1 Introduction. 2 Fundamental Group

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1 1 Introduction When are two spaces X, Y homeomorphic, that is, when is there a continuous map f : X Y with continuous inverse? Let f, g : X Y be continuous maps f is homotopic to g if there exists a homotopy, H : X I Y such that H(x, 0) = f(x) and H(x, 1) = g(x) Put H t : X Y by H t (x) = H(x, t), then H 0 = f, H 1 = g Notation: f, g, f t with f 0 = f, f 1 = g from Hatcher If f is homotopic to g, then we write f g Similarly for spaces Let X, Y be two spaces They have the same homotopy type if there are continuous f : X Y and g : Y X such that g f : X X and f g : Y Y are homotopic to the appropriate identity maps Examples: In R n, we have B n = D n = {x R n : x 1} B n has the homotopy type of a point, for example, {0} Let i : {0} B n be the inclusion, and r : B n {0} by r(x) = 0 Then r i = id {0} and i r : B n B n is homotopic to id B n by H : B n I I by H : B n I B n, H(x, t) = (1 t)x But for n > 1, S n 1 = {x R n : x = 1} are not homotopy equivalent to a point 2 Fundamental Group Definition 21 (Path) Let X be a topological space A path in X is f : I X a continuous map Two paths from x 0 to x 1 are homotopic if there exists H : I I S such that H(s, 0) = f(s), H(s, 1) = g(s), H(0, t) = x 0 and H(1, t) = x 1 In Euclidean space (in fact, any convex subspace of Euclidean space) every path is homotopic to a straight line path, by a straight line homotopy Let f be a path from x 0 to x 1 and g be a path { from x 1 to x 2, then f g is a f(2t) 0 t 1/2 new path from x 0 to x 2 defined by (f g)t) = g(2t 1) 1/2 t 1 If x 0 X, then let c x0 : I X be the constant path c x0 (t) = x 0 Easy: If f 0 f 1 from x 0 to x 1 and g 0 g 1 from x 1 to x 2, then f 0 g 0 f 1 g 1 Definition 22 (Loop) Fix x 0 X and call x 0 the base point A loop at x 0 is a path f in X such that f(0) = f(1) = x 0 Let [f] be all loops at x 0 which are path homotopic to f We denote this relation by f p g Definition 23 We define π 1 (X, x 0 ) = {[f] : f is a loop at x 0 } (We can also define π 0 (X, x 0 ) = all path components of X with base point being path component of x 0 ) Example: O(2) has two path components, the base point is x 0 = I What is π 1 (O(2), I)? SO(2) is the set of all rotations of R 2, and it is homeomorphic to the circle We claim that π 1 (X, x 0 ) is a group, and call it the Fundamental Group of X at x 0 1

2 Theorem 21 π 1 (X, x 0 ) with composition as above forms a group Proof [f][g] = [f g] is the product Take the identity to be [c x0 ] For [f] 1, we take [ f] where f = f(1 t) We must check (f g) g p f (g h), f c x0 p f p c x0 f and f f p f f p c x0 We can do these using reparameterizations If f : I X is any path, and γ : I I is continuous with γ(0) = 0 and γ(1) = 1, then f(γ(t)) is a path which is homotopic to f F (s, t) = f((1 t)γ(s) + ts) Now, see page 27 Lemma 22 (Pasting Lemma) Let X be a topological space such that X = A 1 A l, each A i is closed Let f : X Y be a function to another topological space If i, f Ai : A i Y is continuous, then so is f Proof Let C be a closed set in Y Then f 1 (C) = f 1 A 1 (C) f 1 A l (C) Each term in this union is closed in the appropriate A i, and as A i is closed in X, each term is itself closed in X As the finite union of closed sets is closed, f 1 (C) is closed in X, and so f is continuous Let f, g, h be paths in X such that (f g) h is defined We want to show that this composition is associative That is, that (f g) h p f (g h) The right hand side is also equal to [(f g) h] γ, where 1 2 t 0 t 1 2 γ = t t t 1 4 t 1 f( 4s t+1 ) 0 s (t + 1)/4 And so, we define H(s, t) = g(4s t 1) (t + 1)/4 s (t + 2)/t h( 4s 2 t 2 t ) (t + 2)/4 s 1 Change of Base Point Let X be a topological space, and x 0, x 1 X such that there is a path h from x 0 to x 1 Then β h : π 1 (X, x 1 ) π 1 (X, x 0 ) given by β h ([f]) = [h f h] which is an isomorphism of groups And so, if X is path-connected, we can write π 1 (X) Definition 24 (Simply-Connected) Call X simply connected if X is path connected and π 1 (X) is trivial Quotient Topology I = [0, 1], and we want to identify 0 1 So I/ is a space, and we believe it is homeomorphic ( ) the circle S 1 C So we define p : I S 1 such that p(t) = e 2πit p(0) = p(1) = 1 S 1 C Definition 25 (Quotient Topology) More generally, let X be a topological space and be an equivalence relation Let g : X X/ take x to [x] Call V X/ open iff g 1 (V ) is open in X Then we get a topology on X/ and we get that 2

3 X g f Y h X/ Proposition 23 π 1 (S 1, 1) Z Proof We want to define a homomorphism Φ : Z π 1 (S 1 ) be Φ(n) = [ω n ] where ω n (s) = e 2πins We need to show that this is a homomorphism Let p : R S 1 by p(s) = e 2πis Note that ω n = p ω n, where ω n : I R is given by ω n (s) = ns R ω n ω n p I S 1 Φ(n) = [p f] whenever f is any path in R from 0 to n Then ω n must be homotopic to f And so, Φ is a homomorphism Let τ m : R R be a translation, τ m : x x + m Then ω m τ m ω n is a path in R from 0 (to m) to n + m Thus Φ(n + m) = [p ( ω m τ m ω n ] = [p ω m ] [p(τ m ω n )] = Φ(m) Φ(n) We need to show that Φ is surjective and that ker Φ = {0} We will need two facts 1 If f : I S 1 starting at x 0 X and let x 0 p 1 (x 0 ), then unique lift f : I R starting at x 0 This implies that Φ is surjective, as we let [f] π 1 (S 1, 1) and taking x 0 = 1, x 0 = 0, we apply it and get f Then p( f(1)) = f(1) = 1 So f(1) = m Z, e 2πi f(1) = 1 so Φ(m) = [p f] = [f] 2 Let F : I I S 1 be a homotopy of paths, all starting at x 0 S 1 Choose some x 0 p 1 (x 0 ) Then, a unique lifted homotopy F : I I R of paths starting at x 0 F (R, x 0 ) p F I I (S 1, x 0 ) We note that p F = F, and so p f t = f t The second fact implies that Φ is injective Suppose that Φ(n) = e = [c 1 ] (S 1, 1) Assume that n 0 Then [ω n ] = e = [c 1 ] Then ω n is path homotopic to the constant map, and so, by the second fact, we can lift this homotopy 3

