7 Cell Complexes and Cellular Homology

Autumn Semester 2016–2017 MATH41071/MATH61071 Algebraic topology

7 Cell complexes and cellular homology

Cell complexes In this section we will consider topological spaces constructed inductively out of open and closed cells (instead of simplices), i.e. topological (sub-)spaces being homeomorphic to closed or open unit ball.

7.1 Definition. A finite cell complex (or finite CW complex) is a Haus- ` dorff space X which is a finite disjoint union X = α eα of subspaces that are open cells, together with a a characteristic map for every cell eα: a con- k k tinuous map fα : D → X that maps B homeomorphically to eα and maps k−1 the boundary ∂B = S continuously into a union of cells eβ that all have k smaller dimension than k. The image fα(D ) is denotede ¯α. The union of cells of dimension less or equal then k is called the k-skeleton of X and will be denoted by Xk. A a pair of cell complexes is a pair (X,A) that consists of a finite cell complex X and a subcomplex (i.e. the union of a subset of cells) A.

In the following we will usually drop the word ﬁnite . Cellular complexes are more ﬂexible and easier to handle than simplicial complexes in many situations: If X and Y are cellular complexes, then the product X ×Y is again a cellular complex. If (X,A) is a cellular pair then X/A is again a cellular complex. It has one new vertex that corresponds to A, and one k-cell for each k-cell of X that does not lie in A.

Cellular homology The aim of this section is to provide an effective tool to calculate homology groups on a cell complex. The main tool for this calculation is called cellular homology. Note, that one can generalise the axioms for a reduced homology theory from Definition 6.8 almost literally to the case of cell complexes. Now, given a cell complex X and a reduced homology theory H˜ we define the cellular chain groups as

7.2 Deﬁnition. Given a cell complex X we deﬁne the cellular chain groups as k k−1 Ck(X) = H˜k(X /X )

1 with boundary maps dk : Ck(X) → Ck−1(X) k k−1 given by the composition dk = (pk−1)∗ ◦ ∂k. Where ∂k : H˜k(X /X ) → k−1 H˜k−1(X ) is the corresponding homomorphism from the long exact se- k k−1 k−1 k−1 k−2 quence for the pair (X ,X ) and pk−1 : X → X /X is the contraction map. For the case k = 0 the above deﬁnition still makes sense when setting

X0/X−1 := X0 t ∗ = X0 ∨ S0.

The cellular homology groups are deﬁned by

cell Hk (X) = ker dk/ im dk+1.

To make sense to the last deﬁnition we need to have im dk ⊂ ker dk−1. This follows from

7.3 Lemma. For the boundary maps deﬁned above we have

dk ◦ dk+1 = 0, i.e. im dk ⊂ ker dk−1. Proof. By deﬁnition of d we have the following commutative diagram

k+1 k H˜k+1(X /X )

dk+1 ∂k+1 ( k (pk)∗ k k−1 ∂k k−1 H˜k(X ) / H˜k(X /X ) / H˜k−1(X )

dk (pk−1)∗ ) k−1 k−2 H˜k−1(X /X )

Note, that horizontal line is a part from the long exact sequence for (Xk,Xk−1). In particular, we have ∂k ◦ (pk)∗ = 0 and hence also dk ◦ dk+1 = 0. At a ﬁrst glance the deﬁnition of cellular homology seems to be useless to calculate homology groups, since it itself involves the homology of Xk/Xk−1, W k k W k but these spaces are naturally homeomorphic to α(¯eα/∂eα) =: α Sα a wedge sum of k-spheres. Hence, by Theorem 6.13 we obtain

∼ M ˜ k Ck(X) = Hk(Sα) α where α is running over the index set of k-cells of X. Every component of the k above direct sum gives rise to a generator corresponding to 1 ∈ Hk(Sα) = Z,

2 k which we denote by eα (by abuse of notation). Hence, Ck(X) can be written as M k Z · eα. α It now remains to investigate the boundary maps to make the calculation of cellular homology groups eﬀective. 7.4 Proposition. The boundary maps as deﬁned above are given by their values on generators

k X k−1 dk(eα) = deg(ϕαβ) · eβ . β

Here β is running over all indices of (k − 1)-cells and ϕαβ is the composition

k−1 fα k−1 p k−1 k−2 ∼ _ k−1 p¯β k−1 S −→ X −→ X /X = Sγ −→ Sβ γ withp ¯β being the projection to the β-component of the wedge sum as in Theorem 6.13 and Remark 6.14.

