LECTURE NOTES 1: CELLULAR HOMOLOGY 1. Introduction We

LECTURE NOTES 1: CELLULAR HOMOLOGY

MATTHEW ANDO

1. Introduction

We show how to use a cellular decomposition of a space X to calculate its homology: that is, that, cellular homology equals singular homology. The argument and result are standard one, and for more details the reader may wish to consult the texts of Bredon or Hatcher. The reason for presenting the argument in these notes is to introduce a style of argument which I shall use throughout the course.

2. NDR pairs

Let X be a space, and let A ⊂ X be a subspace. Recall from [?] that (X,A) is an NDR pair if there are maps u :X → I d :X × I → X such that A = u−1(0) d(x, 0) = x x ∈ X d(a, t) = a a ∈ A d(x, 1) ∈ A u(x) < 1. Let B = u−1([0, 1)). The map d induces a deformation retraction of the inclusion of A in B: A,→ B is a homotopy equivalence. Lemma 2.1. In this situation, the inclusion (X,A) ,→ (X,B) induces isomorphisms. ∼ Hq(X,A) = Hq(X,B).

Proof. The long exact sequences of the pairs give a ladder

H¯qA −−−−→ H¯qX −−−−→ H¯q(X,A)    ∼ ∼  =y =y y

H¯qB −−−−→ H¯qX −−−−→ H¯q(X,B) in which the indicated arrows are isomorphisms. The “5-lemma” gives the result.

Date: Version 1.0. c 2007 by the author. You may copy this document for personal study. 1 ` Now suppose that f : A → Y is a map. Let Z = X f Y be the pushout in the diagram

f A −−−−→ Y     y y g X −−−−→ Z. The following is easy to check. Lemma 2.2. If (X,A) is an NDR pair, then the map

g|X−A : X − A → Z − Y is a homeomorphism.

You can use the deﬁnition of NDR pair and the pushout property to prove the following. I also gave a proof in [?].

Lemma 2.3. If (X,A) is an NDR pair, then so is (Z,Y ). Proposition 2.4. In the situation above, the map

Hqg : Hq(X,A) → Hq(Z,Y ) is an isomorphism for all q.

Proof. Let B = u−1[0, 1) be the deformation neighborhood of A. Note that u˜−1[0, 1) = Y ∪ g(B) is the deformation neighborhood of Y . Lemma 2.1 gives isomorphisms i Hq(X,A) −→ Hq(X,B) ∼= j Hq(Z,Y ) −→ Hq(Z,Y ∪ g(B)). ∼= The virtue of the neighborhood B over A is that we can excise A from it: we have a commutative diagram

∼= Hq(X − A, B − A) −−−−→ Hq(X,B)   H g∼ α q y= y ∼= Hq(Z − Y, g(B − A)) −−−−→ Hq(Z,Y ∪ g(B)). Lemma 2.2 gives the left vertical isomorphism. It follows that α is an isomorphism. Now consider the commutative diagram i Hq(X,A) −−−−→ Hq(X,B) ∼=    ∼α y =y j ` Hq(Z,Y ) −−−−→ Hq(Z,Z g B). ∼= Three of the arrows are isomorphisms, so the fourth is, too.

Applying the proposition in the case that Y = ∗, we get Corollary 2.5. If (X,A) is an NDR pair, then the natural map

Hq(X,A) → Hq(X/A, ∗) = H¯q(X/A) is an isomorphism. 2 This is charming, but if you apply the Barratt-Whitehead Lemma to the diagram

H¯qA −−−−→ H¯qX −−−−→ Hq(X,A) −−−−→ H¯q−1A       ∼  y y y= y

H¯qY −−−−→ H¯qZ −−−−→ Hq(Z,Y ) −−−−→ H¯q−1Y you get a long exact sequence

... H¯qA → H¯qY ⊕ H¯qX → H¯qZ → .... (2.6)

3. The effect on homology of attaching some n-cells.

Let Y be a space, and let

f Sn−1 −→α Y be a family of maps, indexed by α ∈ J. We may view this as a single map

‘ a f= fα Sn−1 −−−−−→ Y. α∈J

Suppose that Z is the pushout of the diagram

` n−1 f J S −−−−→ Y     y y ` n J D −−−−→ Z. We say Z is obtained from Y by attaching n-cells, using the map f; it is common to write

a n Z = Y ∪f ( D ) J for the resulting space. About this situation we make the following remarks.

(1) We proved in [?] that ` Sn−1 → ` Dn is an NDR pair. It follows that (2) Y → Z is an NDR pair. It follows that ∼ (3) H∗(Z,Y ) = H¯∗(Z/Y ). (4) Moreover, this group is easily understood, because we have homeomorphisms ! ! a a _ Z/Y =∼ Dn / Sn−1 =∼ Sn. J J J Using these observations, the long exact sequence in reduced homology of the pair (Y,Z) becomes

H¯ Y / H¯ Z (3.1) ∗ e ∗ KKK KKK δ −1K K ∼ ¯ W n = L n H∗( α S ) α Σ Z.

