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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