208 CHAPTER 10. BROUWER FIXED POINT THEOREM Rn∗

Definition 10.0.2 If S is an n simplex. Then it is triangulated if it is the union of smllersub-simplices, the triangulation, such that if S1,S2 are two simplices in the triangulation,with

S1 ≡[z1

0, · · · ,z1m], S2 ≡

[z2

0, · · · ,z2p]

thenS1∩S2 =

[xk0 , · · · ,xkr

]where

[xk0 , · · · ,xkr

]is in the triangulation and{

xk0 , · · · ,xkr

}={

z10, · · · ,z1

m}∩{

z20, · · · ,z2

p}

or else the two simplices do not intersect.

The following proposition is geometrically fairly clear. It will be used without commentwhenever needed in the following argument about triangulations.

Proposition 10.0.3 Say [x1, · · · ,xr] , [x̂1, · · · , x̂r] , [z1, · · · ,zr] are all r−1 simplices and

[x1, · · · ,xr] , [x̂1, · · · , x̂r]⊆ [z1, · · · ,zr]

and [z1, · · · ,zr,b] is an r+1 simplex and

[y1, · · · ,ys] = [x1, · · · ,xr]∩ [x̂1, · · · , x̂r] (10.0.1)

where{y1, · · · ,ys}= {x1, · · · ,xr}∩{x̂1, · · · , x̂r} (10.0.2)

Then[x1, · · · ,xr,b]∩ [x̂1, · · · , x̂r,b] = [y1, · · · ,ys,b] (10.0.3)

Proof: If you have ∑si=1 tiyi + ts+1b in the right side, the ti summing to 1 and nonneg-

ative, then it is obviously in both of the two simplices on the left because of 10.0.2. Thus[x1, · · · ,xr,b]∩ [x̂1, · · · , x̂r,b]⊇ [y1, · · · ,ys,b].

Now suppose xk = ∑rj=1 tk

j z j, x̂k = ∑rj=1 t̂k

j z j, as usual, the scalars adding to 1 and non-negative.

Consider something in both of the simplices on the left in 10.0.3. Is it in the right? Theelement on the left is of the form

r

∑α=1

sα xα + sr+1b =r

∑α=1

ŝα x̂α + ŝr+1b

where the sα , are nonnegative and sum to one, similarly for ŝα . Thus

r

∑j=1

r

∑α=1

sα tαj z j + sr+1b =

r

∑α=1

r

∑j=1

ŝα t̂αj z j + ŝr+1b (10.0.4)

Now observe that

∑j∑α

sα tαj + sr+1 = ∑

α

∑j

sα tαj + sr+1 = ∑

α

sα + sr+1 = 1.