Proof. Note that the characteristic fα is by deﬁnition a map of pairs fα : k k−1 k k−1 (D ,S ) → (X ,X ) hence it gives rise to the commutative diagram (Exactness axiom)

∼ ˜ k ∂ ˜ k−1 ∼ Z = Hk(S ) / Hk−1(S ) = Z

(f¯α)∗ (fα)∗ k k−1 ∂k k−1 H˜k(X /X ) / H˜k−1(X ) k k ∼ W k ¯ k Note, that X /X = γ Sγ and fα is exactly the inclusion of Sα into this ¯ k ˜ k k−1 wedge sum. In particular, (fα)∗(1) equals the generator eα in Hk(X /X ) by Remark 6.14. From Exercise 7.5 we know that the ﬁrst row of the diagram is an isomorphism (which we can assume maps 1 to 1 after choosing the isomorphisms to Z accordingly). By deﬁnition of ϕαβ we may extend the above diagram to the right

(ϕ ) ˜ k ∂ ˜ k−1 αβ ∗ ˜ k−1 Hk(S ) / Hk−1(S ) / Hk−1(Sβ ) O (f¯α)∗ (fα)∗ (¯pβ )∗ (p ) k k−1 ∂k k−1 k−1 ∗ k−1 k−2 H˜k(X /X ) / H˜k−1(X ) / H˜k−1(X /X )

k−1 k−2 ∼ W k−1 Note, that X /X = γ Sγ for this to make sense. Now, by deﬁnition pf dk we have k k ¯ dk(eα) = (pk−1)∗ ◦ ∂k(eα) = (pk−1)∗ ◦ ∂k ◦ (fα)∗(1).

3 By commutativity of the left part of the diagram this gives

k dk(eα) = (pk−1)∗ ◦ (fα)∗(1).

k−1 k−2 When expressing an element of H˜k−1(X /X ) in the basis elements k−1 k−1 eγ the coeﬃcient at eβ is given by the natural projection of the direct k−1 sum to the component Z · eβ

˜ k−1 k−2 ∼ M k−1 k−1 Hk−1(X /X ) = Z · eγ −→ Z · eβ , γ but from Remark 7.16 we know that this projection coincides with (¯pβ)∗. Now, by commutativity of the right part of the diagram it follows that

k (¯pβ)∗ ◦ dk(eα) = ϕαβ(1) = deg(ϕαβ).

7.5 Remark. One need’s to be a bit careful when interpreting the proposition and it’s proof for the case k = 1. Interpreting the statement of the 1 theorem correctly this means d1 sends eα to 0 if the boundary consists of only one point and if the boundary consists of two points the image is the diﬀerence of the corresponding basis elements of C0(X). When taking X0/X−1 = X0/∅ := X0 t ∗ this follows indeed from the above proof.

Using this we are indeed able to calculate the cellular homology groups for cell complexes.

2 ∼ 2 1 7.6 Example (Sphere). Using the identiﬁcation S = D /S one obtains a cell decomposition of the sphere into the image of the interior (a 2-cell) and the complement, which is just a point (a 0-cell). ( k = 0, 2 Hcell(S2) = Z k 0 else, since all the boundary maps are necessarily trivial. 2 Alternatively we may use the identiﬁcation S2 =∼ D /∼ from Exercise 1.7, where (x, z) ∼ (x0, y0) if and only if x = x0 and y = |y0|. This induces a cell 2 1 decomposition into one 2-cell e1, the image of the interior, one 1-cell e2, the 1 0 0 image of S \{(1, 0), (−1, 0)} and two 0-cells e3, e4, the images of (1, 0) and (−1, 0). Hence, the cellular chain complex looks as follows.

d2 d1 2 0 → Z −→ Z −→ Z → 0.

4 1 0 0 From Remark 7.5 we see that d1(e2) = e3 − e4. In particular, the kernel of d1 is 0. For d2, note that ϕ1,2 is given by ( eαi 0 α π ϕ (eαi) = 6 6 1,2 −αi e π 6 α 6 2kπ.

One obtains deg(ϕ1,2) = 0 and d2 ≡ 0. This leads to the same results for the homology as in the above calculation. We saw that for this example the cellular homology groups didn’t depend on the chosen cell decomposition and that they coincide with the simplicial homology groups.

7.7 Example (Torus). We can consider the cell decomposition of the unit square into one 2-cell (the interior which is isomorphic to S2) four 1- cells (the edges) and four 0-cells (vertices) all with the obvious characteristic maps. Via the quotient map the 1-cells get identiﬁed in pairs and 0-cells are mapped to the same point. This induces a cellular decomposition of 2 1 1 0 the torus with one 2-cell e1, two 1-cells e2, e3 and one 0-cell e4. Hence, the cellular chain complex looks as follows

d2 2 d1 0 → Z −→ Z −→ Z → 0.

Since the two endpoints of every edge get identiﬁed under the quotient map. The boundary map d1 sends the corresponding generators to 0, i.e. d1 ≡ 0. For d2 note that ϕ1,2 is given by

S1 =∼ ∂(I × I) → S1, (s, t) 7→ es·2πi.

Which has degree 0 [Exercise]. Similarly we see deg(ϕ1,3) = 0. Hence, d2 ≡ 0 and we obtain  k = 0, 2 Z cell 2 Hk (T ) = Z k = 1 0 else.

We see that cellular and simplicial homology coincide.

