 2-dimensionl CW-complex X G\G = D 0 D 2 = (G\G D 0 ) (G\G D 2 ) {g α} D 1 {r β } {g α}(g\g D 1 ) {r β } with fundmentl group π 1 (X G\G ) = g α r β = G. This is the most simple spce with fundmentl group G so it is nturl to pply Theorem 7.8 to X G\G. So wht re the connected covering spces of X G? There is n equivlence of ctegories X? : O G Cov 0 (X G\G ): H\G (X H\G X G\G )

24 24 J.M. MØLLER nd the Cyley complex of H\G is the 2-dimensionl CW-complex X H\G while the Cyley grph is its 1- skeleton. We now define these CW-complexes more explicitly for ny oject of O G (or for ny right G-spce for tht mtter) reltive to the given presenttion of G. The 0-skeleton of X H\G is the right G-set XH\G 0 = H\G; this is the fire of the covering mp X H\G X H\G s right G-spce. The 1-skeleton of X H\G is the Cyley grph for H\G, the 1-dimensionl H\N G (H)- CW-complex XH\G 1 = (H\G D0 ) (H\G D 1 ) {Hg Hgg α} otined from the 0-skeleton H\G y ttching to ech right coset Hg H\G n rrow from Hg to Hgg α for ech genertor g α ; note tht we hve no other choice since the loop g α in the se spce lifts to pth in the totl spce tht goes from Hg in the fire H\G to Hgg α in the fire. (The Cyley grph is simply grphicl presenttion of the Cyles tle for Cyley tle for group multipliction H\G G H\G.) In this wy, the Cyley tle for H\G is G: H -fold covering spce of the 1-skeleton {g α} S1 of X G. The Cyley grph is connected since ech group element g is product of the genertors which mens tht there is sequence of rrows connecting the 0-cells He nd Hg. Next ttch 2-cells t ech Hg H\G long the loop r β for ech reltion r β. Since the reltion r β is fctoriztion of the neutrl element e in terms of the g α, it defines loops Hg Hgr β = Hg sed t ech 0-cell Hg in the Cyley grph X 1 H\G. The resulting left H\N G(H)-CW-complex X H\G = (H\G D 0 ) {Hg Hgg α} (H\G D 1 ) {Hg Hg} rβ (H\G D 2 ) is the Cyley complex of H\G. The Cyley complex is still connected for ttching 2-cells does not lter the set of pth components (Corollry 4.9). Clerly, every G-mp H 1 \G H 2 \G extends to covering mp X H1\GtoX H2\G. In prticulr, tking H = {e} to e the trivil group, the Cyley complex for the right G-set {e}\g = G, X {e}\g = (G D 0 ) {g α}(g D 1 ) {r β } (G D 2 ) is 2-dimensionl left G-CW-complex, the universl covering spce of X G\G. The 0-skeleton is G, t ech g G there is n rrow from g to gg α for ech genertor g α nd 2-cell ttched y the loop g r β gr β = g. In other words, there is one 0-G-cell G D 0, one G-1-cell G D 1 for ech genertor g α, ttched y the left G-mp tht tkes {e} D 1 = {e} {0, 1} to e nd g α, nd one G-2-cell G D 2 for ech reltion r β ttched y the left G-mp tht on {e} D 2 is the loop r β t e. The orit spce under the left ction of H < G on X G ({e}\g) is the Cyley complex for the orit spce H\G: H\X {e}\g = X H\G. In prticulr, G\X {e}\g = X G\G = D 0 {g α} D 1 {r β } D 2 = X G is point {Ge} with n rrow Ge gα Ge for ech genertor g α nd with one 2-cell ttched long the loop Ge r β Ge for ech reltion r β. It is very instructive to do few exmples. See [4] for informtion out grph theory. Exmple 7.15 (Cyley complexes for cyclic groups). For the infinite cyclic group G = C =, X {e}\g = Z (Z D 1 ) = R nd X G\G = G\R = S 1. For the cyclic group G = C 2 = 2 of order 2, X {e}\g = (C 2 D 0 ) (C 2 D 1 ) (C 2 D 2 ) = S 2 nd X G\G = G\S 2 = RP 2. For the cyclic group G = C m = m of order m, X {e}\g is circle with C m D 2 ttched nd nd X G\G = G\X {e}\g is the mpping cone for S 1 m S 1. Exmple 7.16 (Cyley grphs for F 2 -sets). Let G =, = Z Z e free group F 2 on two genertors. Then X G\G = S 1 S 1 nd O G = Cov 0 (S 1 S 1 ). Since there re no reltions, Cyley complexes for right

