Simplexes, The Simplicial Approximation Theorem

The Simplicial Approximation Theorem

Start with a finite complex K and a complex L, and a continuous function f from K into L. Note that f need not be a simplicial map. It could smear the simplexes of K all over L. We're going to slide the image f(K) about, stretching here and shrinking there, but never pushing f across simplex boundaries. If f(x) lies in a certain simplex of L, it's always going to be in that simplex. In the end we will have a function g, homotopic to f, that is a simplicial map on a barycentric subdivision of K. At a high resolution, g acts on the simplexes of K, squashing f into walls, edges, or corners, or expanding f to fill the destination simplex in L.

Unless otherwise stated, star means open star. I'll say closed star when I mean closed star.

Embed K in E_j, so that K becomes a metric space. Open balls form a base for the topology of K.

For any x in K, let g(x) be a vertex in L. If f(x) is already a vertex then g(x) = f(x). Otherwise g(x) is any vertex in the minimal simplex of L containing f(x). (We don't need the axiom of choice here, because everything is finite. If you like, select the "least" vertex of L that satisfies our criteria.)

The star about g(x) is an open set in L, containing f(x), with an open preimage in K. Let b(x) be a ball in K centered at x that is wholly contained in this open preimage. Thus every x has a ball b(x) around it, that maps, via f, into the star in L about the vertex g(x).

Cut the radius of each ball in half, giving a new ball c(x) inside each b(x). These balls cover K, and since K is compact, we can choose a finite subcover. Let r be the smallest radius.

At this point we are ready to subdivide K into smaller simplexes. Do this again and again, until the diameter of the largest simplex is less than r. Let K′ be the resulting complex. Now each vertex v of K′ lies in some c(x), and its star is wholly contained in b(x). Therefore f maps this star into the star in L about g(x).

If v is a vertex in K′, let g(v) = g(x), as described above. If v is part of several different balls c(x), pick any g(x). This builds a map from the vertices of K′ into the vertices of L, but is it a simplicial map? Do simplexes map to simplexes?

Let σ be a simplex in K′, having vertices v₁ through v_n. Let g map these vertices to w₁ through w_n, which are vertices in L. (Some of these vertices may repeat.) Look at an arbitrary vertex v_i. It doesn't matter which ball c(x) we selected, b(x) contains σ, and f maps σ into the star at w_i. This holds for each v_i. Hence f(σ) lies in the intersection of the stars about w_i. Let p be any point in σ and let q = f(p). Thus q lies in the intersection of the stars about w_i. Place q in its minimal simplex τ in L. Suppose τ does not contain w_i. The star about w_i contains q, but not τ. It must bring in a smaller simplex containing q, but this does not exist. Therefore τ contains each w_i, and this set of vertices determines a simplex, e.g. a face within τ.

The above holds for every simplex in K′, hence g is a valid abstract map. Extend this to a simplicial map from K′ into L.

We showed above that p, in σ, maps to q in its minimal simplex τ, and that g defines a face within τ. The simplicial map carries p to a point on this face, which is still part of τ. The image of f has not left its minimal simplex τ in L. We have pushed things around, but we haven't crossed any borders. Therefore g is a simplicial approximation to f.

As a bonus, f and g are homotopic. Slide f linearly towards g. Since τ is convex, q slides easily to its new destination g(p). This holds for each p in σ, hence f(σ) and g(σ) are homotopic. Apply the same homotopy to each simplex in K′ in parallel. Note that these homotopies agree whenever simplexes intersect. This is because the homotopy is described globally, i.e. a linear map from f to g, across the entire domain.