THE FUNDAMENTAL GROUP AND COVERING SPACES


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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 Gsets 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 endpoints {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 welldefined. 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 pthcomponent of X, there is no reltion t ll etween the fundmentl groups. But if the two se points lie in the sme pthcomponent, 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 nspce. 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 nsphere 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 nspce 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 selfmp 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 fixedpoint 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.
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