Graphics Reference
In-Depth Information
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
()
=
()
+-
(
) ( )
htt
x
j
x
1
tf
x
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
s
1
...
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
0
()
=
()
=
sd
K
K
,
(
)
n
n
-
1
()
sd
K
sd sd
K
,
for n
≥
1
.
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].
