Math 752 Week s 1 1

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1 Math 752 Week 13 1 Homotopy Groups Definition 1. For n 0 and X a topological space with x 0 X, define π n (X) = {f : (I n, I n ) (X, x 0 )}/ where is the usual homotopy of maps. Then we have the following diagram of sets: (I n, I n ) f g (I n / I n, I n / I n ) (X, x 0 ) Now we have (I n / I n, I n / I n ) (S n, s 0 ). So we can also define π n (X, x 0 ) = {g : (S n, s 0 ) (X, x 0 )}/ where g is as above. Remark 1. If n = 0, then π 0 (X) is the set of connected components of X. Indeed, we have I 0 = pt and I 0 =, so π 0 (X) consists of homotopy classes of maps from a point into the space X. Now we will prove several results analogous to the n = 1 case. Proposition 1. If n 1, then π n (X, x 0 ) is a group with respect to the following operation +: f(2s 1, s 2,..., s n ) 0 s (f + g)(s 1, s 2,..., s n ) = 1 g(2s 1 1, s 2,..., s n ) 2 s 1 1 (Note that if n = 1, this is the usual concatenation of paths/loops.) Proof. First note that since only the first coordinate is involved in this operation, the same argument used to prove π 1 is a group is valid here as well. Then the identity element is the constant map taking all of I n to x 0 and inverses are given by f = f(1 s 1, s 2,..., s n ). 1

2 2 Proposition 2. If n 2, then π n (X, x 0 ) is abelian. Intuitively, since the + operation only involves the first coordinate, there is enough space to slide f past g. Proof. Let n 2 and let f, g π n (X, x 0 ). We wish to show f + g g + f. Consider the following figures: We first shrink the domains of f and g to smaller cubes inside I n and map the remaining region to the base point x 0. Note that this is possible since both f and g map to x 0 on the boundaries, so the resulting map is continuous. Then there is enough room to slide f past g inside I n. We then enlarge the domains of f and g back to their original size and get g + f. So we have constructed a homotopy between f + g and g + f and hence π n (X, x 0 ) is abelian. If we view π n (X, x 0 ) as homotopy classes of maps (S n, s 0 ) (X, x 0 ), then we have the following visual representation of f + g (one can see this by collapsing boundaries in the cube interpretation). Now recall that if X is path-connected and x 0, x 1 X, then there is an isomorphism β γ : π 1 (X, x 0 ) π 1 (X, x 1 )

3 3 where γ is a path from x 0 to x 1, i.e. we have γ : [0, 1] X with γ(0) = x 0 and γ(1) = x 1. This β γ is given by β γ ([f]) = [γ 1 f γ] for any [f] π 1 (X, x 0 ). We claim the same result holds for all n 1. Proposition 3. If n 1, then there is β γ : π n (X, x 1 ) π n (X, x 0 ) given by β γ ([f]) = [γ β] where γ f denotes the construction given below and γ is a path from x 1 to x 0. Construction: We shrink the domain of f to a smaller cube inside I n, then insert γ radially from x 1 to x 0 on the boundaries of these cubes. Then we have the following results: 1. γ (f + g) γ f + γ g 2. (γ η) f γ (η f) if η is another path from x 0 to x 1 3. c x0 f f where c x0 denotes the constant path based at x β γ is well-defined with respect to homotopies of γ or f. Now (1) implies β γ is a group homomorphism and (2) and (3) show that β γ is invertible. Indeed, if γ(t) = γ(1 t), then β 1 γ = β γ. So, as with n = 1, if the space X is path-connected, then π n is independent of the choice of base point. Further, if x 0 = x 1, then (2) and (3) also imply that π 1 (X, x 0 ) acts on π n (X, x 0 ): π 1 π n π n (γ, [f]) [γ f]

