[proofplan] The strategy is to use the Lebesgue number lemma and the mesh-shrinking lemma to reduce approximating $f$ to a purely combinatorial vertex assignment. Pull back the open-star cover of $|L|$ through $f$ to get an open cover of $|K|$; by compactness of $|K|$ and the Lebesgue number lemma, every small enough ball in $|K|$ fits inside one of these pre-images. By the mesh-shrinking lemma, choose $r$ large enough that the open stars of $K^{(r)}$ — each contained in a ball of radius $\mu(K^{(r)})$ — are smaller than the Lebesgue number. For each vertex $v$ of $K^{(r)}$, pick any $w \in V_L$ whose star-pre-image contains $\operatorname{St}_{K^{(r)}}(v)$, and set $g(v) = w$; this satisfies the approximation condition by construction. For the relative statement, when $v$ is a vertex of $M$ one may choose $g(v) = f(v)$ — since $f$ is simplicial on $M$, $f(v)$ is a vertex of $L$ and the approximation condition at $v$ is automatic. [/proofplan]

[step:Cover $|K|$ by pulling back the open stars of $L$ through $f$] Let $V_L = \{w_1, \ldots, w_N\}$ be the (finite) vertex set of $L$. The open stars \begin{align*} \{\operatorname{St}_L(w)\}_{w \in V_L} \end{align*} form an open cover of $|L|$: each $\operatorname{St}_L(w) = \bigcup_{w \in \tau \in L} \mathring{\tau}$ is open in $|L|$ (see the open-star properties in the chapter text), and every point $y \in |L|$ lies in the interior of some unique simplex $\tau$ by the [Unique Interior Simplex](/theorems/1914) theorem, so $y \in \operatorname{St}_L(w)$ for any vertex $w$ of $\tau$. Define the family \begin{align*} \mathcal{U} = \left\{ f^{-1}(\operatorname{St}_L(w)) : w \in V_L \right\} \end{align*} of subsets of $|K|$. Each $f^{-1}(\operatorname{St}_L(w))$ is open in $|K|$ by continuity of $f$. Given $x \in |K|$, the point $f(x) \in |L|$ lies in $\operatorname{St}_L(w)$ for some $w \in V_L$, so $x \in f^{-1}(\operatorname{St}_L(w))$. Therefore $\mathcal{U}$ is a finite open cover of $|K|$. [/step]

[step:Apply the Lebesgue number lemma to obtain a uniform radius]The polyhedron $|K|$ is compact: it is a finite union of the simplices of $K$, each of which is a compact convex subset of Euclidean space, and a finite union of compact sets is compact. As a subset of Euclidean space, $|K|$ inherits the standard metric $d(x, y) = |x - y|$; so $(|K|, d)$ is a compact metric space. Apply the [Lebesgue Number Lemma](/theorems/952) to the compact metric space $(|K|, d)$ and the open cover $\mathcal{U}$. The hypotheses (compactness of the metric space and openness of the cover) have been verified above. The conclusion provides a Lebesgue number $\delta > 0$ such that for every $x \in |K|$, the ball $B(x, \delta) \subseteq |K|$ is contained in some element of $\mathcal{U}$; that is, there exists a vertex $w_x \in V_L$ with $B(x, \delta) \subseteq f^{-1}(\operatorname{St}_L(w_x))$.[/step]

[guided]We must promote "every point of $|K|$ has a neighbourhood contained in some $f^{-1}(\operatorname{St}_L(w))$" to the stronger uniform statement "every small enough ball in $|K|$ — regardless of centre — is contained in some $f^{-1}(\operatorname{St}_L(w))$". The tool is the Lebesgue number lemma. Verify the hypotheses of the [Lebesgue Number Lemma](/theorems/952): - **Compact metric space**: $(|K|, d)$ is compact because $|K|$ is a finite union of simplices, each of which is compact (closed and bounded in Euclidean space). The metric is the standard Euclidean restriction. - **Open cover**: $\mathcal{U}$ is an open cover, verified in Step 1. The [Lebesgue Number Lemma](/theorems/952) then yields $\delta > 0$ such that every ball $B(x, \delta)$ in $|K|$ is contained in some $f^{-1}(\operatorname{St}_L(w))$. This $\delta$ is the uniform input we need to beat with the mesh of a sufficiently fine subdivision. Why is uniformity crucial? If $\delta$ were allowed to depend on $x$, we could not hope to cover $|K|$ by balls of a *fixed* radius; but the mesh of $K^{(r)}$ is a single number, bounding every open star simultaneously. Without the uniform Lebesgue number, mesh-shrinking alone would not let us assign $g(v)$ consistently.[/guided]

[step:Choose $r$ so the mesh of $K^{(r)}$ forces open stars inside single pre-images] By the [Mesh Shrinks under Subdivision](/theorems/1918) theorem, $\mu(K^{(r)}) \to 0$ as $r \to \infty$. Choose any integer $r \geq 0$ with \begin{align*} \mu(K^{(r)}) < \delta. \end{align*} We claim: for every vertex $v \in V_{K^{(r)}}$, there exists $w \in V_L$ such that $\operatorname{St}_{K^{(r)}}(v) \subseteq f^{-1}(\operatorname{St}_L(w))$. Fix $v \in V_{K^{(r)}}$. By the property of the mesh, $\operatorname{St}_{K^{(r)}}(v) \subseteq B(v, \mu(K^{(r)}))$: every point of the open star lies in a simplex with $v$ as a vertex, and in such a simplex the distance from $v$ to any other point is bounded by the length of an edge emanating from $v$, which is at most $\mu(K^{(r)})$. (More precisely: if $\sigma = \langle v, b_1, \ldots, b_k \rangle$ is a simplex containing $v$, and $y = t_0 v + \sum t_i b_i \in \sigma$ with $t_0 + \sum t_i = 1$, then $y - v = \sum t_i (b_i - v)$ and $|y - v| \leq \sum t_i |b_i - v| \leq \max_i |b_i - v| \leq \mu(K^{(r)})$.) Since $\mu(K^{(r)}) < \delta$, we have $\operatorname{St}_{K^{(r)}}(v) \subseteq B(v, \mu(K^{(r)})) \subseteq B(v, \delta)$. By the Lebesgue number property of $\delta$, $B(v, \delta) \subseteq f^{-1}(\operatorname{St}_L(w))$ for some $w \in V_L$. Hence $\operatorname{St}_{K^{(r)}}(v) \subseteq f^{-1}(\operatorname{St}_L(w))$, which rewrites as \begin{align*} f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(w). \end{align*} [/step]

[step:Define $g: V_{K^{(r)}} \to V_L$ and invoke the approximation-gives-homotopic theorem] For each vertex $v \in V_{K^{(r)}}$, select one such $w = g(v) \in V_L$ from Step 3 (the axiom of choice applied to a finite vertex set is just a finite sequence of selections, which requires no special principle). This defines a vertex map \begin{align*} g: V_{K^{(r)}} &\to V_L \\ v &\mapsto g(v) \end{align*} satisfying $f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(g(v))$ for every vertex $v$. By the definition of simplicial approximation, $g$ is a simplicial approximation to $f: |K^{(r)}| \to |L|$. By the [Barycentric Subdivision Preserves the Polyhedron](/theorems/1917) theorem, $|K^{(r)}| = |K|$, so $f: |K^{(r)}| \to |L|$ and $f: |K| \to |L|$ are the same continuous map (the identifications coincide on the level of underlying sets and metric structure). By the [Simplicial Approximation Gives Homotopic Maps](/theorems/1915) theorem applied to $g$, the map $g$ is simplicial (hence the notation $g: K^{(r)} \to L$ is justified) and $|g| \simeq f$ as continuous maps $|K| \to |L|$. This establishes the first assertion. [/step]

[step:Handle the relative case by forcing $g(v) = f(v)$ on the subcomplex $M$]Suppose $f$ is simplicial on a subcomplex $M \subseteq K$: there exists a simplicial map $h: M \to L$ with $f\big|_{|M|} = |h|$. For each vertex $v \in V_M$, simpliciality of $h$ gives $f(v) = h(v) \in V_L$. Let $M^{(r)}$ denote the $r$th barycentric subdivision of $M$; this is a subcomplex of $K^{(r)}$, and $V_M \subseteq V_{M^{(r)}} \subseteq V_{K^{(r)}}$ (the original vertices of $M$ persist in its subdivisions, being barycenters of their own $0$-simplices). We modify the construction of $g$ as follows. For each vertex $v \in V_{K^{(r)}}$ we must select $g(v) \in V_L$ with $f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(g(v))$. Two cases: **Case 1** ($v \in V_{M^{(r)}}$): We set $g(v) = h'(v)$, where $h': M^{(r)} \to L$ is the composite $V_{M^{(r)}} \xrightarrow{|h| \circ \hat{\cdot}} V_L$ — more concretely, $h'(v)$ is any vertex of $L$ that serves as a simplicial approximation to $h$ (which is already simplicial, so $h' = h$ on $V_M$, and on $V_{M^{(r)}} \setminus V_M$ we may extend via any simplicial approximation to the identity $|M^{(r)}| \to |M|$ followed by $h$). We verify this choice satisfies the approximation condition at $v \in V_{M^{(r)}}$. For $v \in V_M$ specifically, observe that since $f\big|_{|M|} = |h|$ is simplicial and sends the open star $\operatorname{St}_{K^{(r)}}(v) \cap |M|$ into $\operatorname{St}_L(h(v)) = \operatorname{St}_L(f(v))$; *however*, the open star $\operatorname{St}_{K^{(r)}}(v)$ in $K^{(r)}$ may leave $|M|$, so this direct argument is insufficient unless we also invoke the Lebesgue-number choice. To handle this cleanly, refine $r$ if necessary so that simultaneously: - $\mu(K^{(r)}) < \delta$ (so the approximation condition holds *globally* for some vertex choice), and - for each $v \in V_M$, the vertex choice $g(v) = f(v)$ from Step 3 is admissible, i.e., $f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(f(v))$. The second condition is checked as follows. The open star $\operatorname{St}_{K^{(r)}}(v) \subseteq B(v, \mu(K^{(r)}))$ (Step 3). Since $f$ is continuous at $v$ and $\operatorname{St}_L(f(v))$ is an open neighbourhood of $f(v) = f(v)$ in $|L|$, there exists $\eta_v > 0$ with $f(B(v, \eta_v) \cap |K|) \subseteq \operatorname{St}_L(f(v))$. Setting $\eta = \min_{v \in V_M} \eta_v > 0$ (finite minimum over vertices of $M$), and choosing $r$ further so that $\mu(K^{(r)}) < \min(\delta, \eta)$, the approximation condition \begin{align*} f(\operatorname{St}_{K^{(r)}}(v)) \subseteq f(B(v, \eta)) \subseteq \operatorname{St}_L(f(v)) \end{align*} holds, and we may set $g(v) = f(v)$ for each $v \in V_M$. **Case 2** ($v \in V_{K^{(r)}} \setminus V_{M^{(r)}}$): Set $g(v)$ to be any vertex satisfying $\operatorname{St}_{K^{(r)}}(v) \subseteq f^{-1}(\operatorname{St}_L(g(v)))$, as in Step 4. The vertex map $g$ thus defined is a simplicial approximation to $f$ with $g\big|_{V_M} = f\big|_{V_M}$. By the [Simplicial Approximation Gives Homotopic Maps](/theorems/1915) theorem (part 3), $|g|\big|_{|M|} = f\big|_{|M|}$ and the homotopy $|g| \simeq f$ can be taken stationary on $|M|$. This proves the relative statement. Combining Steps 1-5, the Simplicial Approximation Theorem is established.[/step]

[guided]The relative case is a refinement: we want the same construction but with $g$ pinned to $f$ on the vertices of $M$. The key is to choose the subdivision index $r$ large enough to satisfy *two* smallness conditions: 1. Global: $\mu(K^{(r)}) < \delta$, where $\delta$ is the Lebesgue number from Step 2. 2. Local on $V_M$: for each $v \in V_M$, $\mu(K^{(r)}) < \eta_v$, where $\eta_v > 0$ is a continuity modulus for $f$ at $v$ — specifically, $\eta_v$ is chosen so $f(B(v, \eta_v) \cap |K|) \subseteq \operatorname{St}_L(f(v))$. Such $\eta_v > 0$ exists because $\operatorname{St}_L(f(v))$ is an open neighbourhood of $f(v)$ in $|L|$ (since $f(v)$ is a vertex of $L$, $f(v) \in \mathring{\langle f(v) \rangle} \subseteq \operatorname{St}_L(f(v))$) and $f$ is continuous. Set $\eta = \min_{v \in V_M}\eta_v$; this minimum is positive because $V_M$ is finite. Choose $r$ with $\mu(K^{(r)}) < \min(\delta, \eta)$. With this choice, for each $v \in V_M$: \begin{align*} \operatorname{St}_{K^{(r)}}(v) \subseteq B(v, \mu(K^{(r)})) \subseteq B(v, \eta_v), \end{align*} and consequently $f(\operatorname{St}_{K^{(r)}}(v)) \subseteq f(B(v, \eta_v)) \subseteq \operatorname{St}_L(f(v))$. So $g(v) = f(v)$ satisfies the approximation condition. For $v \in V_{K^{(r)}} \setminus V_M$, we simply use Step 3's assignment. The resulting vertex map $g: V_{K^{(r)}} \to V_L$ is a simplicial approximation to $f$ with $g(v) = f(v)$ for every $v \in V_M$. By the [Simplicial Approximation Gives Homotopic Maps](/theorems/1915) theorem (part 3), this pinning of $g$ to $f$ on $V_M$ forces $|g||_{|M|} = f|_{|M|}$, and the straight-line homotopy is stationary on $|M|$.[/guided]