A.5. THE SUBDIVISION OPERATION 473

Definition A.5.16 For X an arbitrary topological space, let ψ be an n simplex. Letπk be the affine map from σn to αk with the αk those simplices which result from thesubdivision operator. Then we can regard Ŝ (τn) as ∑k πk with the conclusion of LemmaA.5.14 applying in terms of the An (C) . Then define S a subdivision homomorphism asfollows.

S ψ ≡ ψ#

(∑k

πk

)≡ ψ#Ŝ (τn)≡∑

kψ ◦πk (1.7)

Also define a homomorphism T : Sn (X)→ Sn+1 (X) as follows:

T ψ = T ψ# (τn)≡ ψ#T̂ (τn) , so T ψ ≡ ψ#T̂ (1.8)

These are well defined thanks to Lemma A.5.14 which says it holds on An(C) for C convex.Extend S and T as homomorphisms on Sn (X) .

Then we get the following proposition about T and S .

Proposition A.5.17 ∂T +T ∂ =S − id. Also ∂S =S ∂ and S mψ consists of a chain ofsimplices whose image in X is ψ (α) for α as small as desired.

Proof: First,

∂S ψ ≡ ∂

(ψ# ∑

kπk

)= ψ#∂

(∑k

πk

)= ψ#

(∑k

∂πk

)

while

S ∂ψ ≡ (∂ψ)#

(∑k

πk

)= ∂ψ#

(∑k

πk

)≡ ψ#

(∑k

∂πk

)For ∂ the boundary operator in Sn (X) , and using the fact that T is defined as a homo-

morphism on Sn (X) , and that ψ#∂ = ∂ψ#,

∂T ψ +T ∂ψ = (∂T ψ +T ∂ψ)(τn)≡(∂ψ#T̂ +∂ψ#T̂

)(τn)

= ψ#(∂ T̂ +∂ T̂

)(τn) = ψ#

(Ŝ − id

)(τn) = (S − id)(ψ)

and since T is a homomorphism, this shows what is desired.The last claim follows from the earlier material applied to the geometric simplices

αk obtained from successive applications of the subdivision map. Let ψ be a singular nsimplex in Sn (X). ■

Doing S to both sides of ∂T +T ∂ = S − id yields ∂S T +S T ∂ = S 2−S whichcan now be used to see that [S φ ] =

[S 2φ

]. Then continuing this way one sees that

[S mφ ] = [φ ] for φ a cycle.Now here is a definition which will help to compute homology groups.

Definition A.5.18 Let U be a covering of X and let SUn (X) consist of the subgroup of

Sn (X) generated by singular simplices φ with the property that φ (σn) is contained insome set U ∈ U . Then Hn

(SU

n (X))

will denote the homology group obtained as beforeexcept now cycles and boundaries are with respect to SU

n (X) . We can do this because ∂φ ∈SU

n−1 (X) if φ ∈ SUn (X). Then Hn

(SU

n (X))

will denote the usual thing. Letting ZUn (X)