Geometric proof of the Hurewicz Theorem

Appendix: the equivariant case

4.5 Geometric proof of the Hurewicz Theorem

Proposition 4.5.3. (i)The following diagram commutes

· · · //_πn(A, x) ^i# //

h_n

πⁿ(X, x) ^j# //

h_n

πⁿ(X, A, x) ^∂# //

h_n

πⁿ−1(A, x) //

h_n−1

· · ·

· · · //_Hn(A;Z) ^i∗ //_Hn(X;Z) ^j∗ //_Hn(X, A;Z) ^∂∗ //_Hn−1(A;Z) //_{· · ·}

(ii)If f : (X, A, x)→(Y, B, y) is a map, the following diagram commutes:

πn(X, A, x) −−−−→^f# πn(Y, B, y)

⏐⏐

^hn ⏐⏐^hn Hn(X, A;Z) −−−−→^f∗ Hn(Y, B;Z).

It is convenient to prove 4.5.2 first, as one case of that theorem is needed in the proof of 4.5.1.

Proof (of 4.5.2). We first deal with the special case in which (X, x) is a bouquet of n-spheres,

α

S_αⁿ, w

. We are to prove h_n : π_n

α

S_αⁿ, w

→ Hn

α

Sⁿ_α

∼=

α

Zhas trivial kernel, since the rest of the theorem is obvi- ous in this case. We pause for some preparations.

Recall from Sect. 4.4 the definition of anisland onSⁿ; loosely, it is a nicely placedn-ball. Anarchipelago inSⁿ is a finite set of pairwise disjoint islands.

The mapf : Sⁿ →

α∈A

S_αⁿ isconcentrated on the archipelago{Wγ | γ ∈ C}

if (i) for each γ ∈ C there is some α ∈ A such that f(Wγ) ⊂ S_αⁿ, and (ii) f

'

Sⁿ−'

γ

W_γ ((

={w}.

Proposition 4.5.4.For any map f : (Sⁿ, v) →

α∈A

S_αⁿ, w

there is an archipelago {Wγ |γ∈ C}in Sⁿ such that f is homotopic rel{v} to a mapg concentrated on{Wγ}.

Proof. For each α∈ A, pickz_α= v lying in S_αⁿ. By 1.3.7, there is a neigh- borhoodN_α ofz_α ∈S_αⁿ− {w} and a homotopyD^α: S_αⁿ×I→S_αⁿ, rel {v}, such thatD^α₀ = id,D₁^α(V_α) ={w}andD^α₁(N_α) =S_αⁿ, whereV_α:=S_αⁿ−N_α.

4.5 Geometric proof of the Hurewicz Theorem 121

LetU =∪{V_α|α∈ A}. ThenU :={U} ∪

%

α

S_αⁿ− {v}

&

is an open cover of

α∈A

S_αⁿ. The homotopies D^α give a homotopyD :

α∈A

S_αⁿ

×I →

α∈A

S_αⁿ, rel{v}, such thatD₀ = id,D_t(S_αⁿ) =S_αⁿ for all α∈ A and 0 ≤t≤1, and D₁(U) ={w}.

We will use I^•n+1 as the domain of f. We give

α∈A

Sⁿ_α the CW complex structure consisting of one vertex,w, and ann-cell,S_αⁿ, for eachα. By 2.4.13, there existsk such that each cell ofI^•n+1_k lies in some element off⁻¹U. f is homotopic toD1◦f which maps eachn-cell ofI^•n+1_k intoS_αⁿ for someα. By 1.4.4,D1◦f is homotopic to a cellular map f such thatf takes the entire (n−1)-skeleton ofI^•n+1 tow, and maps eachn-cell ofI^•n+1_k into someS_αⁿ.

The n-cells of I^•n+1_k are not pairwise disjoint, so they do not form an archipelago. We remedy this by passing toI^•n+1_k+2. The effect is to partition each n-celleⁿ_γ ofI^•n+1_k into 4ⁿn-cells, 2ⁿ of which are disjoint frome^•n_γ. The union of these 2ⁿcells is ann-cell ˜eⁿ_γ, and there⁸is a homotopyH:eⁿ_γ×I→eⁿ_γ such thatH₀= id,H₁(eⁿ_γ −e˜ⁿ_γ)⊂^•eⁿ_γ andH₁(˜eⁿ_γ) =eⁿ_γ. The required archipelago is{e˜ⁿ_γ |γ∈ C}whereC indexes then-cells of I^•n+1_k . The required mapg is of

f◦H1.

Proposition 4.5.5. Let n ≥ 2, let A be finite and let the map f : Sⁿ →

α∈A

S_αⁿ be concentrated on the archipelago {Wγ |γ∈ C}. Thenf is homotopic to a mapgconcentrated on an archipelago{V_α|α∈ A}such thatg(V_α)⊂S_αⁿ for eachα∈ A.

Proof. Pick an archipelago {Vα | α ∈ A} in Sⁿ. By a simple application of 2.4.15, there is a homeomorphismh : Sⁿ → Sⁿ which is homotopic to the identity map, such that whenever f is non-constant on the island W_γ and mapsW_γ intoS_αⁿ thenh(W_γ)⊂V_α. The required mapg isf◦h⁻¹. Proof (of 4.5.2 (concluded)). Let f : (Sⁿ, v) →

α∈A

S_αⁿ, w

represent an element of ker(hn). Since Sⁿ is compact we may proceed as ifAwere finite.