4 to ω n p ω 0 = c 0, but these have different endpoints, and so cannot be path homotopic, which is a contradiction All that remains is to prove these two facts Observe the property of p : R S 1 : there exists an open cover {U α } of S 1 such that p 1 (U α ) = i V αi where the V αi are disjoint open sets and p Vαi : V αi U α is a homeomorphism One such cover is the covering of S 1 by two open sets with appropriate overlap We will first prove fact 1: (R, x 0 ) f p f (I, 0) (S 1, x 0 ) The Lebesgue Covering Lemma states that if X is a compact metric space and {W α } is an open cover of X, then there exists λ > 0 such that every subset of X with diameter less than λ lies in some W α So, there exists a partition 0 = t 0 < t 1 <, t l = 1 such that f([t i 1, t i ]) some U α We construct a lifting f on [0, t k ] by induction on k We start with t 0, [0, 0], f(0) = x0 Assume k < l, have f such that p f = f We need to define f on [t k, t k+1 ] We know that p f(t k ) = f(t k ), f(t k ) U α for some α, so f(t k ) V αi for some unique i Put f(t) for t k t t k+1, equal to p Vαi f(t) V α,i R, so existence is done Now we prove uniqueness If p : X X is a covering projection, that is, p is onto and if given a continuous map X, x 0 h, g p f Y, y 0 X, x 0 and if g, h : (Y, y 0 ) ( X, x 0 ) are two lifting maps of f, then g = h provided that Y is connected, Y is locally connected We prove this by putting A = {y Y : g(y) = h(y)} so that y 0 A Put B = {y Y : g(y) h(y)} Both A and B are open We now must prove fact 2 R, x 0 F F p I I, (0, 0) S 1, x 0 Proof by picture: Divide the unit square into a smaller grid F (lower left square) U α, and so 4

5 F (segment I) V αi, and you can extend the map by continuing it from square to adjacent square Or see Hatcher Theorem 24 (Brouwer Fixed Point Theorem) Any continuous f : D 2 D 2 has a fixed point Usually this is proved with the no retraction theorem, though can be proved the other way The no retraction theorem states that there is no continuous r : D 2 S 1 such that r(x) = x for x S 1 Theorem 25 (Borsuk-Ulam Theorem) If f : S n R n is continuous, then there exists x S n such that f(x) = f( x) For n = 1, intermediate value theorem proves it For n = 2, use the fundamental group, see page 32 For n 2, need homology and transfer or cohomology Look at page 38, exercise 9 π 1 as a functor (X, x 0 ) gets π 1 (X, x 0 ) the fundamental group Also, (X, x 0 ) φ (Y, y 0 ) induces a homomorphism of groups π 1 (X, x 0 ) φ π 1 (Y, y 0 ) define by φ ([f]) = [φ f] Note φ = id X φ = id π1(x,x 0) and (X, x 0 ) φ (Y, y 0 ) ψ (Z, z 0 ) gives ψ φ : π 1 (X, x 0 ) π 1 (Z, z 0 ) equal to (ψ φ) Proposition 26 π 1 (X Y ) π 1 (X) π 1 (Y ), and the isomorphism is obtained by the induced maps from the projection maps of X Y to X, Y π 1 (S 1 S }{{} 1 ) Z n n times S 1 S 1 is two circles intersecting at a point Then π 1 (S 1 S 1 ) = F 2 We will need the Seifert-Van Kampen theorem If we take C \ { 1, 1}, we see that it retracts onto S 1 S 1 Proposition 27 π 1 S n = 0 for n > 1 Corollary 28 π 1 (R n \ {0}) = 0 if n > 2 S n 1 R n \ {0} is a homotopy equivalence by r(x) = tx + (1 t) x x x x, and r t(x) = Definition 26 (Retraction) If A X, then a retraction of X onto A is r : X A such that r(a) = a for all a A If i : A X be the inclusion, then r i = id A Choose x 0 A X Then π 1 (A, x 0 ) i π 1 (X, x 0 ) and r in the other direction, we now have that r i = id π1(a) And so, i is injective no retraction r : D 2 S 1 Why? If there were one, then there would be a homomorphism π 1 (S 1 ) π 1 (D 2 ) which is injective, that is, an injective map Z 0, which is impossible Notation: X Y is used to denote homeomorphic, X Y is homotopic If f g : (X, x 0 ) (Y, y 0 ) then f = g : π 1 (X, x 0 ) π 1 (Y, y 0 ) 5

6 Corollary 29 If f : (X, x 0 ) (Y, y 0 ) is a homotopy equivalence of spaces with base points, then f : π 1 (X, x 0 ) π 1 (Y, y 0 ) is an isomorphism The reason is that g : (Y, y 0 ) (X, x 0 ) such that g f id X, f g id Y as based maps, and so π 1 (X, x 0 ) π ( Y, y 0 ) has an inverse homomorphism, g, and so they are inverse isomorphisms Definition 27 (Deformation Retraction) Suppose A X is a subspace A deformation retraction of X onto A is a homotopy H : X I X such that for all x X, H(x, 0) = x, H(x, 1) A and H(a, t) = a for all t if a A Then i : A X is a homotopy equivalence, and r = H(x, 1) : X A is a homotopy inverse Note S n R n+1 \ {0} shows, by x tx + (1 t) x x that Sn R n+1 \ {0} Proposition 210 (Prop 114) π 1 (S n ) = 0 for n 2 Proof S n = U V where U, V are homeomorphic to R n U V is path connected and homotopic to S n 1 ( ɛ, ɛ) We now show that π 1 (S n, x 0 ) = {1} Claim: If X = U V, U, V open, x 0 U V, and U V path connected, consider: π 1 (X, x 0 ) i j π 1 (U, x 0 ) π 1 (V, x 0 ) Then each element of π 1 (X, x 0 ) is a finite product of elements in Im i or Im j Proof of claim: Let f : I X be a loop at x 0, f : [0, 1] X Use Lebesgue covering lemma to get partition 0 = t 0 < t 1 < < t l 1 < t l = 1 such that f([t i 1, t i ]) is contained in U or V Assume that f([t t 1, t i ]) U, then f([t i, t i+1 ]) V We call the restriction to each of these f i, and so f p f 1 f l, and it is homotopic to (f 1 ḡ 1 ) (g l 1 f l ) f p f 1 f 2 f 3 p f 1 ḡ 1 g 1 f 2 ḡ 2 g 2 f 3, that is, a loop in U, one in V, and then one in U Theorem 211 (Easy Part of Van Kampen Theorem) Let X be A α with A α path connected and open in X, x 0 A α for all α Assume that Aα A β is aways path connected Then i α X the inclusion induces i α : π 1 (A α, x 0 ) π 1 (X, x 0 ) so that each Im(i α ) is a subgroup of π 1 (X, x 0 ) Then, each element of π 1 (X, x 0 ) is a product of elements in Im i α Proposition 212 Let φ : X Y be a homotopy equivalence, and x 0 X Then φ : π 1 (X, x 0 ) π 1 (Y, φ(x 0 )) is an isomorphism 6

7 Lemma 213 Let φ t : X Y be a homotopy, let x 0 X, and let h(t) = φ t (x 0 ), a path in Y from φ 0 (x 0 ) to φ 1 (x 0 ) Then here s a commutative triangle π 1 (Y, φ 1 (x 0 )) β h φ 1 φ 0 π 1 (X, x 0 ) π 1 (Y, φ 0 (x 0 )) Proof Put h t (s) = h(ts) Let f be a loop in X at x 0 Inspect h t (φ t f) h t, a loop at φ 0 (x 0 ) If t = 0, get φ 0 ([f]) If t = 1, then get β h (φ 1 ([f])) Now we prove the prop: Proof φ : X Y, ψ : Y X and ψ φ id X, φ ψ id Y The Lemma gives that ψ φ is an isomorphism, so φ is injective Also, it says that ψ is an isomorphism, and so A α B β C are groups, with α, β injective and β α isomorphism, then α is onto, and so an isomorphism We now go back to chapter zero to define CW-Complexes Definition 28 (CW-Complexes) Let X be a Hausdorff Space and assume that X has an increasing filtration by closed subspaces, that is, there are X 0 X 1 X n, such that 1 X = n 0 X n We call X n the n-skeleton 2 X 0 is discrete called the vertices or 0-cells 3 X 1 the 1-skeleton adjoins the edges, the 1-cells, gives a graph 4 X n is obtained from X n 1 by adjoining n-cells as follows: X n = (X n 1 α Dn α)/ if x Dα n = Sα n 1, then x φ α (x) X n 1 for φ α : Dα n X n 1 continuous Further, a subset A of X is open (closed) in X iff A X n is open (closed) in X n Note: When is A X n closed? True iff A X n 1 is closed and under the characteristic maps Φ α : Dα n X n 1 β Dn β Xn, Φ 1 α (A) is closed in Dα n Also, e n α = Φ α (int of Dα) n X n is an open n-cell Φ α (Dα, n Sα n 1 ) (X n, X n 1 ), Φ α S n 1 = φ α α, X n = X n 1 α en α Third, e n α = Φ α (Dα) n may be identified with e n α = e n α \ e n α = φ α (Sα n 1 ) Examples 1 Any graph, even infinite 2 α S1 α for any index set 7