25 THE FUNDAMENTAL GROUP AND COVERING SPACES 25 G-sets re Cyley grphs. In prticulr, X {e}\g = (G D 0 ) (G D 1 G D 1 ) is the G-grph g g 1 g g with vertex set G nd two edges from g to g nd g for every vertex g G, nd X G\G = G\X {e}\g is the grph, S 1 S 1, with one vertex G\G nd two edges. In generl, for ny sugroup H of G, the Cyley grph, X H\G, for H\G is the covering spce of X G\G = S 1 S 1 chrcterized y ny of these three properties: X H\G is the Cyley tle for H\G G H\G reltive to the genertors nd the fire of X H\G X G\G is the right G-set H\G the imge of the monomorphism π 1 (X H\G ) π 1 (X G\G ) = G is conjugte to H Here re some exmples: g If H = 2,, 2 then the Cyley tle nd the Cyley grph of the G-set H\G = {He, H} re He H H H He He He H ecuse He H, He H = H 2 = H, H H = He, nd H H = He. The sugroup H is norml since it hs index two. Note tht H is free of rnk 3. If H = [G, G] is the commuttor sugroup of G then the Cyley grph gives tiling of the the plne y squres with edges leled 1 1. If H = G 2 is the smllest sugroup contining ll squres in G, the right cosets re H\G = {He, H, H, H} nd the Cyley grph is the grph of the Cyley tle. If H is the smllest norml sugroup contining 3, 3, nd () 3, then the Cyley grph gives tiling of the plne y hexgons, with edges, nd tringles with edges or. Oserve tht Hx 3 y = Hx 3 x 1 x 3 y = Hxy. It is, in generl, difficult prolem to enumerte the cosets of H in G. Exercise [5, (3) p 58] Let G =<, > e the free group on two genertors nd H = 2, 2, 1, 1. Drw the Cyley grph for H\G with the help of the informtion provided y this mgm session: > G<,>:=FreeGroup(2); > H:=su<G ^2, ^2, **^-1, **^-1>; > Index(G,H); 3 > T,f:=RightTrnsversl(G,H); > T; {@ //The vertices of the Cyley grph > E:={@ <v,(v*)@f,(v*)@f> : v in > E; {@ <Id(G),, >, <, Id(G), >, <,, //The edges > Exercise Let G e free group of finite rnk nd H sugroup of G. Show tht H is free nd tht G: H (rk(g) 1) = rk(h) 1. (This exercise is most esily solved y using the Euler chrcteristic.) Exmple When G = C m = g g m is the cyclic group of order m > 0, the Cyley complex X G ({e}\g) = (G D 0 ) (G D 1 ) (G D 2 ) is the universl covering spce of the mpping cone for the degree m mp on the circle. It is the left G- CW-complex consisting of circle with m 2-discs ttched. (When m = 2, this is the 2-sphere which is the

26 26 J.M. MØLLER Figure 5. H\X {e}\g for H = () 3,, 2, 2 = G universl covering spce of the mpping cone RP 2 for the degree 2-mp of the circle.) Wht is the covering spce ction of G on X G ({e}\g)? Exmple [5, Exmple 1.48, Exercise ] Let G =, 2, 2 = Z/2 Z/2 4.2 = Z Z/2 e the free product of Z/2 with itself. Then X G\G = RP 2 RP 2 nd Cov 0 (RP 2 RP 2 ) = O C2 C 2. The totl spce X {e}\g of its universl covering spce X {e}\g X G\G is n infinite string of S 2 s. Indeed, the 0-skeleton is G, the 1-skeleton otined y ttching two 1-discs to ech 0-cell, is e nd the 2-skeleton is otined y ttching two 2-discs t ech 0-cell long the mps 2 nd 2. The left ction of G which swps e,, etc is the ntipodl mp on the sphere contining e nd. The sugroup H = () 3 = 3Z Z is norml in G so tht the orit set H\G = {He, H, H, H, H, H} = 3Z\Z 2Z\Z is ctully group; it is the dihedrl group of order 6, isomorphic to Σ 3. The quotient spce H\X {e}\g = X H\G is necklce of six S 2 s formed from the 1-skeleton He H H H H H y ttching 2-discs t ech vertex long the loops 2 nd 2. The fundmentl group of H\ X G is H nd the deck trnsformtion group is H\N G (H) = H\G since H is norml. The dshed rrows show the covering spce left ction y H H\G; the orit spce for this ction is the Cyley complex of the next exmple. The element H H\G cts y rotting the grph two plces in clockwise direction. For nother exmple, tke H = () 3, = 3Z Z/2; H is not norml for N G (H) = H, H\G = {He, H, H} hs 3 elements, nd He H H is the Cyley grph for H\G. The Cyley complex, otined y ttching six 2-discs long the mps 2 nd 2 t ech vertex, is RP 2, S 2, S 2, RP 2 on string s shown ove. The deck trnsformtion group H\N G (H) = H\H is trivil.

