Math 752 Week 13

1 Homotopy Groups

Definition 1. For n ≥ 0 and X a topological space with x0 ∈ X, define

n n πn(X) = {f :(I , ∂I ) → (X, x0)}/ ∼ where ∼ is the usual homotopy of maps.

Then we have the following diagram of sets:

f (In, ∂In) / (X, x ) 6 0 g  (In/∂In, ∂In/∂In)

n n n n n Now we have (I /∂I , ∂I /∂I ) ' (S , s0). So we can also define

n πn(X, x0) = {g :(S , s0) → (X, x0)}/ ∼ where g is as above.

0 Remark 1. If n = 0, then π0(X) is the set of connected components of X. Indeed, we have I = pt 0 and ∂I = ∅, so π0(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 πn(X, x0) is a group with respect to the following operation +:   1 f(2s1, s2, . . . , sn) 0 ≤ s1 ≤ 2 (f + g)(s1, s2, . . . , sn) =  1 g(2s1 − 1, s2, . . . , sn) 2 ≤ s1 ≤ 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 π1 is a group is valid here as well. Then the identity element is the constant map n taking all of I to x0 and inverses are given by −f = f(1 − s1, s2, . . . , sn).

1 2

Proposition 2. If n ≥ 2, then πn(X, x0) 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 ∈ πn(X, x0). 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 In and map the remaining region to the base point x0. Note that this is possible since both f and g map to x0 on the boundaries, so the resulting map is continuous. Then there is enough room to slide f past g inside In. 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 πn(X, x0) is abelian.

n If we view πn(X, x0) as homotopy classes of maps (S , s0) → (X, x0), 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 x0, x1 ∈ X, then there is an isomorphism

βγ : π1(X, x0) → π1(X, x1) 3

where γ is a path from x0 to x1, i.e. we have γ : [0, 1] → X with γ(0) = x0 and γ(1) = x1. This βγ −1 is given by βγ([f]) = [γ · f · γ] for any [f] ∈ π1(X, x0).

We claim the same result holds for all n ≥ 1.

Proposition 3. If n ≥ 1, then there is βγ : πn(X, x1) → πn(X, x0) given by βγ([f]) = [γ · β] where

γ · f denotes the construction given below and γ is a path from x1 to x0.

Construction: We shrink the domain of f to a smaller cube inside In, then insert γ radially from x1 to x0 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 x0 to x1

3. cx0 · f ' f where cx0 denotes the constant path based at x0.

4. βγ is well-defined with respect to homotopies of γ or f.

Now (1) implies βγ is a group homomorphism and (2) and (3) show that βγ is invertible. Indeed, −1 if γ(t) = γ(1 − t), then βγ = βγ. So, as with n = 1, if the space X is path-connected, then πn is independent of the choice of base point. Further, if x0 = x1, then (2) and (3) also imply that

π1(X, x0) acts on πn(X, x0):

π1 × πn → πn

(γ, [f]) 7→ [γ · f] 4

Definition 2. We say X is an abelian space if π1 acts trivially on πn for all n ≥ 1.

In particular, this means π1 is abelian, since the action of π1 on π1 is by inner-automorphisms, which must all be trivial.

2 2 Exercise 1. Compute π3(S ) and π2(S ).

In general, computing the homotopy groups of spheres is a difficult problem. It turns out that we 2 2 have π3(S ) = Z and π2(S ) = Z. We will discuss this further at a later point, and for now only 2 2 2 remark that since any map f : S → S has an integral degree, we have a map π2(S ) → Z given 3 2 2 by [f] 7→ deg f. Also, the Hopf map f : S → S is an element of π3(S ).

We next observe that πn is a functor.

Claim 1. A map φ : X → Y induces group homomorphisms φ∗ : πn(X, x0) → πn(Y, φ(x0)) given by [f] 7→ [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 ψt is a homotopy between f and g, then φ ◦ ψt 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. (idX )∗ = idπn(X,x0)

Corollary 1. If φ :(X, x0) → (Y, y0) is a homotopy equivalence, then φ∗ is an isomorphism.

Proposition 4. If p : Xe → X is a covering map, then p∗ : πn(X,e xe) → πn(X, x) (where x = p(xe)) is an isomorphism for all n ≥ 2.

We will delay the proof of this proposition and first consider several examples.

n n n Example 1. Consider R (or any contractible space). We have πi(R ) = 0 for all i ≥ 1, since R is homotopy equivalent to a point.

1 1 2πit Example 2. Consider S . We have the usual covering map p : R → S given by p(t) = e . We 1 1 already know π1(S ) = Z. Now if n ≥ 2, then we have πn(S ) = πn(R) = 0.

n 1 1 1 n n Example 3. Consider T = S × S × · · · × S n-times. We have π1(T ) = Z . Now there is a n n n n covering map p : R → T , so we have πi(T ) = πi(R ) = 0 for i ≥ 2. 5

Definition 3. If πn(X) = 0 for all n ≥ 2, then X is called aspherical.