8 3 If X is a finite CW-complex, then χ(x) = i 0 ( 1)i α i (X) where α i (X) is the number of i-cells For T 2, α 0 = 1, α 1 = 2, α 2 = 1 so χ = 0 4 S n start with S 1 the circle in C χ(s 1 ) = 0 because we get n edges and n vertices For S 2, we see the half spheres are 2-cells, and then notice that we get 2 i-cells for each, which gives χ(s 2 ) = 2 S n is a CW complex with 2 i-cells for 0 { i n More economically, 2 n 2Z S n = e 0 e n, and so χ(s n ) = 1 + ( 1) n = 0 n / 2Z 5 RP n = S n / where x x It is also the set of lines through the origin in R n+1 6 CP n = S 2n+1 C n+1 with x y if x = λy for λ S 1 RP n = RP n 1 e n? Well, RP n = S n / identifying antipodal points, which is equal to D n /, identification of antipodal points on the boundary φ : D n D n / gives us the attaching map, because S n 1 S n 1 / = RP n 1 are contained in D n Example: {S n }, S n 1 S n Then S = S n is contractible RP = RP n and CP = CP n Definition 29 (Subcomplex) Let X be a CW-complex Call a subspace A X if A is closed, and if e n α A then e n α A Put A n = A X n for all n Then A is a CW-complex For example, each X n is a subcomplex of X We call (X, A) a CW-pair if X is a CW-complex and A is a subcomplex of X Then X/A is what you get when you identify all points of A X/A can be made naturally into a CW-complex FACT: If (X, A) is a CW-pair and A is contractible to a point, then g : X X/A is a homotopy equivalence Homotopy Extension Property: Given A X, A closed and f : A Y, can you extend f to f : X Y? Sometimes, given i : A X and f : A Y, there exists f g such that g does have an extension g : X Y Want to be able to conclude that f has extension f : X Y Definition 210 (Homotopy Extension Property: Provisional Def) (X, A), A closed in X, has the HEP iff whenever f g : A X and g has an extension f : X Y so does g Definition 211 (Homotopy Extension Property) (X, A) has HEP if any map F : A I X 0 Y has extension F : X I Y When does (X, A) have HEP? Answer is iff A I X 0 is a retract of X I Example: (D n, S n 1 ) has HEP Proof: D n I is a solid cylinder, S n 1 I D n 0 is the boundary, minus the top It is a retraction because if we take 8

9 a point higher up and hold it still, we draw a line through it to any point on the sides or bottom, and everything on that line gets projected to that point This implies that any CW-pair (X, A) has HEP (prop 016) Prop 017 says that if (X, A) has HEP and A is contractible, then X X/A is a homotopy equivalence Topological Properties of CW-complexes: X is path connected iff X 1 is path connected X is locally contractible X is normal X is compactly generated If K is a compact subset of X, then n such that K X n, even better, K is contained in a finite union of closed cells, best, K is contained in a finite subcomplex And now, we return to chapter 1 We need an algebraic aside so that we can do VK Problem: Given groups {G α } α A and homomorphisms φ α : G α H, H another group, want some group G together with homomorphisms i α : G α G such that Φ : G H unique subject to G α i α φ α Φ G G Then (G, i α ) is the free product or coproduct of the G α s Make G = α G α as all words h 1 h 2 h l with h i G αi \ {1} We call it a reduced word if it cannot be simplified further G is the reduced words under concatenation Theorem 214 (Seifert-van Kampen) Let X be a topological space, x 0 X such that X = α I A α with A α open, path-connected and x 0 A α for all α If A α A β A γ is always path-connected, then we have a homomorphism Φ : α π 1 (A α ) π 1 (X) which is surjective, and N = ker Φ is the normal subgroup of the free product generated by all elements i αβ (ω)i 1 βα (ω) where i αβ : A α A β A α is the inclusion That is, π 1 (X) α (A α )/N A universal property holds: G φ α π 1 (X) π 1 (A α ) π 1 (A β )! φ β π 1 (A α A β ) Example: π 1 ( α X α) with x α X α and α, x α U α X α, U α open and U α deformation retracts to x α Then π 1 (X) π 1 (X α ), where X = α X α Also, assume each A α is path connected, then so is X 9

10 Proof Take A α = X α ( β U β) X α Then for α β, A α A β = γ U γ x 0 Now we apply VK Theorem trivially Using this, we find that π 1 (S 1 S 1 ) = Z Z, the free group on two generators π 1 (S 1 S 2 ) Z In fact, if G is any connected graph, then a maximal tree can be contracted to a point, and so has fundamental group free, and so is a bouquet of circles π 1 (R 2 \N pts) =free group on n generators This space contracts to N i=1 S1 So π 1 (S 2 \ N) free group on n 1 generators Suppose R n \ K is path connected, K is closed and bounded π 1 (R n \ K) π 1 (S n \ K) is an isomorphism when n > 2 Proof Take U = R n \ K and V = S n \ B R (0) where K B R (0) Note that U V = R n \ B R (0), which is path connected (ADD DETAILS LATER) Let A be a circle in the plane Then R 3 \ A S 1, so π 1 (R 3 \ A) = Z Proposition 215 Inspect S p+q 1 P p+q It has S p 1, S q 1 as subspheres Then there exists a deformation retraction from S p+q 1 \S q 1 onto S p 1 Thus, π 1 (S 3 \ A) π 1 (S 1 ) Proof Define f t : S p+q 1 \ S q 1 S p+q 1 \ S q 1 by f t (x, y) = ((1 t)(x, y) + ), 0)/( T OP ) t( x x R 3 \ (A B) where A, B are unknotted circles contracts to S 1 S 1 R 3 \ (A B) where A, B are linked, then R 3 \ (A B) S 1 S 1, and so the fundamental group is Z Z What is π 1 (R 3 \ (A B)), the complement of linked circles? It is isomorphic to π 1 (S 3 \ (A B)) Now, S 3 = {(z, w) C 2 : z 2 + w 2 = 1} Put T z = {(z, w) S 3 : w 2 1/2} and T w = {(z, w) S 3 : z 2 1/2} Note that {(z, 0) : z = 1} T z and {(0, w) : w = 1} T w Now, S 3 = T z T w and T z T w = {(z, w) S 3 : z 2 = w 2 + 1/2} S 1 S 1 So T z S 1 D 2 the solid torus T z T w is a deformation retract of S 3 \ (A B) T z = {(z, w) : z 2 + w 2 = 1, w 2 1/2} z = z e iθ and z 2 = 1 w 2 so (z, w) (e iθ = z/ z, w) S D 2 S 1 D 2 π 1 (K m,n ) where (m, n) = 1 is the group a, b a m b n Attaching 2-cells creates relations by how they are attached Let X be a path connected space, x 0 X Attach 2-cells via ϕ α : (S 1, s 0 ) (X, x 0 ) and put Y = (X α D2 α)/ where ϕ α (x) x for all x S 1 α Proposition 216 Then π 1 (Y, x 0 ) π 1 (X, x 0 )/N, N is the smallest subgroup containing all [ϕ α ] π 1 (X, x 0 ) Proof Case 1: a Single 2-cell Φ : D 2 e 2 Y, y = Φ(0) Y = (Y \{y}) e 2 = e 2 \ {y} S 1 VK implies that π 1 (Y, Φ(d 1 )) π 1 (X \ {y}, Φ(d 1 ))/minimal subgroup generated by [f] π 1 (Y, Φ(d 1 )) π 1 (Y, x 0 ) π 1 (X, x 0 )/[ϕ α ] 10