4 4 Definition 2. We say X is an abelian space if π 1 acts trivially on π n for all n 1. In particular, this means π 1 is abelian, since the action of π 1 on π 1 is by inner-automorphisms, which must all be trivial. Exercise 1. Compute π 3 (S 2 ) and π 2 (S 2 ). In general, computing the homotopy groups of spheres is a difficult problem. It turns out that we have π 3 (S 2 ) = Z and π 2 (S 2 ) = Z. We will discuss this further at a later point, and for now only remark that since any map f : S 2 S 2 has an integral degree, we have a map π 2 (S 2 ) Z given by [f] deg f. Also, the Hopf map f : S 3 S 2 is an element of π 3 (S 2 ). We next observe that π n is a functor. Claim 1. A map φ : X Y induces group homomorphisms φ : π n (X, x 0 ) π n (Y, φ(x 0 )) given by [f] [φ f] for all n 1. Proof. From the definition of +, it is clear that we have φ (f + g) = (φ f) + (φ g). So φ ([f + g]) = φ ([f]) + φ ([g]). Further, if f g, then φ f φ g. Indeed, if ψ t is a homotopy between f and g, then φ ψ t is a homotopy between φ f and φ g. Hence φ is a group homomorphism. Remark 2. Induced homomorphisms satisfy the following two properties: 1. (φ ψ) = φ ψ 2. (id X ) = id πn(x,x0 ) Corollary 1. If φ : (X, x 0 ) (Y, y 0 ) is a homotopy equivalence, then φ is an isomorphism. Proposition 4. If p : X X is a covering map, then p : π n ( X, x) π n (X, x) (where x = p( x)) is an isomorphism for all n 2. We will delay the proof of this proposition and first consider several examples. Example 1. Consider R n (or any contractible space). We have π i (R n ) = 0 for all i 1, since R n is homotopy equivalent to a point. Example 2. Consider S 1. We have the usual covering map p : R S 1 given by p(t) = e 2πit. We already know π 1 (S 1 ) = Z. Now if n 2, then we have π n (S 1 ) = π n (R) = 0. Example 3. Consider T n = S 1 S 1 S 1 n-times. We have π 1 (T n ) = Z n. Now there is a covering map p : R n T n, so we have π i (T n ) = π i (R n ) = 0 for i 2.

5 5 Definition 3. If π n (X) = 0 for all n 2, then X is called aspherical. We now return to the previous proposition. Proof. (of Proposition 4) First we claim p is surjective. Let f : (S n, s 0 ) (X, x). Now since n 2, we have π 1 (S n ) = 0, so f (π 1 (S n, s 0 )) = 0 p (π 1 ( X, x)) and thus f satisfies the lifting criterion. So there is f : (S n, s 0 ) ( X, x) such that p f = f. Then [f] = [p f] = p ([ f]). Thus p is surjective. (S n, s 0 ) f f (X, x) p ( X, x) Now we claim p is injective. Suppose [ f] ker p. So p ([ f]) = [p f] = 0. Then p f c x via some homotopy φ t : (S n, s 0 ) (X, x 0 ) with φ 1 = f and φ 0 = c x. Let p f = f. Again, by the lifting criterion, there is a unique φ t : (S n, s 0 ) ( X, x) with p φ t = φ t. (S n, s 0 ) φ t φ t (X, x) p ( X, x) Then we have p φ 1 = φ 1 = f and p φ 0 = c x, so by the uniqueness of lifts, we must have φ 1 = f and φ 0 = c x. Then φ t is a homotopy between f and c x. So [ f] = 0. Thus p is injective. Proposition 5. Let {X α } α be a collection of path-connected spaces. Then ( ) π n X α = π n (X α ) α α for all n. Proof. First note that a map f : Y α X α is a collection of maps f α : Y X α. For elements of π n, take Y = S n (note that since all spaces are path-connected, we may drop the reference to base points). For homotopy, take Y = S n I. Exercise 2. Find two spaces X, Y such that π n (X) = π n (Y ) for all n but X and Y are not homotopy equivalent.