I have written the long exact sequence itself as a triangle, so as to be able to display several of these at once in the next section. 3 4. CW complexes

S (n) (0) (n) Now suppose that K = n K is a CW complex: so K is a discrete space, and for n ≥ 1, K is obtained from K(n−1) as a pushout f ` Sn−1 −−−−→n K(n−1) α∈Jn     y y ` Dn −−−−→ K(n) (4.1) α∈Jn    π y y W Sn K(n)/K(n−1) α∈Jn You should check that you can prove the following. Lemma 4.2. (1) For all q, the natural map (n) colim H¯qK → H¯qK n is an isomorphism. (2) For q > n we have (n) H¯qK = 0. (3) In degree n we have ¯ (n) ¯ (n+1) ∼ ¯ (n+2) ∼ HnK HnK = HnK = .... In particular the natural map (n+1) H¯nK → H¯nK is an isomorphism. This Lemma together with the exact triangle (3.1) justiﬁes cellular homology. The CW structure of K gives us a collection of exact triangles (n−2) (n−1) (n) (n+1) ... H¯∗K / H¯∗K / H¯∗K / H¯∗K / (4.3) Qh Q Si S Si S QQQ I SSS II SSS III QQ H¯ π SS H¯ π SS H¯ π Q−1 ∗ SS−1 ∗ SS−1 ∗ δ QQQ δ SSSS δ SSSS (n−1) (n−2) (n) (n−1) (n+1) (n) H¯∗(K /K ) H¯∗(K /K ) H¯∗(K /K )

∼= ∼= ∼= L Σn−1 L Σn L Σn+1 Jn−1 Z Jn Z Jn+1 Z

Deﬁnition 4.4. The group of cellular chains of degree n of K(∗) is cell (∗) def ¯ (n) (n−1) ∼ M Cn K = Hn(K /K ) = Z. Jn

For n ≥ 1 we have a map ¯ cell (∗) cell (∗) ∂ = Hn−1(π)δ : Cn K → Cn−1K . Lemma 4.5. ∂∂ = 0.

Proof. We have δH¯∗(π) = 0 whenever this makes sense (the triangles are exact). It follows that

∂∂ = H¯n−2(π)δH¯n−1(π)δ = 0. 4 (∗) cell (∗) Deﬁnition 4.6. The cellular homology of K is the homology of the chain complex (C∗ K , ∂). In other words Ker ∂ : CcellK(∗) → Ccell K(∗) HcellK(∗) = n n−1 . n cell (∗) cell (∗) Img ∂ : Cn+1K → Cn K Theorem 4.7. There is a canonical isomorphism ¯ ∼ cell (∗) H∗K = H∗ (K ).

We give the proof in three steps. First we construct a map cell c : Zn K → HnK.

Second, we show that c is surjective. Third, we show that its kernel is ∂nK. For the ﬁrst part, suppose that cell z ∈ Zn K. That is, (n) (n−1) z ∈ H¯nK /K , and ∂z = Hn−1πδz = 0. (n) I claim that there is a unique w ∈ H¯nK such that

Hnπw = z. (n−1) First, w is unique if it exists because the triangle labeled II is exact, and HnK = 0, so Hnπ is injective.

Second, w exists if δz = 0. We know that Hn−1πδz = 0. We also know that the triangle labeled I is exct, (n−2) and that Hn−1K = 0, so Hn−1π is injective. Thus δz = 0, and w exists. We can now deﬁne c(z) = j(w), (n) where j is the natural map HnK → HnK.

Now we show that c is surjective. Suppose that x ∈ HnK. We know that (n) HnK → HnK (n) is surjective, so there is a w ∈ HnK such that j(w) = x. If we set z = Hnπw, then it’s easy to see that ∂z = 0 and c(z) = x. Finally, suppose that c(z) = j(w) = 0. Since (n+1) ∼ HnK = HnK, (n+1) we know that w maps to zero in HnK . The exactness of the triangle III shows that there is a y ∈ ¯ (n+1) (n) cell Hn+1(K /K ) = Cn+1(K) such that δy = w and so ∂y = z. This concludes the proof of Theorem 4.7.

We end with an explicit formula for the cellular boundary map ∂. For α ∈ Jn and β ∈ Jn−1 let

n−1 _ n−1 fn (n−1) iα :S → S −→ K

Jn (n−1) (n−2) ∼ _ n−1 n−1 pβ :K /K = S → S

Jn−1

5 be respectively the attaching map of the α cell and the pinch out to the β cell. Let n−1 n−1 gαβ = pβiα : S → S . For a proof of the following, see for example Hatcher or Bredon’s discussion of cellular homology. Proposition 4.8. The map ¯ M ∼ ¯ _ n−1 ¯ _ n−1 ∼ M Hn−1(πfn): Z = Hn−1( S ) → Hn−1( S ) = Z, α∈Jn Jn Jn−1 β∈Jn−1 and so the map M ∼ cell (∗) cell (∗) ∼ M ∂ : Z = Cn K → Cn−1K = Z, α∈Jn β∈Jn−1 is given on the α summand of the source and β summand of the target by deg(gαβ).