11 Case 2: Induction for the case of a finite number of 2-cells Case 3: Any number at all: Let ϕ α : S 1 X Y = X α A e2 α, y α = Φ α (0) U α = X β (e β \ y β ) E α = Y = β {y β } e α U α will be open and path connected subsets Thus, α β U α U β = Y β A {y β } X Let F be all finite subsets of A Put for F F, U F = α F U α X α F e α We know π 1 (U F ) = π 1 (X)/normal subgroup generated by [ϕ α ] for α F VK Theorem: Note that U F1 U F2 = U F1 F 2, and also for triple intersections Thus, G π 1 (Y )! π 1 (U F1 ) π 1 (V F2 ) π 1 (U F1 F 2 ) If F is the set of all finite subsets of A and we order F by inclusion, we obtain an directed ordered set Given F 1, F 2 F, note that F 1 F 1 F 2 and F 2 F 1 F 2 Thus, π 1 (Y ) is the colimit (or direct limit) of {π 1 (U F )} F F along with the homomorphisms from inclusions in F An algebraic comment: If D is a directed set α D with G α a group and d d implies there is a homomorphism i d,d : G d G d and d = d implies it is the identity, with d d d gives G d G d G d Then lim = colim(g α, i α,α ) = d G α /N where N is the normal subgroup generated by {gα 1 i α,α (g α )} with g α G α and α α NEXT: Covering Spaces, read the section Definition 212 (Covering Space) A covering map p : X X is a map such that X is covered by evenly covered open sets U, that is, p 1 (U) = α V α, V α s open and disjoint in X with p : V α U a homeomorphism Remarks: p is a local homeomorphism, meaning for all x X, there is an open nbhd V of x such that p V : V p(v ) is a homeomorphism p(v ) open in X p is an open mapping p is a quotient mapping, provided p is surjective 11

12 Examples: and any X, id : X X The number of sheets is the index of the subgroup that is the fundamental group of the covering space It is normal if you get the same subgroup (rather than a conjugate one) from a change of basepoint Lifting Theorems Theorem 217 (Homotopy Lifting Property) Given f 0 Y X inc p {f t } Y I X Then there exist a unique { f t } : Y I X Theorem 218 If p : X X is any covering map and p( x0 ) = x 0 then 1 p : π 1 ( X, x 0 ) π 1 (X, x 0 ) is injective 2 Im p = {[f] : f based at x 0 such that the lifting f such that f(0) = x 0 is a loop} Proof 1 how that ker p = 1 Suppose that [ f] π 1 ( X, x 0 ) such that [p f] = 1 π 1 (X, x 0 ) So p f = f c x0 The HLT says that we can lift to get f p c x0, and so [ f] = 1 2 is clear is also clear Proposition 219 If X and X are path connected, then p 1 (x), x X is constant and equal to [π 1 (X, x 0 ) : p (π 1 ( X, x 0 ))] Theorem 220 (Lifting Theorem) Assume that Y is path connected and locally path connected, with f : (Y, y 0 ) (X, x 0 ) and p : ( X, x 0 ) (X, x 0 ) is a covering map A lift f : Y X exists iff f (π 1 (Y, y 0 )) p (π 1 ( X, x 0 )) Proof It f exists, the inclusion is clear Assume that the inclusion holds Take y Y run a path γ y = γ from y 0 to y Then f γ y is a path in X from x 0 to f(y) Lift to f γ y which is a path in X from x 0 to a point we shall name f(y) f(y) = f γ y (1) is set to f(y) by p Must check the independence of path from y 0 to y, that f : Y X is continuous and f(y 0 ) = x 0, though the last part is trivial So is f γ y (1) = f δ y (1) for δ, γ paths from y 0 to y? (f γ y ) f δ y f (γ y δ y ) p p ɛ where ɛ is a loop at x 0 in X Thus f γ y (p ɛ) (f δ y ), and so f γ y p ɛ f δ y Thus, the endpoints are the same Why must f : Y X be continuous? Let y Y, and check continuity at y f(y) U evenly covered, f(g) Vα, p : V α U homeo Show that there is a 12

13 neighborhood W of y such that f(w ) V α Since U is open and f(y) U, there is an open nbhd W of y such that f(w ) U hypotheses on Y let us assume that W is path connected Recall, f(y) = f γ y (1) V α Let w W, run a path in W, ζ w from y to w Then f(w) = f (γ y ζ w )(1) = (f γ y ) (f ζ w )(1) The first part ends at f(y), and the second part happens in U And so, this equals f γ y p 1 (f ζ w )(1) V α And so, the map is continuous Next we construct the universal cover If H π 1 (X, x 0 ) then we construct X H X such that p α ( X H ) = H We will also attempt to classify all covering spaces of some good X up to isomorphism Graphs For a graph, χ(x) = V E, and if X is an n-sheeted cover of X, then χ( X) = nχ(x) If we take X X/T S 1 is a homotopy equivalence, and so χ(x) = χ( S 1 ) = 1 k, where k is the number of circles Thus, χ( X) = n(1 k) = 1 l and l = 1 + (n 1)n If X is some nice space, path conn and locally path conn, then an isomorphism of spaces over X is f : X1 X 2 such that p 2 f = p 1 We define Cov(X) =all isomorphism classes of coverings of X Ỹ = p 1 ( X) X p p φ Y X then Ỹ = {(y, x) Y X : φ(y) = p( x)} maps down to Y by π Y Thus, we get a map φ : Cov(X) Cov(Y ) Assume that X is as before and also that for each x X, there is an open U containing x such that every loop in U is homotopic to the constant loop in X Then we call it SLSC (Semi-Locally Simply Connected) [Losternik-Schrilmann Category of a Space]: RP n = S n /(x x) Let U i RP n be given by x i 0 So RP n is a union of n + 1 contractible sets In general, X = V 0 V n is an open cover such that ι i : V i X is homotopic to a constant map This is similar to SLSC Theorem 221 There exists a simply connected covering space if SLSC holds for X, and conversely Let p u : X u X be simply connected covering space of X and let X X be any other covering space, X also path connected Then the diagram commutes: X u, x u p X, x 0 p u φ X, x 0 13

