462 APPENDIX A. HOMOLOGICAL METHODS∗

One sees the pattern from this. Now

∂0φ (t0, ..., tn−1) ≡ φ (0, t0, ..., tn−1) ,

∂1φ (t0, ..., tn−1) ≡ φ (t0,0, t1, ..., tn−)

Thus

T ∂0φ (t0, ..., tn)≡

{(1− t0)φ

(0, t1

1−t0, ..., tn

1−t0

)+ t0x if t0 < 1

x if t0 = 1

T ∂1φ (t0, ..., tn) =

{(1− t0)φ

(t1

1−t0,0, t2

(1−t0), ..., tn

1−t0

)+ t0x if t0 < 1

x if t0 = 1

etc. That is, T ∂0φ = ∂1T φ and in general, T ∂i−1φ = ∂iT φ until T ∂nφ = ∂n+1T φ .Then ∂T φ +T ∂φ = φ +∑

n+1i=1 (−1)i

∂iT φ +∑ni=0 (−1)i T ∂iφ

= φ +n+1

∑i=1

(−1)i∂iT φ +

n+1

∑i=1

(−1)i−1 T ∂i−1φ

= φ +n+1

∑i=1

(−1)i∂iT φ +

n+1

∑i=1

(−1)i−1∂iT φ = φ

Thus ∂T +T ∂ is the identity. It follows that if φ ∈ Zn (X) , then φ = ∂T φ +T ∂φ = ∂T φ

and so φ is a boundary. Therefore, Zn (X)/Bn (X) = 0 because [φ ] = 0 for all φ ∈ Zn (X).Boundaries and cycles are the same thing for n > 0 and X convex. ■

Definition A.2.10 A set is star shaped if there is a special point x called the star centersuch that segments from x to other points are contained in the set.

An examination of the proof of the above shows the following corollary.

Corollary A.2.11 Let X ⊆Rp be star shaped with star center x. Then if n > 0,Hn (X) = 0.

x

The above picture is of a star shaped set which is definitely not convex.

A.3 HomotopyThis is about homology groups of homotopic maps.

Lemma A.3.1 Let X ,Y be convex and let g0,g1 : X → Y be given continuous functions.Then for each n > 0 there exists T̂ : Sn (X)→ Sn+1 (Y ) such that

g0#−g1# = T̂ ∂ +∂ T̂