Simplicial Structures in Topology
Simplicial Structures in Topology
Simplicial Structures in Topology
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II.2 Abstract <strong>Simplicial</strong> Complexes 49<br />
simplicial complex K ′ ⊂ R n associated with K. We shall see <strong>in</strong> a short while that<br />
K ′ is isomorphic to the geometric realization |K| (actually, there exists an isometry<br />
between these two metric spaces; furthermore, the set of all functions X → R≥0<br />
co<strong>in</strong>cides with the positive quadrant of R n ).<br />
The follow<strong>in</strong>g statement holds true: two po<strong>in</strong>ts p,q ∈|K| co<strong>in</strong>cide if and only if<br />
they have the same barycentric coord<strong>in</strong>ates.<br />
The geometric realization functor<br />
| |: Csim −→ Top<br />
is def<strong>in</strong>ed over an object K ∈ Csim as the geometric realization |K|, and over a<br />
morphism f ∈ Csim(K,L) as<br />
| f |: |K|→|L|, | f |(∑αixi)=∑αi f (xi) .<br />
To prove that ||is <strong>in</strong>deed a functor, we need the follow<strong>in</strong>g result.<br />
(II.2.7) Theorem. The function | f | <strong>in</strong>duced from a simplicial function f : K → Lis<br />
cont<strong>in</strong>uous.<br />
Proof. It is enough to prove that, for every p ∈|K|, there exists a constant c(p) > 0<br />
which depends on p and such that, for every q ∈|K|, d(| f |(p),| f |(q)) ≤ c(p)d(p,q).<br />
Assume that<br />
s(p)={x0,...,xn} and s(q)={y0,...,ym}<br />
and also that p(xi)=αi for i = 0,...,n,andq(y j)=β j for j = 0,...,m. We consider<br />
three cases.<br />
Case 1: s(p) ∩ s(q)=/0 - In this situation<br />
d(p,q)=<br />
� n<br />
∑ α<br />
i=0<br />
2 i +<br />
m<br />
∑<br />
j=0<br />
β 2 j ≥<br />
�<br />
n<br />
∑ α<br />
i=0<br />
2 i ;<br />
because ∑ n i=0 αi = 1, ∑ n i=0 α2 i has its m<strong>in</strong>imum value only when αi = 1/(n + 1),for<br />
every i = 0,...,n. It follows that d(p,q) ≥ 1/ √ n + 1and<br />
d(| f |(p),| f |(q))<br />
d(p,q)<br />
(recall that d(| f |(p),| f |(q)) ≤ √ 2); so,<br />
≤<br />
√ 2<br />
1/ √ n + 1<br />
d(| f |(p),| f |(q)) ≤ � 2(n + 1)d(p,q);<br />
thus, we def<strong>in</strong>e c(p)= � 2(n + 1).<br />
Case 2: s(p) ∩ s(q) �= /0, but s(p) �⊂ s(q) and s(q) �⊂ s(p) –