By 4.5.4 and 4.5.5 there is an archipelago{Vα} so thatf(Vα)⊂S_αⁿ for each α∈ A. The mapping degree [Sⁿ :Sⁿ_α:f] is the “degree” with which f maps V_α ontoS_αⁿ, namely 0 since [f] ∈ker(h_n). By 2.4.5,f | V_α is homotopic rel bdV_α to the constant map atv, for eachα. Thus the special case is proved.

8 The details here just involve elementary games with product cells inI_kⁿ₊₂⁺¹.

We use this to prove the general case of 4.5.2. By induction onn, we may assumeX is (n−1)-connected whenn≥2; this is trivial whenn= 1. Clearly hn :πn(X, x)→Hn(X;Z) is surjective (exercise). We are to prove ker(hn) is trivial. Consider the commutative diagram

πn(Xⁿ, x) −−−−→^i# πn(X, x)

∼=

⏐⏐ ^hn

⏐⏐ ^hn Hn(Xⁿ;Z) −−−−→^i∗ Hn(X;Z)

whereidenotes the inclusion map. Here, we have renamed one of the Hurewicz homomorphismsh_n to distinguish it fromhn. Applying 4.2.1 inductively to the (n−1)-connected pair (X, x), we may assume Xⁿ⁻¹ = {x}. The n- cells [resp. (n+ 1)-cells] of X are {eⁿ_α | α ∈ A} [resp. {eⁿ⁺¹_β | β ∈ B}].

Thus Xⁿ =

α

eⁿ_α =

α

S_αⁿ. Let gβ : Sⁿ → Xⁿ be the attaching map for eⁿ⁺¹_β . We may assume g_β(v) = x, so g_β defines [g_β] ∈ π_n(Xⁿ, x). We have h_n([gβ]) =

%

α

[eⁿ⁺¹_β :eⁿ_α]eⁿ_α

&

. Letf : (Sⁿ, v)→(X, x) represent an element [f] ∈ker(hn)≤πn(X, x). Write f : (Sⁿ, v)→ (Xⁿ, x) for the corestriction of f. In the above diagram, hni#([f]) = hn([f]) = 0, so h_n([f]) ∈ keri_∗. Thus there are integers n_β such that h_n([f]) =

⎧⎨

⎩∂

⎛

⎝

β

n_βeⁿ⁺¹_β

⎞

⎠

⎫⎬

⎭ =

⎧⎨

⎩

β

nβ

α

[eⁿ⁺¹_β :eⁿ_α]eⁿ_α

⎫⎬

⎭=

β

nβh_n([gβ]). Buth_n is an isomorphism by the special case. So [f] =

β

nβ[gβ]. And i#([gβ]) = 0 since gβ extends across

eⁿ⁺¹_β . Soi#([f]) = [f] = 0.

Proof (of 4.5.1). The induction onnbegins withn= 2. By 4.5.3 (i) we have a commutative diagram

π2(A, x) −−−−→ π2(X, x) −−−−→ π2(X, A, x) −−−−→ 0

⏐⏐ ^h2

⏐⏐ ^h2

⏐⏐ ^h2

H2(A;Z) −−−−→ H2(X,Z) −−−−→ H2(X, A;Z) −−−−→ 0

with exact horizontal lines. By 4.5.2, the Hurewicz homomorphismsh₂ and h₂ are isomorphisms. Soh2 is an isomorphism.

Now we proceed by induction, assuming n≥3. As before, the only diffi- culty is in proving ker(hn) = 0. The proof of this is totally analogous to the proof of 4.5.2. We leave the details as an instructive exercise.

4.5 Geometric proof of the Hurewicz Theorem 123 Example 4.5.6.Consider X = S¹∨S² with vertexv. Its universal cover ˜X is a line with a 2-sphere adjoined at each integer point. By 4.5.2 and 4.4.10, H2( ˜X;Z)∼=

∞ n=−∞

Z∼=π2( ˜X,v)˜ ∼=π2(X, v). ThusX is an example of a finite CW complex whose second homotopy group is not finitely generated (as an abelian group). However, since the group of covering transformations of ˜X is infinite cyclic, H2( ˜X;Z) can be regarded as a module over the group ring⁹ ZC where C denotes the infinite cyclic group of covering transformations (∼= π₁(X, v)). This homology group is finitely generated as a ZC-module, indeed cyclic, so the same is true ofπ₂( ˜X,˜v)∼=π₂(X, v). It is often desirable to view higher homotopy groups as modules over the group ringZGwhereG is the fundamental group. In particular, whenπ_n(X, v) is finitely generated as aZG-module one can kill it by attaching finitely many (n+ 1)-cells toX.

Historical Note:The Relative Hurewicz Theorem appeared first in [88].

Exercises

1. Prove thathnis a homomorphism.

2. Prove that under the hypotheses of 4.5.2hn is onto.

3. Fill in the missing details in the proof of 4.5.1.

4. LetXandY be path connected CW complexes. Prove that iff: (X, x)→(Y, y) induces isomorphisms on fundamental group and on Z-homology of universal covers, thenfis a homotopy equivalence. (Hint: See Exercise 1 of Sect. 4.4.) 5. Let X = A∪B be a CW complex where A, B and A∩B are contractible

subcomplexes. Prove thatXis contractible using the Van Kampen theorem, the Mayer-Vietoris sequence, the Hurewicz Theorem and the Whitehead Theorem.

(Compare Exercise 5 in Sect. 4.1.)

6. Find an example of a path connected CW complex X where h₁ is an isomor- phism buth₂ is not surjective.

9 See Sect. 8.1 for more on group rings.

5