0.7. Dimension theory 15
Corollary 0.7.2. Let X be a compact subset of the PL manifold W and let
k = dim X. For every 0 there exists a k-dimensional polyhedron K W
and an onto map g : X K such that d(x, g(x)) for every x X.
Proof. Since X is compact, we can replace W with a compact ∂-manifold
W that contains X in its interior. Now W is an ANR, so it is n-LC
at x for every n and for every x. The compactness of W together with
a Lebesgue number argument establishes the following uniform version of
local connectivity: for each nonnegative integer n and for each 0 there
exists δ 0 such that any map of
∂In
into a δ-subset of W extends to a
map of
In
into an -subset of W .
Let 0 be given. Choose δk such that any map of
∂Ik
into a δk-subset
of W extends to a map of
Ik
into an (/2)-subset of W . Then recursively
choose δk−1,δk−2,...,δ1 such that any map of
∂Ij
into a δj-subset of W
extends to a map of
Ij
into a (δj+1/2)-subset of W . Set δ = δ1.
By the Alexandroff Theorem there is a δ-mapping f : X L, where
L is a compact polyhedron of dimension k. The idea is to use the local
connectivity of W to construct a map h : L W that is an approximate
inverse for f. Then we can define K = h(L) and g = h f.
If d(x, x ) δ, then f(x) = f(x ). Compactness gives a positive number
η such that d(x, x ) δ implies that the distance from f(x) to f(x ) is η
in L. Let T be a triangulation of L such that each simplex in T has diameter
η; then diam f
−1(σ)
δ for every σ T.
For each vertex v T, define h(v) to be some point in f
−1(v).
Let σ be
a 1-simplex in T. We have already defined h|∂σ and the choice of δ1 allows
us to extend h to σ in such a way that diam h(σ) δ2/2. Now consider
a 2-simplex τ T. Note that h|∂τ is already defined and diam h(∂τ)
δ2. Hence the choice of δ2 allows us to extend h to τ in such a way that
diam h(τ) δ3/2. This process is continued inductively and results in a
map h : L W such that diam h(σ) /2 for every σ T. We may
assume that h is a PL map in general position.
Define K = h(L) and define g : X K by g(x) = h(f(x)). Fix x X.
We must check that d(x, g(x)) . Locate a simplex σ T such that
f(x) σ and choose a vertex v of σ. Then
d(x, g(x)) d(x, h(v)) + d(h(v),g(x)) = d(x, h(v)) + d(h(v),h(f(x))).
Now x and h(v) are both in f
−1(σ),
so d(x, h(v)) δ /2. In addi-
tion, both h(v) and h(f(x)) are in h(σ), so d(h(v),h(f(x))) /2. Thus
d(x, g(x)) .
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