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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) –

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