We now return to the previous proposition.

n Proof. (of Proposition 4) First we claim p∗ is surjective. Let f :(S , s0) → (X, x). Now since n n n ≥ 2, we have π1(S ) = 0, so f∗(π1(S , s0)) = 0 ⊂ p∗(π1(X,e xe)) and thus f satisfies the lifting n criterion. So there is fe :(S , s0) → (X,e xe) such that p ◦ fe = f. Then [f] = [p ◦ fe] = p∗([fe]). Thus p∗ is surjective.

f (Sn, s ) / (X, x) 0 O p fe $ (X,e xe)

Now we claim p∗ is injective. Suppose [fe] ∈ ker p∗. So p∗([fe]) = [p ◦ fe] = 0. Then p ◦ fe ' cx via n some homotopy φt :(S , s0) → (X, x0) with φ1 = f and φ0 = cx. Let p ◦ fe = f. Again, by the n lifting criterion, there is a unique φet :(S , s0) → (X,e xe) with p ◦ φet = φt.

φ (Sn, s ) t / (X, x) 0 O p φet $ (X,e xe)

Then we have p ◦ φe1 = φ1 = f and p ◦ φe0 = cx, so by the uniqueness of lifts, we must have φe1 = fe and φ = c . Then φ is a homotopy between f and c . So [f] = 0. Thus p is injective. e0 xe et e xe e ∗

Proposition 5. Let {Xα}α be a collection of path-connected spaces. Then ! Y ∼ Y πn Xα = πn(Xα) α α for all n.

Q Proof. First note that a map f : Y → α Xα is a collection of maps fα : Y → Xα. For elements of n πn, take Y = S (note that since all spaces are path-connected, we may drop the reference to base points). For homotopy, take Y = Sn × I.

∼ Exercise 2. Find two spaces X,Y such that πn(X) = πn(Y ) for all n but X and Y are not homotopy equivalent. 6

Recall that Whitehead’s Theorem states that if a map of CW complexes f : X → Y induces isomorphisms on all πn, then f is a homotopy equivalence. So we must find X,Y such that there 2 3 is no continuous map f : X → Y inducing the isomorphisms on πn’s. Consider X = S × RP and 2 3 2 3 2 3 3 3 Y = RP × S . Then πn(X) = πn(S × RP ) = πn(S ) × πn(RP ). Now S is a covering of RP , 2 3 2 3 so for n ≥ 2, we have πn(X) = πn(S ) × πn(S ). We also have π1(X) = π1(S ) × π1(RP ) = Z2. 2 3 2 3 2 2 Similarly, we have πn(Y ) = πn(RP × S ) = πn(RP ) × πn(S ). Now S is a covering of RP , so 2 3 2 3 for n ≥ 2, we have πn(Y ) = πn(S ) × πn(S ). We also have π1(Y ) = π1(RP ) × π1(S ) = Z2. So

πn(X) = πn(Y ) for all n. By considering homology groups, however, we see X 6' Y . Indeed, we 3 2 have H5(X) = Z while H5(Y ) = 0 (since RP is oriented while RP is not).

n n Theorem 1. (Hurewicz) There is a map πn(X) → Hn(X) given by [f : S → X] 7→ f∗[S ] where n n [S ] is the fundamental class of S . Then if πi(X) = 0 for all i < n where n ≥ 2, then Hi(X) = 0 ∼ for i < n and πn(X) = Hn(X).

Corollary 2. If X and Y are CW complexes with π1(X) = π1(Y ) = 0 and f : X → Y induces isomorphisms on all Hn, 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 6' Y ).

2 Relative Homotopy Groups

Given a triple (X, A, x0) where x0 ∈ A ⊂ X, we wish to define the relative homotopy groups.

n−1 n Definition 4. Let X be a space and let A ⊂ X and x0 ∈ A. Let I = {(s1, . . . , sn) ∈ I : sn = 0} and let J n−1 = ∂In\In−1. Then define

n n n−1 πn(X, A, x0) = {f :(I , ∂I ,J ) → (X, A, x0)}/ ∼ where, as before, ∼ refers to homotopy equivalence. 7

n n−1 Alternatively, we can take πn(X, A, x0) = {g :(D ,S , s0) → (X, A, x0)}/ ∼.

Proposition 6. If n ≥ 2, then πn(X, A, x0) forms a group under the usual operation. Further, if n ≥ 3, then πn(X, A, x0) is abelian.

Remark 3. Note that the proposition fails in the case n = 1. Indeed, we have

π1(X, A, x0) = {f :(I, {0, 1}, {1}) → (X, A, x0)}/ ∼ .

Then π1(X, A, x0) consists of paths starting anywhere A and ending at x0, so we cannot always concatenate two paths.

As before, a map φ :(X, A, x0) → (Y, B, y0) induces homomorphisms φ∗(X, A, x0) → (Y, B, y0) for all n.