27 THE FUNDAMENTAL GROUP AND COVERING SPACES 27 Exmple Let G = Z/2 Z/3 4.2 = PSL(2, Z) e the free product of cyclic group of order two nd cyclic group of order three. This grph e is the eginning of the Cyley complex for G. Descrie the left G-CW-complex X G (G)! Norml covering mps. Let p: Y X e covering mp etween pth connected spces. Definition The covering mp p: Y X is norml if the group Cov(X)(Y, Y ) of deck trnsformtions cts trnsitively on the fire p 1 (x) over some point of X. If the ction is trnsitive t some point, then it is trnsitive t ll points. Why re these covering mps clled norml covering mps? Corollry Let X e pth connected, loclly pth connected nd semi-loclly simply connected spce nd p: Y X covering mp with Y pth connected. Then The covering mp p: Y X is norml The sugroup π 1 (Y ) is norml in π 1 (X) Proof. The ction H\N G (H) H\G H\G of the group of covering mps on the fire is trnsitive iff nd only if H is norml in G. All doule covering mps re norml since ll index two sugroups re norml Sections in covering mps. A section of covering p: E X is (continuous) mp s: X E such tht s(x) lies ove x, ps(x) = x, for ll x X. In other words, section is lift of the identity mp of the se spce. Ech section trces out copy of the se spce in the totl spce (nd tht is why it is clled section). Lemm Let p: E X e covering spce over connected, loclly pth connected nd semi-loclly simply connected se spce X. Then the evlution mp s s(x) {sections of p: E X} p 1 (x) π1(x,x) is ijection. Proof. Since X is connected, sections re determined y their vlue t single point (2.12), so the mp is injective. It is lso surjective ecuse ny π 1 (X, x)-invrint point corresponds (under the clssifiction of covering spces over X) to the trivil covering mp X X which oviously hs section. In fct, E contins the trivil covering p 1 (x) π1(x,x) X s sucovering. If either of Y 1 X or Y 2 X is norml, then { π 1 (Y 2 )\π 1 (X) π 1 (Y 2 ) π 1 (Y 1 ) Cov(X)(Y 1, Y 2 ) = otherwise for the trnsporter N π1(x)(π 1 (Y 1 ), π 1 (Y 2 )) equls π 1 (X) if π 1 (Y 1 ) π 1 (Y 2 ) nd otherwise. 8. Universl covering spces of topologicl groups Suppose tht G is connected, loclly pth connected, nd semi-loclly simply connected topologicl group (for instnce, connected Lie group) nd let G 1 e the universl covering spce (7.4) of G. We cn use the group multipliction in G to define multipliction in G 1 simply y letting the product [γ] [η] of two homotopy clsses of pths [γ], [η] G 1 equl the homotopy clss [γ η] G 1 of the product pth (γ η)(t) = γ(t) η(t) whose vlue t ny time t is the product of the vlues γ(t) G nd η(t) G.