14 Note: 13 Ex 16 says that φ is also a covering map Can a sc X have covering space no isom to id : X X? p : X X given, then isomorphic as it must be a one-sheeted covering space Easy: If 1-sheet, then p is a homeomorphism Also if s is a section of the covering map (p s = id X ) and s(x) p 1 (x) for all x X, then p is a homeomorphism Proposition 222 If H is a subgroup of π 1 (X, x 0 ) then there exists a covering X X such that the induced map p (π 1 ( X, x 0 )) = H We define an automorphism of a covering space to be an isomorphism f : X X We also call these Deck Transformations We name the group of these as G( X, p) = G( X/X) If for any two points x 1, x 2 such that p( x 1 ) = p( x 2 ) X, there exists φ G( X, p) such that φ( x 1 ) = x 2, call p normal Proposition 223 Let p : ( X, x 0 ) (X, x 0 ) be a covering with X, X path conn and X locally path conn Then 1 p is normal iff p π 1 ( X, x 0 ) is a normal subgroup of π 1 (X, x 0 ) 2 Assuming p is normal, there exists an isomorphism G( X, p) π 1 (X, x 0 )/p π 1 ( X, x 0 ) So for the universal covering, G( X, p) π 1 (X, x 0 ) Assume X X is a universal cover So G( X, p) acts transitively on p 1 (x 0 ) Define φ : π 1 (X, x 0 ) G( X, p) by if [γ] π 1 (X, x 0 ) lift γ to γ a path in X with γ(0) = x 0 and γ(1) p 1 (x 0 ) Then there exists g G( X, p) such that g( x 0 ) = γ(1) Put φ([γ])( x 0 ) = γ(1), and φ([γ]) G( X, p) This element is unique, as if g 1, g 2 G( X, p) and if x 0 such that g 1 x 0 = g 2 x 0 then g 1 = g 2 by uniqueness of lifting Then we can show easily that φ is a homomorphism of groups and that φ is surjective φ is also injective Change point of view: If Y is a given space and G is a group, G Y Y an action such that g : Y Y is a homeomorphism of Y Then y gy for g G gives an equivalence relation, and Y Y/G is the space of orbits Favorite Example: Y = S n, G = {±1} Z 2 by (+1)y = y and ( 1)y = y, the antipodal map Y/G = S n /Z 2 RP n Also π 1 (RP n ) G Z 2 Let G act on Y Call the action even if g Y, open neighborhood V of y such that g h gv hv = Lemma 224 If G acts evenly on Y, then p : Y Y/G is a covering map Proof We know p is continuous and open since if W Y is open then p 1 (p(w )) = g G gw In fact, this is a disjoin union Claim: For V as in definition, then p(v ) is evenly covered Note that p gv : gv V is bijective [Clearly surjective, for injective, we say gv 1, gv 2 are in the same orbit, so h G such that gv 1 = hgv 2, disjointness implies that hg = g, so h = 1, hence v 1 = v 2 ] 14

15 We call such a covering space a G-covering Isomorphism of G-coverings is φ : Y 1 Y 2 over Y 1 /G X Y 2 /G such that φ is a homeo and φ(gy) = gφ(y) Example, Y = S 2n 1 C n, G = {z C : z k = 1}, then G S 2n 1 S 2n 1 by v zv Then S 2n 1 S 2n 1 /G gives us a lens space Trivial G-covering over a given X is X G X by projection where G has the discrete topology on it h(x, g) = (x, hg) Lemma 225 Any G-covering Y X is locally trivial as a G-covering ie, X can be covered by open sets U such that gv p 1 (U) U is isomorphic to U G U by projection Proposition 226 Let p : Y X be a G-covering Then we have a homomorphism G Aut(Y, p) which is injective The the case where Y is connected, this is an isomorphism Proof Pick y 0 Y, then for ϕ Aut(Y, p) = G(Y, p), ϕ(y 0 ) satisfies p(ϕ(y 0 )) = py 0 so g G such that ϕ(y 0 ) = gy 0 So ϕ and y gy 0 are 2 elements of Aut(Y, p) which agree on y 0 Uniqueness props for lifting implies that φ sends y gy Proposition 227 Let p : Y X be a covering, Y connected, then Aut(Y, p) acts evenly on Y If Aut(Y, p) acts transitively on a fiber of p, then the covering is a G-covering for G = Aut(Y, p) Chapter 11 and 13 of Fulton Proposition 228 Let G act evenly on a simply connected and locally path connected space Y Then for X = Y/G, π 1 X = G Proof We ve seen that π 1 (X) Aut(Y, p) for p : Y X and prop above implies that G Aut(Y, p), so π 1 (X) Aut(Y, p) G Chapter 14 - The Van Kampen Theorem x 0 X = U V, x 0 U V, U, V, U V path connected, then we can define the fundamental group and a map π 1 (X) G which is unique to each G that makes a commutative diagram commute What is the meaning of a π 1 (X, x 0 ) G? Answer: G-coverings of X up to isomorphism preserving basepoints Suppose given a homomorphism ρ : π 1 (X, x 0 ) G Construct a G-covering p ρ : Y ρ X with y ρ Y ρ st p ρ (y ρ ) = x 0 Give G the discrete topology Consider X G and a left action of π 1 (X, x 0 ) on it by [σ] (z, g) = ([σ] z, g ρ([σ]) 1 ) Then, put Y ρ = ( X G)/π 1 (X) 15

16 3 Homology Let X be a topological space We want to define H i (X) the homology groups i 0 with some good properties We expect it to be functorial and to be useful -complexes We look at T 2 v a v a c b v b v And calling the inside of the triangles U and L Let v 0,, v n R m, assume that they are affinely independent That is, no three are collinear, no four coplanar, etc Equivalently, if c 0 v c n v n = 0 and c c n = 0 then c i = 0 for all i So, let v 0,, v n be affinely independent Then define [v 0,, v n ] = all n i=0 t iv i where t i 0 and t i = 1 1 The Barycenter is n+1 vi The standard n-simplex is n = [e 0,, e n ] R n+1 If v 0,, v n are affinely independent points, a homeomorphism n [v 0,, v n ], as any point in n is of the form (t 0,, t n ), and we can send it to ti v i We can orient the edges by the edge between v i and v j points to the greater one We define simplicial homology for a -complex We look at X = e n α open n-simplexes and this union is disjoint We have for each e n α σ α : n [v 0,, v n ], which gives a map from int n e n α We define n (X) to be the free abelian group generated by all open n- simplexes We can identify e n α σα n So the elements look like finite sums of simplices or of these maps We call each of these a simplicial chain We now attempt to define a boundary operator ([v 0, v 1 ]) = v 1 v 0, ([v 0, v 1, v 2 ]) = [v 0, v 1 ] + [v 1, v 2 ] [v 0, v 2 ] = [ˆv 0, v 1, v 2 ] [v 0, ˆv 1, v 2 ] + [v 0, v 1, ˆv 2 ] In general ([v 0,, v n ]) = ( 1) i [v 0,, ˆv i,, v n ] And so, we have for -complex X, n (X) n 1 (X) Lemma 31 = 0 We define Z n (C) = ker( n (X) n 1 (X)) to be the cycles and B n (X) = Im( n+1 (X) n (X)) B n Z n by the lemma We define H n (X) = Z n (X)/B n (X) For the torus above, 2 (T 2 ) has basis U, L, 1 (T 2 ) has basis a, b, c and 0 (T 2 ) has basis v (U) = (V ) = a + b c, (a) = (b) = (c) = 0 H 0 = 0 /0 Z H 1 = 1 /(a + b c) Z Z H 2 = Z(U L)/0 Z 16

