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)) .