Proposition 7. The relative homotopy groups of (X, A, x0) fit into a long exact sequence

∂ · · · → πn(A, x0) → πn(X, x0) → πn(X, A, x0) −→ πn−1(A, x0) → · · · → π0(A, x0) → π0(X, x0) → 0

(note that the final terms of the sequence are not groups, just sets) and the map ∂ is defined by

∂[f] = [f|In−1 ] and all others are induced by inclusions.

Now we wish to describe 0 ∈ πn(X, A, x0). Recall that for [f] ∈ πn(X, x0), we had [f] = 0 iff f is n homotopic to cx0 . Equivalently, for f :(S , s0) → (X, x0), we have [f] = 0 iff f extends to a map n+1 D → X. (Note that the proof is the same as for π1).

Lemma 1. Let [f] ∈ πn(X, A, x0). Then [f] = 0 iff f ' g for some map g with im g ⊂ A. 8

Proof. (⇐) Suppose f ' g for some g with im g ⊂ A.

n n−1 Then we can deform I to J as above and so g ' cx0 . Since homotopy is a transitive relation, we have f ' cx0 . n+1 (⇒) Suppose [f] = 0 in πn(X, A, x0). So f ' cx0 via some homotopy F : I → X. Then we may deform In inside In+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 π1(X) acts on πn(X) for all n and πn(X) is independent of our choice of base point. Now we can relax to the following if x0, x1 ∈ A and im γ ⊂ A: 9

Lemma 2. If A is path-connected, then βγ : πn(X, A, x0) → πn(X, A, x1) is an isomorphism, where

γ is a path in A from x0 to x1.

Remark 5. If x0 = x1, we get an action of π1(A) on πn(X,A).

Definition 5. We say (X,A) is n-connected if πi(X,A) = 0 for i ≤ n and X is n-connected if

πi(X) = 0 for i ≤ n.

Example 4. 0-connected ⇒ connected and 1-connected ⇒ simply-connected ⇒ connected.

3 Homotopy Groups of Spheres

n Now we turn our attention to computing πi(S ). 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 n n n i = n, we should have πn(S ) = Z by associating to each map f : S → S its degree. For i < n, n i n n i we have πi(S ) = 0. If f : S → S is not surjective, i.e. there is y ∈ S \f(S ), and so f factors n n through R , which is contractible. So we compose f with the retraction R → x0 and so f ' cx0 . BUT there are surjective maps Si → Sn 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(Xn) ⊂ Y n for all n (here Xn denotes the n-skeleton of X).

Theorem 2. Any map between CW-complexes can be homotoped to a cellular map.

n Corollary 3. For i < n, we have πi(S ) = 0.

Proof. Choose the standard CW-structure on Si and Sn. We may assume f : Si → Sn is cellular. Then f(Si) ⊂ (Sn)i. But (Sn)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 πi(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 :(Di,Si−1) → (X,A) to g with g(Di) ⊂ Xi. But Xi ⊂ A, so im g ⊂ A and thus [f] = 0.

n Example 5. πi(X,X ) = 0 for i ≤ n

n Corollary 5. For i < n, we have πi(X) = πi(X ). 10

Proof. Consider the long exact sequence of the pair (X,A).

Theorem 3. (Suspension Theorem) Let f : Si → Sn and consider its suspension,

Σf :ΣSi = Si+1 → ΣSn = Sn+1.

n n+1 Then from [f] ∈ πi(S ), we can create [Σf] ∈ πi+1(S ). Then this association creates an n ∼ n+1 isomorphism πi(S ) = πi+1(S ) for i < 2n − 1 and a surjection for i = 2n − 1.

n Corollary 6. πn(S ) 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:

1 2 ∼ 3 ∼ 4 ∼ π1(S )  π2(S ) = π3(S ) = π4(S ) = ···

1 ∼ 2 To show π1(S ) = π2(S ), 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  o

E

 B

· · · → πi(F ) → πi(E) → πi(B) → πi−1(F ) → · · ·

f Applying the above to the Hopf fibration S1 ,→ S3 −→ S2, we have

1 3 2 1 3 · · · → π2(S ) → π2(S ) → π2(S ) → π1(S ) → π1(S ) → · · ·

1 2 1 ··· π2(S ) → 0 → π2(S ) → π1(S ) → 0 → · · ·

2 ∼ 1 Thus π2(S ) = π1(S ) = Z.

Remark 6. The above long exact sequence also yields that

2 ∼ 2 ∼ π3(S ) = π2(S ) = Z

Remark 7. Unlike the homology and cohomology groups, the homotopy groups of a finite CW- complex can be infinitely generated. 11

1 n 1 n 1 n 1 n Example 6. For n ≥ 2, consider S ∨ S . We have πn(S ∨ S ) = πn(S^∨ S ), where S^∨ S is the universal cover of S1 ∨ Sn, depicted below:

By contracting the segments between integers, we have S^1 ∨ Sn ' W Sn. So π (S1 ∨ Sn) = k∈Z k n π (W Sn), which is free abelian generated by the inclusions Sn ,→ W Sn. n k∈Z k k k∈Z k

Note: Several images taken from Hatcher