17 Singular Homology [Eilenberg] Let X be any space A singular n-simplex in X is a continuous map σ : n X Put C n (X) = the free abelian group generated by all singular n-simplexes, so elements look like n σ σ where σ is a map n X where n σ Z and only finitely many are nonzero : C n (X) C n 1 (X) is defined by, if σ : n X, σ = n i=0 ( 1)i σ i where σ i : n 1 X is the inclusion of the i th face, φ i followed by σ φ i (t 0,, t n 1 ) = (t 0,, t i 1, 0, t i,, t n 1 ) Lemma 32 2 = 0 If X = Y, then H n (X) = H n (Y ) Proposition 33 If X = {pt}, then H n (X) = { Z n = 0 0 else Proof For n 0, σ n : n {pt} is unique Thus, σ n is the alternative sum of σ n 1 s, which is 0 if n is odd, 1 if n is even Thus, we get the complex Z Z Z Z In each, either everything is a cycle and everything is a boundary, or else none of either is Except in dimension zero, as everything is a cycle, but nothing is a boundary, so H 1 (X) = Z Proposition 34 If X has path components X α, then H n (X) = α H n (X α ) Proposition 35 If X is path connected and nonempty, then H 0 (X) = Z Proof Choose x 0 X Consider C 0 (X) = x X n xx C 1 (X) C 0 (X) 0, but we can augment it, so instead we get C 0 (X) ɛ Z ɛ( n x x) = n x Note: ɛ = 0 since ɛ σ 1 = 1 1 = 0 Claim: ker ɛ = Im We know that Im ker ɛ We need to show the converse Let c = n 1 x 1 + +n l x l C 0 (X) with ɛ(c) = 0 so n 1 + +n l = 0 Choose paths f i from x 0 to x i (1 i l) so f i = σ i : 1 X with σ i = x i x 0 for all i Then σ( n i σ i ) = σn i (x i x 0 ) = n i x i n i x 0 = c Definition 31 (Reduced Homology) Let X be { out space We have C 2 (X) C 1 (X) C 0 (X) ɛ Z Define H Hn (X) n > 0 n (X) = ker ɛ/ Im n = 0 Fact: If X is path connected, x 0 X, then there exists a homomorphism π 1 (X, x 0 ) H 1 (X) called the Hurewicz homomorphism which is surjective and has kernel the commutator subgroup of π 1 Definition 32 (Induced Homomorphism) If f : X Y is a continuous map, then f : C n (X) C n (Y ) by composing f with σ : n X to get σ : n Y 17

18 For c C n (X), f (c) = f (c) Note: f (Z n (X)) Z n (Y ) and f (B n (X)) B n (Y ) induces f : H n (X) H n (Y ) 0 B n (X) Z n (X) H n (X) 0, f gives an homomorphism of short exact sequences to 0 B n (Y ) Z n (Y ) H n (Y ) 0 Theorem 36 f, g : X Y and f g, then f = g : H n (X) H n (Y ) Proof Let F : X I Y continuous with F (x, 0) = f(x) and F (x, 1) = g(x) for all x X We want to show that f = g : H n (X) H n (Y ) Consider n I n {0} = [v 0,, v n ], n {1} = [w 0,, w n ] We have additional n-simplices [v 0,, v i, w i+1,, w n ] Note: n I = n i=0 [v 0,, v i, w i,, w n ] is an n + 1-simplex Define P : C n (X) C n+1 (Y ) by σ n i=0 ( 1)i F (σ id) [v0,,v i,w i,,w n] Check: P = g f P as hom C n (X) C n (Y ) or P +P = g f (P is a chain homotopy between f and g ) Let z Z n (X) Then z H n (X) Then f z = f z and g z = g z, so to show that f = g, we must show that f z g z is a boundary f z g z = ( P z + P z) = ( P z) Corollary 37 X Y implies H n (X) H n (Y ) What is H n (S k ) when k, n > 0? Want relative homology long exact sequence, H n+1 (X, A) H n (A) i H n (X) j H n (X, A) Given a space X, A X, Z A such that Z int(a) Have (X \ Z, A \ Z) (X, A) Theorem 38 i : H n (X \ Z, A \ Z) H n (X, A) is an isomorphism Use the above to determine the homology groups of S k by induction on k; later (or read) Definition 33 (Relative Homology) We define the relative chain complex of a pair (X, A) where A X as C n (X, A) = C n (X)/C n (A) The relative homology is the homology of this chain complex Proof of Theorem: Proof We have, easily, i, j, we need : H n (X, A) H n 1 (A) Let z Z n (X, A) with z H n (X, A) Pick x C n (X) with x z Then there exists a unique a C n 1 (A) such that a x Note that a = 0 so a Z n 1 (A) Put ( z) = ā H n 1 (A) Must check that is well-defined Theorem 39 For (X, A), there exists a long exact sequence H n+1 X H n+1 (X, A) H n (A) H n (X) H n (X, A) δ H n 1 (A) 18

19 Proof Must verify exactness at H n (A), H n (X), H n (X, A) At H n (A): i z = 0 is clear from the definition of z Suppose ā H n (A) a Z n (A) and i ā = 0 in H n (X) Then i a = x, x C n+1 (X) Exactness follows by diagram chasing At H n (X): Clearly j i = 0 Suppose given z H n (X) such that z 0 in H n (X, A), there exists a C n (A) and n C n+1 X sch that z = n + i a check that a is a cycle, i n = z Proposition 310 Let f, g : (X, A) (Y, B) be homotopic as maps of pairs Then natural f : H n (X, A) H n (Y, B) agrees with g Proof P : C n (X) C n+1 (Y ) for all n such that P + P = g f a chain homotopy Observe that since f(a) B, P : C n (A) C n+1 (B), we pass to quotients to get P : C n (X, A) C n+1 (Y, B) still a chain homotopy between f and g Thus, on relative homology, f, g are the same Naturality of Long Exact Sequence: Suppose f : (X, A) (Y, B) H n+1 (X, A) H n (A)H n (X)H n (X, A) f f f f H n+1 (Y, B) H n (B)H n (Y )H n Y, B commutes Reformulation of Excision Theorem X A, Z A put B = X \ X, so A B = A \ Z Now, Z = X \ int(x \ Z) So Z int(a) means that X \ int(b) int(a), thus A = (A) int(b) Theorem 311 If U = {A, B} covers X so that X = int(a) int(b), then H n (B, A B) H n (X, A) is an isomorphism Application: X is a space SX is a suspension of X,m that is, X I/ with (x, 1) (x, 1) and (x, 0) (x, 0) for all x, x X Then there exists isomorphisms H n+1 (SX) H n (X) Corollary 312 Hk (S n ) is Z for k n, 0 else Proof We now prove the suspension theorem A = SX \ {S}, B = SX \ {N} A B X, so 0 H n+1 (SX) H n+1 (SX, A) 0 exact (thus middle is an isomorphism), and also H n+1 (SX, A) isomorphic to H n+1 (B, A B), and 0 H n+1 (B, A B) H n (A B) 0 Generalized Excision Theorem: X is given Let U = {U α } be a cover of X such that {int(u α )} is a cover of X Let Cn U (X) C n (X) be a subgroups generated by all U-small singular n-complexes σ : n X meaning σ( n ) U α for some α Observe that σ U-small implies that σ is U-small Thus, maps Cn U (X) Cn 1(X) U So get ι : Hn U (X) H n (X) is an isomorphism 19

