7.2.2.7. Lemma. Let f : ΩKΩÆΩLΩ be a continuous map and suppose that j:K Æ L
is a simplicial approximation to f.
(1) The map ΩjΩ:ΩKΩÆΩLΩ is homotopic to f.
(2) If f = ΩyΩ, where y:K Æ L is a simplicial map, then y=j.
Proof. To prove (1), define a homotopy h:ΩKΩ¥[0,1] ÆΩLΩ between f and ΩjΩ
by
for x ŒΩKΩ and t Œ [0,1]. That h(x,t) actually lies in ΩLΩ follows from the fact that
ΩjΩ(x) and f(x) lie in a simplex of L, which means that the segment [ΩjΩ(x),f(x)] is
contained in ΩLΩ because simplices are convex.
To prove (2), let v be a vertex of K. Then w = f(v) is a vertex of L. Since a vertex
is also a 0-simplex, the definition of a simplicial approximation implies that j(v) = w.
This proves the lemma.
Part (2) of Lemma 7.2.2.7 means that the only simplicial approximation to a sim-
plicial map is the map itself. An arbitrary continuous map does not have a unique
simplicial approximation, however.
If K is a simplicial complex, define a new simplicial complex, denoted by sd(K),
as follows:
(1) The vertices of sd(K) are the barycenters b(s) of the simplices s in K.
(2) The q-simplices of sd(K), q > 0, are all the q-simplices of the form b(s
0
)b(s
1
)
···b(s
q
), where the s
i
are distinct simplices of K and s
0
Ɱ s
1
Ɱ ...Ɱ s
q
.
It is easy to show that sd(K) is a simplicial complex (Exercise 7.2.2.3) that is a sub-
division of K. Clearly, Ωsd(K)Ω=ΩKΩ.
Definition. The simplicial complex sd(K) is called the (first) barycentric subdivision
of K. The nth barycentric subdivision of K, denoted by sd
n
(K), is defined inductively
by
Figure 7.7 shows a simplex and its barycentric subdivision.
7.2.2.8. Theorem. (The Simplicial Approximation Theorem) Let K and L be sim-
plicial complexes and suppose that f :ΩKΩÆΩLΩ is a continuous map. Then there is
an integer N ≥ 0 such that for each n ≥ N, f admits a simplicial approximation j:
sd
n
(K) Æ L.
Proof. See [AgoM76].
,.