6 6 Recall that Whitehead s Theorem states that if a map of CW complexes f : X Y induces isomorphisms on all π n, then f is a homotopy equivalence. So we must find X, Y such that there is no continuous map f : X Y inducing the isomorphisms on π n s. Consider X = S 2 RP 3 and Y = RP 2 S 3. Then π n (X) = π n (S 2 RP 3 ) = π n (S 2 ) π n (RP 3 ). Now S 3 is a covering of RP 3, so for n 2, we have π n (X) = π n (S 2 ) π n (S 3 ). We also have π 1 (X) = π 1 (S 2 ) π 1 (RP 3 ) = Z 2. Similarly, we have π n (Y ) = π n (RP 2 S 3 ) = π n (RP 2 ) π n (S 3 ). Now S 2 is a covering of RP 2, so for n 2, we have π n (Y ) = π n (S 2 ) π n (S 3 ). We also have π 1 (Y ) = π 1 (RP 2 ) π 1 (S 3 ) = Z 2. So π n (X) = π n (Y ) for all n. By considering homology groups, however, we see X Y. Indeed, we have H 5 (X) = Z while H 5 (Y ) = 0 (since RP 3 is oriented while RP 2 is not). Theorem 1. (Hurewicz) There is a map π n (X) H n (X) given by [f : S n X] f [S n ] where [S n ] is the fundamental class of S n. Then if π i (X) = 0 for all i < n where n 2, then H i (X) = 0 for i < n and π n (X) = H n (X). Corollary 2. If X and Y are CW complexes with π 1 (X) = π 1 (Y ) = 0 and f : X Y induces isomorphisms on all H n, then f is a homotopy equivalence. Exercise 3. Find X, Y with the same homology groups, cohomology groups, and cohomology rings, but with different homotopy groups (thus implying X Y ). 2 Relative Homotopy Groups Given a triple (X, A, x 0 ) where x 0 A X, we wish to define the relative homotopy groups. Definition 4. Let X be a space and let A X and x 0 A. Let I n 1 = {(s 1,..., s n ) I n : s n = 0} and let J n 1 = I n \I n 1. Then define π n (X, A, x 0 ) = {f : (I n, I n, J n 1 ) (X, A, x 0 )}/ where, as before, refers to homotopy equivalence.

7 7 Alternatively, we can take π n (X, A, x 0 ) = {g : (D n, S n 1, s 0 ) (X, A, x 0 )}/. Proposition 6. If n 2, then π n (X, A, x 0 ) forms a group under the usual operation. Further, if n 3, then π n (X, A, x 0 ) is abelian. Remark 3. Note that the proposition fails in the case n = 1. Indeed, we have π 1 (X, A, x 0 ) = {f : (I, {0, 1}, {1}) (X, A, x 0 )}/. Then π 1 (X, A, x 0 ) consists of paths starting anywhere A and ending at x 0, so we cannot always concatenate two paths. As before, a map φ : (X, A, x 0 ) (Y, B, y 0 ) induces homomorphisms φ (X, A, x 0 ) (Y, B, y 0 ) for all n. Proposition 7. The relative homotopy groups of (X, A, x 0 ) fit into a long exact sequence π n (A, x 0 ) π n (X, x 0 ) π n (X, A, x 0 ) π n 1 (A, x 0 ) π 0 (A, x 0 ) π 0 (X, x 0 ) 0 (note that the final terms of the sequence are not groups, just sets) and the map is defined by [f] = [f I n 1] and all others are induced by inclusions. Now we wish to describe 0 π n (X, A, x 0 ). Recall that for [f] π n (X, x 0 ), we had [f] = 0 iff f is homotopic to c x0. Equivalently, for f : (S n, s 0 ) (X, x 0 ), we have [f] = 0 iff f extends to a map D n+1 X. (Note that the proof is the same as for π 1 ). Lemma 1. Let [f] π n (X, A, x 0 ). Then [f] = 0 iff f g for some map g with im g A.

8 8 Proof. ( ) Suppose f g for some g with im g A. Then we can deform I n to J n 1 as above and so g c x0. Since homotopy is a transitive relation, we have f c x0. ( ) Suppose [f] = 0 in π n (X, A, x 0 ). So f c x0 via some homotopy F : I n+1 X. Then we may deform I n inside I n+1 (while fixing the boundary) to J n. Composing with F, we get a homotopy from f to g with im g A. Remark 4. The above shows that when proving [f] = 0, it is sufficient to show f g for such a g. We showed that if X is path-connected, then π 1 (X) acts on π n (X) for all n and π n (X) is independent of our choice of base point. Now we can relax to the following if x 0, x 1 A and im γ A:

9 9 Lemma 2. If A is path-connected, then β γ : π n (X, A, x 0 ) π n (X, A, x 1 ) is an isomorphism, where γ is a path in A from x 0 to x 1. Remark 5. If x 0 = x 1, we get an action of π 1 (A) on π n (X, A). Definition 5. We say (X, A) is n-connected if π i (X, A) = 0 for i n and X is n-connected if π i (X) = 0 for i n. Example 4. 0-connected connected and 1-connected simply-connected connected. 3 Homotopy Groups of Spheres Now we turn our attention to computing π i (S n ). For i n, i = n + 1, n + 2, n + 3 and a few more cases, this is known. In general, however, this is a very difficult problem. As noted before, for i = n, we should have π n (S n ) = Z by associating to each map f : S n S n its degree. For i < n, we have π i (S n ) = 0. If f : S i S n is not surjective, i.e. there is y S n \f(s i ), and so f factors through R n, which is contractible. So we compose f with the retraction R n x 0 and so f c x0. BUT there are surjective maps S i S n for i < n, in which case the above proof fails. In this case, we can alter f to make it cellular. Definition 6. Let X, Y be CW-complexes. Then a map f : X Y is called cellular if f(x n ) Y n for all n (here X n denotes the n-skeleton of X). Theorem 2. Any map between CW-complexes can be homotoped to a cellular map. Corollary 3. For i < n, we have π i (S n ) = 0. Proof. Choose the standard CW-structure on S i and S n. We may assume f : S i S n is cellular. Then f(s i ) (S n ) i. But (S n ) i is a point, so f is a constant map. Corollary 4. Let A X and suppose all cells of X \ A have dimension > n. Then π i (X, A) = 0 for i n. Proof. Note that there is a similar cellular approximation result for the relative case. Then we may homotope f : (D i, S i 1 ) (X, A) to g with g(d i ) X i. But X i A, so im g A and thus [f] = 0. Example 5. π i (X, X n ) = 0 for i n Corollary 5. For i < n, we have π i (X) = π i (X n ).

10 10 Proof. Consider the long exact sequence of the pair (X, A). Theorem 3. (Suspension Theorem) Let f : S i S n and consider its suspension, Σf : ΣS i = S i+1 ΣS n = S n+1. Then from [f] π i (S n ), we can create [Σf] π i+1 (S n+1 ). Then this association creates an isomorphism π i (S n ) = π i+1 (S n+1 ) for i < 2n 1 and a surjection for i = 2n 1. Corollary 6. π n (S n ) is either Z or a finite quotient of Z (for n 2), generated by the degree map. Proof. By the Suspension Theorem, we have the following: π 1 (S 1 ) π 2 (S 2 ) = π 3 (S 3 ) = π 4 (S 4 ) = To show π 1 (S 1 ) = π 2 (S 2 ), we can use the long exact sequence for the homotopy groups of a fibration. (Note: Covering maps are a good example of a fibration for F discrete). F E B π i (F ) π i (E) π i (B) π i 1 (F ) Applying the above to the Hopf fibration S 1 S 3 f S 2, we have π 2 (S 1 ) π 2 (S 3 ) π 2 (S 2 ) π 1 (S 1 ) π 1 (S 3 ) π 2 (S 1 ) 0 π 2 (S 2 ) π 1 (S 1 ) 0 Thus π 2 (S 2 ) = π 1 (S 1 ) = Z. Remark 6. The above long exact sequence also yields that π 3 (S 2 ) = π 2 (S 2 ) = Z Remark 7. Unlike the homology and cohomology groups, the homotopy groups of a finite CWcomplex can be infinitely generated.

11 11 Example 6. For n 2, consider S 1 S n. We have π n (S 1 S n ) = π n ( S 1 S n ), where S 1 S n is the universal cover of S 1 S n, depicted below: By contracting the segments between integers, we have S1 S n k Z Sn k. So π n(s 1 S n ) = π n ( k Z Sn k ), which is free abelian generated by the inclusions Sn k k Z Sn k. Note: Several images taken from Hatcher