20 More precisely, ι : C U n (X) C n (X) is an inclusion which has chain homotopy inverse ρ : C n (X) C U n (X), ι and ρ are both chain maps and ρ ι c id and ι ρ c id where c is chain homotopy Exercise: From this, you can get the excision theorem Proof of the generalized excision theorem Proof We will prove that ι : H U n (X) H n (X) is an isomorphism We will prove it in four parts 1 Subdivide simplexes, with iterated barycentric subdivision If [w 0,, w n ] is an n-simplex in R N and b R N, then b [w 0,, w n ] = [b, w 0,, w n ], an n + 1 simplex Aim: Barycentric Subdivision of an n-simplex [v 0,, v n ] We define the 1 simplex to be [ ] We will define this by induction on n For n = 1, nothing For n = 0, also no change For n > 0, assume that each (n 1)-face of [v 0,, v n ] has been subdivided into n! (n 1)-faces, which we will call [w 0,, w n 1 ] Then we let b = 1 n+1 (v 0+ +v n ) be the barycenter of [v 0,, v n ] and consider all n-simplexes [b, w 0,, w n 1 ] = b [w 0,, w n 1 ] giving a grand total of n!(n + 1) = (n + 1)! n-simplexes Lemma 313 diam[b, w 0,, w n 1 ] n n+1 diam[v 0,, v n ] 2 Y R N a convex set Let LC n (Y ) C n (Y ) be the subgroup generated by all linear maps σ : n Y Note: LC n (Y ) LC n 1 (Y ), so the linear chains give us a chain complex If b Y, we get b : LC n LC n+1 by b [w 0,, w n ] = [b, w 0,, w n ] Check that b(α) = α b( α) for all α LC n (Y ) That is, b+b = id 0 on LC n (Y ) Take α = [w 0,, w n ], then b[w 0,, w n ] = [b, w 0,, w n ] = [w 0,, w n ] [b, w 1,, w n ] +, so it works out Define S : LC n (Y ) LC n (Y ) by induction on n, let λ : n Y, λ = [w 0,, w n ] and let b λ = the barycenter of [w 0,, w n ] Put S([ ]) = [ ] and S([w 0 ]) = [w 0 ] In general, S(λ) = b λ (S λ) in LC n (Y ) Check S = S Next we construct a chain homotopy T : LC n (Y ) LC n+1 (Y ) between S and id We do this inductively by setting T = 0 on LC 1 (Y ) and T λ = b λ (λ T λ) for n 0 3 We will now subdivide general chains S : C n X C n X by Sσ = σ S [e 0,, e n ], where S[e 0,, e n ] LC n Check Sσ = S σ and putting T σ = σ T n, then T + T = id S 20

21 4 Iterated subdivision Recall that we have S, T such that S = S and T + T = id S This says that S i is a chain map, and that S i is even chain homotopic to id D m = 0 i m T Si = T ( S i ), check that D m + D m = id S m For each σ : n X, there exists m such that S m (σ) C U n (X) Put m(σ) =least m that works Define D : C n (X) C n+1 (X) by Dσ = D m(σ) σ, then Dσ + Dσ = σ ρ(σ) We discover that ρ(σ) C U n (X) Check ρ : C n (X) C U n (X) is a chain map, ρ = ρ, and moreover D + D = id ι ρ where ι is the inclusion of C U into C Note that ρι = id on C U, and so we are done Theorem 213 Definition 34 (Good Pair) H n (X, A) H n (X/A) isomorphism if A, A closed, A V open, A a deformation retract of V Then we call (X, A) a good pair Prop 221 C U (X) C (X) the inclusion, p in the opposite direction, with pi c id and ip c id Relative version, (X, U ), (A, U A ) 0 C U A (A) C U (X) C U (X, A) 0 0 C (A) C (X) C (X, A) 0 implies that H U (X, A) H (X, A) is an isomorphism Pf of excision theorem in form X = (A) (B) implies that H (B, A B) = H (X, A) Proof Let U = {A, B} 0 C U A C U (X) C U (X, A) 0 C (A) C (A) C (B) with vertical columns equivalences Then C U (X, A) = C A+C B C A C B C (A) C (B) = C B C (A B) = C (B, A B) 21

22 (X, A) a good pair implies that q : (X, A) (X/A, A/A) induces an isomorphism H (X, A) H (X/A, A/A) H (X/A) A closed subset of V open, A deformation retract of V implies that A/A closed subset of V/A, A/A a def retract of V/A (X, A, B) triple, X A B, then 0 C (A, B) C (X, B) C (X, A) 0 is a short exact sequence of chain complexes, and so we get a long exact sequence on relative homology groups H (X, A) H (X, V ) q H (X/A, A/A)H (X/A, V/A In fact, we can excise A and get isomorphisms to H (X \ A, V \ A) and to H (X/A \ A/A, V/A \ A/A) i : A V is a homotopy equivalence, so H n (A) H n (V ) H n (V, A) = 0 H n 1 (A) H n 1 (V ) Excision holds for CW-pairs X = A B, X a CW complex and A, B CWsubcomplexes implies that H (B, A B) H (X, A) induces isomorphisms? If X = α X α where x α X α for all α, (X, {x α }) is a good pair for all α, then ι α : X α X the inclusions induce α i α : H (X α ) H (X) Theorem 314 If U R m, V R n are open and V homeomorphic to U, then m = n Proof Let x U Excision implies that H m (U, U \ {x}) H m (R m, R m \ {x}), is an isomorphism, as are H m (S m 1 ) to H m (R m, S m 1 ) to it, and so we get δ lm Z = H m (S m 1 ) Homeomorphisms induce isomorphisms on homology, so m = n, as δ lm Z = δ ln Z Theorem 315 If X is a -complex, then there exists a natural isomorphism H H (X) In fact, if A is a subcomplex of a -complex X, then get isomorphism H (X, A) = H (X, A) Proof Take dim X <, A = Let X k be the union of all simplexes in X of dim less than or equal to k, call this the k-skeleton Induction on k, X 0 X n 1 X n = X, dim X = n See diagram on bottom of 128 for (X k, X k 1 ) By the 5-lemma, if the outside maps are isomorphisms, then the inside one is Degree of a map f : S n S n Definition 35 (Degree) Hn (S n ) Z for all n 0 f induces f : H n (S n ) H n (S n ), so there exists d Z such that f (α) = dα for all α H n (S n ) Set deg f = d 22

23 Here are some basic properties: 1 deg(id) = 1 2 f g f = g deg f = deg g 3 f constant deg f = 0 So if f : S n S n not surjective, then deg f = 0 4 deg fg = deg f deg g 5 R : S n S n a reflection implies deg R = 1 (Why?) 6 deg A = ( 1) n+1, where Ax = x for all x 7 If f : S n S n has no fixed points, then deg f = ( 1) n 1 8 If A O(n + 1), then deg A = det A 9 f n : z z n on S 1 has deg f n = n 10 f, g : S n S n and deg f = deg g imply that f g 11 f : S n S n has suspension Sf : S n+1 S n+1 indeed, SX for any f : X Y induces Sf : SX SY Then deg Sf = deg f [Giver f : S n+1 S n+1, it must be homotopic to a suspension [S n+1, S n+1 ] Z, there exists a map, the Hopf map, S 3 h S 2 such that h constant map] 12 f : S n S n, y S n, f 1 (y) = {x 1,, x k }, deg xi f = ±1, then deg f = degxi f Cellular Homology X a CW-complex, X 0 X n, X = i Xi H k (X) Hk CW (X), the cellular homology given by Cn+1(X) CW d Cn CW (X) d Cn 1(X) CW, d 2 = 0 Put Cn CW (X) = H n (X n, X n 1 ) is free abelian, a generator for each n cell of X Why? (X n, X n 1 ) is a good pair, X n \ X n 1 = α e n α where e n α is homeomorphic to the open ball Let x α correspond to the origin in the homeomorphism Put V = X n \ {x α } open and there exists a retraction of V onto X n 1 Hence, H n (X n, X n 1 ) H n (X n /X n 1 ) = H n ( α Bα/s n n 1 α ) Z with generators e n α We need d : H n (X n, X n 1 ) H n 1 (X n 1, X n 2 ), we get it by j through H n 1 (X n 1 ) where j : (X n 1, ) (X n 1, X n 2 ) is the inclusion but as a map of pairs We must check that d 2 = 0, d = j : H n (X n, X n 1 ) H n 1 (X n 1, X n 2 ), which is true as j = 0 Thus H (C CW (X)) = H CW (X) Lemma H k (X n, X n 1 ) = 0 if k n 23

