Simplicial Structures in Topology

50 II Simplicial Complexes
Let us rewrite the indices of the elements of s(p) and s(q) to have the following
common elements:
xr = y0,xr+1 = y1,...,xn = yn−r.
Notice that the set s(p)∪s(q) has exactly m+r+1 common elements. Now consider
the elements
⎧
⎨ xi, 0 ≤ i ≤ r − 1
zi = xi = yi−r, r ≤ i ≤ n
⎩
yi−r, n + 1 ≤ i ≤ m + r
together with the real numbers
⎧
⎨ −αi, 0 ≤ i ≤ r − 1
γi = −αi + βi−r, r ≤ i ≤ n
⎩
βi−r, n + 1 ≤ i ≤ m + r .
Notice that γi < 0fori = 0,...,r − 1andγi > 0fori = n + 1,...,m + r, because
αi > 0foreveryi = 0,1,...,n and βi > 0fori = 1,...,m. Let us order the numbers
γi in such a way that γ0 ≤ γ1 ≤ ... ≤ γm+r (if necessary, we make a permutation of
the indices). Let l be the largest index for which γl < 0 (because of the assumptions
we made, such a set of indices cannot be empty - thus l exists - nor can it be the set of
all indices - thus r ≤ l ≤ n); moreover, the vertices (viewed as functions) z0,z1,...,zl
are summands of p (the numbers γi are negative), while the vertices zl+1,...,zm+r
are part of q (the corresponding numbers γi are non-negative). At this point, take
λ = ∑ l i=0 γi < 0 and the two finite successions of real positive numbers
The elements
are in |K| because
{ γ0
λ ,...,γl
γl+1
} and {
λ −λ ,...,γm+r
−λ }.
p ′ =
l
γi
∑
i=0
l
γi
∑
i=0
λ zi and q ′ =
λ =
m+r
∑
i=l+1
γi
m+r
∑
i=l+1
= 1;
−λ
γi
−λ zi
from what we proved above, it follows that s(p ′ ) ⊂ s(p) and s(q ′ ) ⊂ s(q). But
s(p ′ ) ∩ s(q ′ )=/0 andso,byCase 1,
The equalities
d(| f |(p ′ ),| f |(q ′ )) ≤ � 2(l + 1)d(p ′ ,q ′ ).
d(p ′ ,q ′ )= 1
−λ d(p,q),
d(| f |(p ′ ),| f |(q ′ )) = 1
d(| f |(p),| f |(q)),
−λ
and the fact that � 2(l + 1) ≤ � 2(n + 1) allow us to conclude that