Simultaneous simplicial approximation of continuous maps (Exercise 31, Chapter 8, Armstrong's "Basic Topology")

Question

I come to you, yet again, with another question regarding the proof of the homotopy-type invariance of simplicial homology groups. As an intermediate step, Armstrong's "Basic Topology" proposes the following problem:

Let $f,g:|K|\to|L|$ be continuous maps, such that the sets $f^{-1}($star$(v,L))\cap g^{-1}($star$(v,L))$ with $v$ a vertex of $L$ form an open covering of $|K|$. Find an integer $m$ and a simplicial map $s:|K^m|\to|L|$ which simplicially approximates both $f$ and $g$.

Here are the definitions involved:

• By $|K|$ we denote the underlying space of the simplicial complex $K$, i.e., the set of points which belong to some simplex of $K$.
• The open star of a vertex $v$ in $K$, represented by star$(v,K)$, is the set of points which are interior to some simplex of $K$ that has $v$ as a vertex.
• A simplicial map is a map which takes simplexes linearly into simplexes, i.e., given a simplex $(v_0,\dots, v_n)$, we have:
  1. $(s(v_0),\dots,s(v_n))$ is also a simplex (not necessarily of the same dimension)
  2. if $x=\sum_{i=0}^n\lambda_iv_i$ with $\lambda_i\geqslant 0$ for each $i$ and $\sum_{i=0}^n\lambda_i=1$ then $s(x)=\sum_{i=0}^n\lambda_is(v_i)$
• A simplicial approximation of a given continuous map $f:|K|\to|L|$ is a simplicial map $s:|K|\to|L|$ such that $s(x)$ belongs to the only simplex which contains $f(x)$ in its interior.
• The barycentric subdivision of a simplicial complex is explained in more detailed in one of my other questions.
• The Simplicial Approximation Theorem guarantees the existence of a simplicial approximation for any continuous map if we take a big enough barycentric subdivision of the simplicial complex of the domain.

I thought maybe the idea was to associate to each vertex of $v_0\in K$ the vertex $v_1\in L$ such that $v_0\in f^{-1}($star$(v_1,L))\cap g^{-1}($star$(v_1,L))$. However, I don't know which barycentric subdivision to take in order for this to be a simplicial map or a simplicial approximation.

Answer 1

I think I got it figured now. Here's my proof, in case it's helpful to someone else:

Take the Lebesgue number $\delta$ of the open covering. We know the mesh of the barycentric subdivisions (the diameter of the biggest simplex) goes to zero as $n$ grows: $\mu(K^n)\overset{n\to\infty}{\longrightarrow}0$. Therefore, we can take $m\in\mathbb{N}$ such that $\mu(K^m)<\frac{\delta}{2}$.

That way, if we take a vertex $u\in K^m$, we have diam$($star$(u,K^m))<2\frac{\delta}{2}=\delta$ and there is a vertex $v_1\in L$ such that star$(u,K^m)\subseteq f^{-1}($star$(v_1,L))\cap g^{-1}($star$(v_1,L))$.

That way, the inclusions $f($star$(u,K^m))\subseteq $star$(v_1,L)$ and $g($star$(u,K^m))\subseteq $star$(v_1,L)$ are true. We take $s(u)=v_1$ and the proof of the Simplicial Appoximation Theorem lets us extend $s:|K^m|\to |L|$ such that it is a simplicial map and a simplicial approximation of both $f$ and $g$.