28 28 J.M. MØLLER Lemm 8.1. G 1 is topologicl group nd G 1 G is morphism of topologicl groups whose kernel is the sugroup {[ω] ω(0) = ω(1)} = π 1 (G, e) of homotopy clsses of loops sed t the unit e G. The set π 1 (G, e) is here equipped with the group structure it inherits from G 1 where multipliction of pths is induced from group multipliction in G. However, we hve lso defined group structure on π 1 (G, e) using composition of loops. It turns out tht these two structures re identicl. Lemm 8.2. Let ω 1 nd ω 2 e two loops in G sed t the unit element e. Then the loops ω 1 ω 2 (group multipliction) nd ω 1 ω 2 (loop composition) re homotopic loops. Proof. There is homotopy commuttive digrm S 1 S 1 S 1 ω 1 ω 2 G G S 1 S 1 G G ω 1 ω 2 G where is the digonl nd the folding mp. The loop defined y the top edge from S 1 to G is the composite loop ω 1 ω 2 nd the loop defined y the ottom edge is the product loop ω 1 ω 2. One cn lso show tht in this sitution π 1 (G, e) must e elin. Let H = R1 Ri Rj Rk e the quternion lger where the rules i 2 = j 2 = k 2 = 1, ij = k = ji, jk = i = kj, ki = j = ki define the multipliction. Let Sp(1) denote the topologicl group of quternions of norm 1. Sp(1) cts in norm preserving wy on the rel vector spce H = R 4 y the rule α v = αvα 1 for ll α Sp(1) nd v R 4 = H. This give homomorphism π : Sp(1) SO(4). Since R1 is invrint under this ction, it tkes R = Ri Rj Rk = R 3 to itself, so there is lso group homomorphism π : Sp(1) SO(3) [2, I.6.18, p 88]. The kernel is R Sp(1) = {±1}. Convince yourself tht π is surjective (see the computtion elow nd recll tht n element of SO(3) is rottion round fixed line), so tht π : Sp(1) SO(3) = {±1}\Sp(1) is doule covering spce. Let ( ) cos θ sin θ R(θ) = sin θ cos θ e the mtrix for rottion through ngle θ. Lemm 8.3. The mp π : Sp(1) SO(3) is the universl covering mp of SO(3). The fundmentl group π 1 (SO(3), E) = {±1} is generted y the loop ( ) R(2πt) 0 ω(t) = πα(t) =, 0 t 1, 0 1 Proof. The topologicl spce Sp(1) = S 3 is simply connected, so Sp(1) SO(3) is the universl covering spce of SO(3). (We hve seen this doule covering efore: It is the doule covering S 3 RP 3.) The fundmentl group π 1 (SO(3), E) = C 2 is generted y the imge loop ω(t) = πα(t) of pth α(t) in Sp(1) from +1 to 1. If we tke then the imge in SO(3) is the loop This follows from the computtion α(t) = cos(πt) + sin(πt)k, 0 t 1, ω(t) = πα(t) = α(t)iα(t) 1 = (cos(πt) + k sin(πt))i((cos(πt) k sin(πt)) ( ) R(2πt) 0, 0 t 1, 0 1 = cos 2 (πt)i + cos(πt) sin(πt)j + cos(πt) sin(πt)j sin 2 (πt)i = cos(2πt)i + sin(2πt)j nd similrly for α(t)jα(t) 1 = sin(2πt)i + cos(2πt)j nd α(t)kα(t) 1 = k.

29 THE FUNDAMENTAL GROUP AND COVERING SPACES 29 It is lso known tht the inclusion SO(3) SO(n) induces n isomorphism on π 1 for n 3. We conclude tht the fundmentl group π 1 (SO(n), E) hs order two for ll n 3 nd tht it is generted y the loop ω(t) in SO(n). Thus the topologicl groups SO(n), n 3, hve universl doule covering spces tht re topologicl groups. Definition 8.4. For n 3, Spin(n) = SO(n) 1 is the universl covering spce of SO(n) nd π : Spin(n) SO(n) is the universl covering mp. The elements of Spin(n) re homotopy clsses of pths in SO(n) strting t E nd, in prticulr, Spin(3) = Sp(1). The kernel of the homomorphism π : Spin(n) SO(n) is {e, z} where e is the unit element nd z = [ω] is the homotopy clss of the loop ω. Proposition 8.5. The center of Spin(n) is C 2 = {e, z} n odd Z(Spin(n)) = C 2 C 2 = {e, x} {e, z} n 0 mod 4 C 4 = {e, x, x 2, x 3 } n 2 mod 4 for n 3. Proof. From Lie group theory we know tht the center of Spin(n) is the inverse imge of the center of SO(n). Thus the center of Spin(n) hs order 2 when n is odd nd order 4 when n is even. Suppose tht n = 2m is even. Then Z(Spin(2m)) = {e, z, x, zx} where x = [η] is the homotopy clss of the pth η(t) = dig(r(πt),..., R(πt)) from E to E. Note tht x 2 is (8.2) represented y the loop η(t) 2 = dig(r(2πt),..., R(2πt)) Conjugtion with permuttion mtrix from SO(2n) tkes ω(t) = dig(r(2πt), E, E,..., E) to dig(e, R(2πt), E,..., E) nd since inner utomorphisms re