7.8 Example (Pinched torus). Consider X = S2/{N,S} the sphere with north and south pole being identiﬁed. This is called a pinched torus. The second cell decomposition of the sphere in Example 7.6 induces a cell decomposition of X into one 2-sphere, one 1-sphere and one 0-sphere (via identiﬁcation of the two 0-cells). Hence, the cellular chain complex looks as follows. d2 d1 0 → Z −→ Z −→ Z → 0.

5 Now, d1 ≡ 0. Similar to Example 7.6 it follows that d2 is trivial, as well, and we obtain. ( k = 0, 1, 2 Hcell(X) = Z k 0 else. Note, that we can obtain the pinched torus also as T/S1, with S1 being the image of one of the edge pairs in the standard construction. Remember that this cycle was one of the generators of the ﬁrst homology. We now see how this cycle “vanishes” in homology after the contraction.

2 ∼ 2 7.9 Example (Projective plane). Using the construction P = D /sim 2 with x ∼ −x for points on the boundary we obtain a decomposition of P into one 2-cell e2, one 1-cell e1 and one 0-cell e0. The 2-cell e2 is given as the image 2 2 of the interior under the quotient map q : D → P and the characteristic map f2 is just the quotient map itself. The one 1-cell e1 coincides with the 1 2 image of S \{(1, 0), (−1, 0)}. A characteristic map f1 :(−1, 1) → P is given t·πi 1 2 2 by t 7→ [e ] ∈ S /∼ ⊂ D /∼ = P . The unique 0-cell e0 is the image of (1, 0) and (−1, 0) and comes with the obvious characteristic map f0.

Since the boundary of e1 consists only of one point we have d1 ≡ 0. For d2 1 0 1 1 ∼ 1 1 note that X /X = X consists just of S /∼ = S and the map ϕ2,1 : S → 1 0 1 1 2 1 ∼ 1 X /X = S is just the quotient map S = ∂D → S /∼ = S (which is given by z 7→ z2). We have seen, that this is a degree-2 map. Hence, we obtain  Z k = 0 cell  Hk (X) = Z/2Z k = 1 0 else. This again coincides with the simplicial homology.

Since usually we have much less cells in our decompositions then simplices in the corresponding the triangulations, the cellular chain groups are much smaller then the simplicial ones and hence the actual calculations are much simpler. Moreover, the results from the examples suggest the cellular homology coincides with simplicial homology. The following theorem shows that this is indeed the case.

7.10 Theorem. For the cellular homology groups one has

( ˜ cell ∼ Hk(X) × Z , k = 0 Hk (X) = H˜k(X) , else, where H˜ is the underlying reduced homology theory.

7.11 Lemma. If X is a ﬁnite cellular complex, then

n n−1 (a) H˜k(X /X ) = 0 for k 6= n,

6 ˜ n (b) Hk(X ) = 0 for k > n > −1, n n (c) the map H˜k(X ) → H˜k(X) corresponding to the inclusion X ,→ X is an isomorphism for k < n and a surjection for k = n. Proof. Claim (a) follows again from the fact that Xn/Xn−1 is just a wedge sum of copies of Sn. Now, consider the long exact sequence for (Xn,Xn−1):

n n−1 n−1 n n n−1 H˜k+1(X /X ) → H˜k(X ) → H˜k(X ) → H˜k(X /X ) Since for k 6= n the last term vanishes, the middle map must be surjective in this case. Similarly, for k 6= n − 1 the map must be injective. Now, consider the homomorphisms induced by inclusions Xk−1 ⊂ Xk

0 1 k−1 k k+1 H˜k(X ) → H˜k(X ) → · · · → Hk(X ) → Hk(X ) → Hk(X ) → · · · We have just seen, that all the homomorphisms are actually isomorphisms. k Except the homomorphism to Hk(X ) might not be surjective and the one k from Hk(X ) might not be injective. From the left part of the sequence 0 follows (b), since H˜k(X ) = 0 for k > 0. From the right hand side follows (c).

Proof of the Theorem 7.10. Assume ﬁrst that k > 0. By using Lemma 7.11 we extend the commutative diagram from Lemma 7.3 with horizontal and vertical sequences being exact as follows.

k+1 k k−2 H˜k+1(X /X ) H˜k−1(X ) = 0

dk+1 ∂k+1 ) k (pk)∗ k k−1 ∂k k−1 0 / H˜k(X ) / H˜k(X /X ) / H˜k−1(X )

dk (pk−1)∗ ) k+1 ∼ k−1 k−2 H˜k(X ) = H˜k(X) H˜k−1(X /X )

Since, (pk)∗ is injective and because of the exactness of the sequence it maps k H˜k(X ) isomorphically onto ker ∂k, where im ∂k+1 gets send to im dk+1. On the other hand, because of the injectivity of (pk−1)∗ we have ker dk = k ∼ ker ∂k. Hence, we obtain H˜k(X )/ im ∂k+1 = ker k/ im dk+1. Now, H˜k(X) = k H˜k(X )/ im ∂k+1 implies the claim. ˜ 0 −1 ˜ 0 0 ˜ 0 For the case k = 0 note, that H0(X /X ) is H0(X ∨ S ) = H0(X ) ⊕ Z by deﬁnition. Then the result follows.