24 2 H k (X n ) = 0 if k > n and if dim X <, H k (X) = 0 if k > dim(x) 3 i : X n X induces isomorphism i : H k X n H k X if k < n Proof Proof of b: Use induction on n For n = 0, clear Suppose n > 0, result is known for smaller skeleton Then H k (X n 1 ) = 0 H k (X n ) H k (X n, X n 1 ) = 0, so H k (X n ) = 0 Proof of c: First, suppose dim X <, X = X N, N = dim X X n X n+1 X N = X If n = N, good and simple Suppose that n < N for k < n, we get 0 = H k+1 (X n+1, X n ) H k (X n ) H k (X n+1 ) H k (X n+1, X n ) = 0 so we get isomorphism for all k Need the diagram on page 139 Proof by diagram chasing We will need the next result to do this If k < n, H k (X n ) H k (X n+1 ) H k (X N ) isomorphisms, but does not necessarily reach H k (X) Lemma 317 If K is a compact subset of a CW-complex X, then K X n for some n Proof Suppose not, then for each n, choose x n K with x n / X n Then K 0 = {x n } K is an infinite set Note: K 0 X p is finite for all p, so K 0 X is closed Thus, K 0 is compact Note that each subset of K 0 is closed, and so K 0 has the discrete topology, contradiction Easiest Applications 1) Suppose there are no n-cells at all for some n Then H n (X) = 0 2) Suppose there are k n-cells Then Cn CW (X) Z k, so Zn CW (X) Z l, l k, so H n (X) has k generators In particular, rank H n (X) k 3) Suppose there are n-cells but no n + 1 or n 1-cells Then d = 0 in C CW (X), so C CW (X) = H n (X) free abelian group on the n-cells 4) H 2k (CP n ) Z for 0 k n, others are 0 Application of Degree Theorem 318 Z 2 is the only nontrivial group that can map freely on S 2n Proof Any homeomorphism h : S 2n S 2n has degree ±1, so we get a homomorphism deg : G {±1} If g G, g 1, then gx x for all x So we x gx is homotopic to A : x x, so deg g = ( 1) 2n+1 = 1 Hence, ker deg = {1}, so deg G {±1} is injective Thus G 2 Let f = f d : S n S n be a degree d map X = S n f e n+1, find H i (X) Have Φ : (D n+1, S n ) (X, S n ), induces isometried on homology, so H n+1 (D n+1, S n ) Hn (S n ) Φ f H n+1 (X) H n+1 (X, S n ) Hn (S n ) Hn (X) 24

25 With the first map on the bottom injective and the last surjective So we conclude that H n (X) Z/dZ, H n+1 (X) = 0 and H i (X) = 0 if i n, n + 1 If f = f 2 : S 1 S 2 is a map of degree two, we can see that RP n has homology H n = Z/2 iff n = 1 and 0 otherwise IF X, Y are CW-Complexes and f : X Y call f cellular if f(x n ) Y n for all n 0 FACT: Cellular Approximation Theorem: Any continuous f : X Y between CW-complexes is homotopic to a cellular map Ex: Any f : S m S n with m < n is homotopic to a constant map Then f : (X n, X n 1 ) (Y n, Y n 1 ) for all n, so get a chain map H n (X n, X n 1 ) H n (Y n, Y n 1 ), so we get Hn CW square H n (Y ) f H n (X) (X) H CW n (Y ), thus we get a commutative H CW n (Y ) Hn CW f CW (X) Cellular Boundary Formula Prelude: Choose gen of H n (S n ) compatibly under isomorphisms H n+1 (S n 1 ) H n (S n ) starting with S 0 Then also H n (D n, S n 1 ) H n 1 (S n 1 ) Use {e n α} as gens for Cn CW (X) Then d n (e n α) = β d αβe n 1 β, d αβ Z the degree of S n 1 α X n 1 Xn 1 X n 1 \e n 1 β S n 1 β Homology with Coefficients If R is a commuative ring with identity (say a field) then we take C n (X; R) to be the free R-module generated by the singular simplexes, which is, in fact, R Z C n (X), and every works out with H n (X; R) So for example, Hn (X; R) R for X = S n Let X be a finite complex, and α i the number of cells in each dimension Then χ(x) = dim X i=0 ( 1) i α i H i (X) = Z βi + F i where F i is finite β i is called the ith Betti number [H i (X) Z Q has dim β i ] Fact: For finitely generated abelian groups, if 0 A B C 0 is exact, then rank A rank B + rank C = 0 Theorem 319 χ(x) = dim X i=0 ( 1) i β i Proof 0 Bi CW (X) Zi CW (X) Hi CW (X) 0 is ses Hi CW (X) H i (X), and 0 Zi CW (X) Ci CW (X) Bi 1 CW (X) 0 ses So ( 1) u rank H i (X) = ( 1) i rank Hi CW (X) = ( 1) i [rank Zi CW (X) rank Bi CW (X)] = ( 1) i (rank Ci CW (X) rank Bi 1 CW (X) rank BCW i (X)) = ( 1) i α i ( 1) i Bi CW (X) ( 1) i rank Bi 1 CW (X) = ( 1) i α i = χ(x) Mayer-Vietoris Sequence 25

26 (Vietoris) Let A, B X, int(a) int(b) = X Our aim is a les H n (A B) Φ H n (A) H n (B) Ψ H n (X) H n 1 (A B) Recall that C n (A)+C n (B) C n (X) for all n gives C (A)+C (B) C (X) which induces an isomorphism on homology groups Next: 0 C (A B) C (A) C (B) C (A) + C (B) C (X) where + is the( join ( x x x, x + y This is a short exact sequence This gives: x) y) H n (A B) H n (A) H n (B) H n (C (A) + C (B)) H n (X) Snake Lemma and some applications 4 Cohomology Why bother with Cohomology? Because it has a ring structure, given by a cup product H p (X) H q (X) H p+q (X) In fact, it is a graded ring This is like differential forms (they are a primary example) A cellular chain, if X is a CW-complex, looks like 0 C n C n 1 C 1 C 1 0 Cochains: If A is an abelian group, so is hom(a, Z) A f B, then we get hom(a, Z) f hom(b, Z), so this is a contravariant functor Similarly, if we have A f B g C, we get a similar sequence of homs If A = Z k, B = Z l, then f : A B a function What is f : hom(b, Z) hom(a, Z)? Well hom(z k, Z) Z k, but this is not a natural isomorphism So a cochain is 0 hom(c n, Z) hom(c n 1, Z) hom(c 1, Z) hom(c 0, Z) 0 Traditionally, we write the map as δ = and call it the coboundary We have δδ = 0 And so, we define H k (X) = ker δ/ Im δ We often denote hom(c n, Z) as C n Why did we do this? This comes from differential forms on manifolds A p-form is useful, because it can be integrated over a p-simplex to obtain a real number p α That is, α C p (M, R) For differential forms, there is a map d from p-forms to (p + 1)-forms called the exterior differential The big property is that dα = α, called Stokes Theorem If α Cp is not necessarily a α p-form, then what is δα? We have C p+1 Cp R So δα = α That is, δα( ) = ( 1) i α( i ) We hope that H k (X) = hom(h k (X), Z) Example: X = RP 2 We have 0 Z x2 Z 0 Z 0, the chain complex C (X) C (X) is 0 Z x2 Z 0 Z 0, so H 0 (X) = Z H 1 (X) = 0 and H 2 (X) = Z 2 So our hopes failbut we did get the same groups, just in the wrong places For S 2, we have C is 0 Z 0 Z 0, so C is 0 Z 0 Z 0, so the cohomology groups are the same as the homology groups In fact, this is a general phenomenon, for finitely generated abelian groups, the free part stays when switching to cohomology, but the torsion part switches 26