sed homotopic to identity mps, oth the ove loops represent the generting loop ω. It follows tht { x 2 = [η(t) 2 ] = [ω(t) m ] = z m e m even = z m odd Thus Z(Spin(2m)) = {z} {x} = C 2 C 2 if m is even nd Z(Spin(2m)) = {x} = C 4 if m is odd. Wht is the fundmentl group π 1 (PSO(2n)) of the topologicl group PSO(2n) = SO(2n)/ E? When will two digonl mtrices in SO(n) commute in Spin(n)? Let D = {dig(±1,..., ±1} e the elin sugroup of digonl mtrices in SO(n). By computing commuttors nd squres in Spin(n) we otin functions [, ]: D D {e, z}, q : D {e, z} given y q(d) = ( d) 2 nd [d 1, d 2 ] = [ d 1, d 2 ] where π( d) = d, π( d 1 ) = d 1, π( d 2 ) = d 2. They re relted y formul q(d 1 + d 2 ) = q(d 1 ) + q(d 2 ) + [d 1, d 2 ] which sys tht [, ] records the devition from q eing group homomorphism (using dditive nottion here). It suffices to compute q in order to nswer the question out commuttivity reltions. Proposition 8.6. q(d) = e iff the numer of negtive entries in the digonl mtrix d D is divisile y 4. [d 1, d 2 ] = e iff the numer of entries tht re negtive in oth d 1 nd d 2 is even. Proof. Note tht two elements of D re conjugte iff they hve the sme numer of negtive entries. Use permuttion mtrices nd, if necessry, the mtrix dig( 1, 1,..., 1). Consider for instnce d 1 = dig( 1, 1, 1,..., 1), d 2 = dig( 1, 1, 1, 1, 1,..., 1) with two, respectively four, negtive entries. The pths d 1 (t) = dig(r(πt), 1,..., 1), d2 (t) = dig(r(πt), R(πt), 1,..., 1)

30 30 J.M. MØLLER represent lifts of d 1 nd d 2 to Spin(n). Then ( d 1 ) 2 = z = q(d 1 ) nd ( d 2 ) 2 = e = q(d 2 ). Computtions like these prove the formul for q nd the formul for [, ] follows. The numer of negtive entries in d 1 + d 2 is the numer of negtive entries in d 1 plus the numer of negtive entries in d 2 minus twice the numer of entries tht re negtive in oth d 1 nd d 2. Exercise 8.7. Let D n Spin(n) e the inverse imge of D SO(n). How mny elements of order 4 re there in D n? Cn you identify the group D n? Show tht there is homomorphism SU(m) Spin(2m). When m is even, wht is the imge of E SU(m)? Wht is the imge of the center of SU(m)? Descrie the covering spces of U(n). The inclusions SO(n) SO(n + 1), n > 2, nd SO(m) SO(n) SO(m + n), m, n > 2, of specil orthogonl groups lift to inclusions Spin(n) Spin(n + 1) Spin(m) (z1,z 2) Spin(n) Spin(m + n) SO(n) SO(n + 1) SO(m) SO(n) SO(m + n) of doule coverings. (Here, Spin(m) (z1,z 2) Spin(n) stnds for (z 1, z 2 ) \(Spin(m) Spin(n))). The inclusion U(n) SO(2n), tht comes from the identifiction C n = R 2n, lifts to n inclusion of doule covering spces s shown in the following digrms. SU(n) Ck U(1) Spin(2n) SU(n) C n U(1) Spin(2n) (A,z) (A,z) SU(n) Cn U(1) = U(n) SO(2n) (A,z) (A,z 2 ) SU(n) Cn U(1) = U(n) SO(2n) To the left, n = 2k is even, nd to the right, n = 2k + 1 is odd; C n = {(ζe, ζ 1 ) ζ n = 1} nd C n = {(ζe, ζ k ) ζ n = 1} re cyclic groups of order n nd C k = {(ζe, ζ 1 ) ζ k = 1} C 2k = C n is cyclic of order k. The isomorphism SU(n) Cn U(1) U(n) tkes (A, z) to za. When n is divisile y 4, z = ( E, 1) nd x = (E, 1) hve order two; when n is even nd not divisile y 4, x = (E, i) hs order four nd x 2 = (E, 1) = z. This explins the computtion of the center of Spin(2n). (Is the group in the upper left corner of the right digrm isomorphic to U(n)? See [1] for more informtion.) There is doule covering mp pin(n) O(n) otined s the pullck of Spin(2n) SO(2n) long the inclusion homomorphism O(n) SO(2n). Exmple 8.8. The inclusion U(2) SO(4) lifts to n inclusion SU(2) U(1) Spin(4). Let G 16 SU(2) U(1) Spin(4) e the group (( ) ) (( ) ) i 0 0 i G 16 =, i,, i 0 i i 0 G 16 hs order 16, center Z(G 16 ) = {x, z} = C 2 C 2 = Z(Spin(4)), nd derived group [G 16, G 16 ] = {xz} = C 2. Its imge under the covering mps Spin(4) Spin(4)/ z = SO(4) Spin(4)/ xz Spin(4)/ x, z = PSO(4) Spin(4)/ x = SSpin(4) is dihedrl D 8 in SO(4), elin C 4 C 2 in Spin(4)/ xz, quternion Q 8 in the semi-spin group SSpin(4) = Spin(4)/ x, nd elementry elin C 2 C 2 in PSO(4). (All proper sugroups of G 16 re elin ut itself nd some of its quotient groups re nonelin.)

