A.2. THE HOMOLOGY GROUPS 461

values in a single Xα . If c, c̃ differ by a boundary, so c− c̃= ∂b, then b= b1+ · · ·+bm whereeach

∣∣b j∣∣ is contained in a single Xα j , j = 1,2, ...,m. It follows that the ci, c̃i just described,

those supported in Xα i satisfy c j− c̃ j = ∂b j and so we can define Hn (X) = ∑α∈A Hn (Xα).In other words, if ci is a boundary supported on Xα i then it is the boundary of some bisupported on Xα i since the boundaries of the other b j will not intersect Xα i . ■

Next is something pretty interesting in the case of a convex topological space mean-ing that if x,y ∈ X , then it makes sense to form xt + y(1− t) ∈ X for t ∈ [0,1] and + iscontinuous from X×X to X . Here is a picture which suggests the construction used.

x θ(t0, ..., tn+1) φ(t0, ..., tn)

Theorem A.2.9 Let X be convex. For all n > 0,Hn (X) = 0. Also, there exists a mapT : φ → θ where φ is a singular n simplex and θ is a singular n+1 simplex. This map canbe obtained from a simple formula for n≥ 0. For X convex cycles and boundaries are thesame for n > 0. Also for φ a simplex, φ = ∂T φ +T ∂φ where T denotes the homomorphismfrom extending T to all of Sn (X).

Proof: For n > 0 and φ ∈ Sn (X). Then define θ in Sn+1 (X) as follows.

θ (t0, ..., tn, tn+1)≡

{(1− t0)φ

(t1

(1−t0), t2(1−t0)

, ...,tn+1(1−t0)

)+ t0x if t0 < 1

x if t0 = 1

Note that we assume ∑n+1j=0 t j = 1 each t j ≥ 0. Therefore, this makes perfect sense because

∑n+1j=1 t j = 1− t0 and so ∑

n+1j=1

t j(1−t0)

= 1. The thing which might not be clear is that thisθ is continuous. There is clearly no problem at any point of σn+1 where t0 < 1 so let(tn0 , ..., t

nn , t

nn+1)→ (1, t1, ..., tn+1) . Then, since the sum is always 1, it follows that tn

0 → 1and ∑

n+1j=1 tn

j → ∑n+1j=1 t j = 0. Also each t j ≥ 0 so they each converge to 0. The set φ (σn) is

bounded and so

(1− t0)φ

(t1

(1− t0),

t2(1− t0)

, ...,tn+1

(1− t0)

)→ 0

and tn0 x→ x. Thus this is indeed continuous.

Let T φ ≡ θ and T a homomorphism. Thus ∂0T φ = φ . Also ∂1θ (t0, ..., tn) =

θ (t0,0, t1, ..., tn) ≡

{(1− t0)φ

(0, t1

(1−t0), ..., tn

(1−t0)

)+ t0x, t0 < 1

x for t0 = 1

and ∂2T φ (t0, ..., tn)≡ θ (t0, t1,0, t2, ..., tn)

=

{(1− t0)φ

(t1

(1−t0),0, t2

(1−t0), ..., tn

(1−t0)

)+ t0x, t0 < 1

x for t0 = 1