31 THE FUNDAMENTAL GROUP AND COVERING SPACES 31 Exmple 8.9. There exists covering spce homomorphisms of topologicl groups U(1) SU(n) U(n): (z, A) za with kernel C n = {(z, z 1 E) z n = 1} = (ζ, ζ 1 E) where ζ n = e 2πi/n. The universl covering spce homomorphism is R SU(n) U(n): (t, A) ζna t with kernel C = (1, ζn 1 E). Any covering spce of U(n) is of the form (k, ζn k ) \(R SU(n)) for some integer k 0. Similrly, let S(U(m) U(n)) denote the closed topologicl sugroup (U(m) U(n)) SU(m + n) of U(m + n). There exists covering spce homomorphisms of topologicl groups U(1) U(m) U(n) S(U(m) U(n)): (z, A, B) dig(z n1 A, z m1 B) with kernel C lcm(m,n) = {(z, z n1 E, z m1 E) z lcm(m,n) = 1} = (ζ lcm(m,n), ζm 1, ζ n ) where m 1 = m/ gcd(m, n) = lcm(m, n)/n nd n 1 = n/ gcd(m, n) = lcm(m, n)/m. The universl covering spce homomorphism of S(U(m) U(n)) is R SU(m) SU(n) S(U(m) U(n)): (t, A, B) (ζma, t ζnb) t with kernel C = (1, ζ 1 m, ζ n ). Any covering spce of S(U(m) U(n) is of the form (k, ζm k, ζn) k \(R SU(m) SU(n)) for some integer k 0. All finite covering spces of U(n) re covered y U(1) SU(n). To see this, let n nd k e integers nd put k 1 = k/ gcd(n, k). Then there is commuttive digrm (z,a) (z k 1,A) of covering spce homomorphisms. U(1) SU(n) R (k,ζ k ) SU(n) U(1) SU(n) (ζ t n,a) (t,a) R (1,ζ 1 ) SU(n) = U(n) References [1] P.F. Bum, Locl isomorphism of compct connected Lie groups, Pcific J. Mth. 22 (1967), [2] Theodor Bröcker nd Tmmo tom Dieck, Representtions of compct Lie groups, Grdute Texts in Mthemtics, vol. 98, Springer-Verlg, New York, MR 86i:22023 [3] Kenneth S. Brown, Cohomology of groups, Grdute Texts in Mthemtics, vol. 87, Springer-Verlg, New York, MR 83k:20002 [4] Reinhrd Diestel, Grph theory, third ed., Grdute Texts in Mthemtics, vol. 173, Springer-Verlg, Berlin, MR MR (2006e:05001) [5] Allen Htcher, Algeric topology, Cmridge University Press, Cmridge, MR 2002k:55001 [6] Roger C. Lyndon nd Pul E. Schupp, Comintoril group theory, Springer-Verlg, Berlin, 1977, Ergenisse der Mthemtik und ihrer Grenzgeiete, Bnd 89. MR 58 #28182 [7] Sunders Mc Lne, Ctegories for the working mthemticin, second ed., Grdute Texts in Mthemtics, vol. 5, Springer- Verlg, New York, MR 2001j:18001 [8] A. D. Mednykh, On the numer of sugroups in the fundmentl group of closed surfce, Comm. Alger 16 (1988), no. 10, MR 90:20076 [9] Derek J. S. Roinson, A course in the theory of groups, second ed., Springer-Verlg, New York, MR 96f:20001 [10] Edwin H. Spnier, Algeric topology, Springer-Verlg, New York, 1981, Corrected reprint. MR 83i:55001 Mtemtisk Institut, Universitetsprken 5, DK 2100 Køenhvn E-mil ddress: [email protected] URL: