[proofplan] The strategy is to produce a single simplicial approximation $s$ at some subdivision level $K^{(r)}$ that simultaneously approximates both $f$ and $g$. Once $s$ approximates each of them, the independence of $f_*$ and $g_*$ from the choice of approximation (established in [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939)) forces $f_* = s_* \circ \nu_{K,r}^{-1} = g_*$. The construction of $s$ proceeds in two steps: first, choose $\varepsilon = \varepsilon(L) > 0$ so that every $2\varepsilon$-ball in $|L|$ lies inside some open star of $L$ (via the [Lebesgue Number Lemma](/theorems/???) applied to the open-star cover of $|L|$); second, subdivide $K$ enough times so that the $f$-image of each open star $\operatorname{St}_{K^{(r)}}(v)$ has diameter less than $\varepsilon$, whence both $f$ and $g$ map that star into a common open star of $L$ (using $\|f - g\|_\infty < \varepsilon$). The shared target vertex defines the simplicial approximation. [/proofplan] [step:Produce a Lebesgue number $\varepsilon(L)$ for the open-star cover of $|L|$] The open stars $\{\operatorname{St}_L(w) : w \in V_L\}$ form an open cover of $|L|$, since every point of $|L|$ lies in the interior of some simplex and hence in the open star of any vertex of that simplex. Since $L$ is a finite simplicial complex, $|L| \subseteq \mathbb{R}^N$ is a compact metric space (with the Euclidean metric inherited from $\mathbb{R}^N$). By the [Lebesgue Number Lemma](/theorems/???) applied to this open cover, there exists $\delta_L > 0$ such that every subset of $|L|$ of diameter less than $\delta_L$ is contained in $\operatorname{St}_L(w)$ for some vertex $w \in V_L$. Set \begin{align*} \varepsilon := \varepsilon(L) := \delta_L / 2 > 0. \end{align*} We claim every open ball $B(y, 2\varepsilon) \cap |L|$ of radius $2\varepsilon = \delta_L$ in $|L|$ centred at a point $y \in |L|$ lies in some star: in fact, every subset of $|L|$ of diameter $< \delta_L$ is contained in such a star by choice of $\delta_L$, and $B(y, 2\varepsilon) \cap |L|$ has diameter at most $2 \cdot 2\varepsilon = 2\delta_L$; replacing $\varepsilon$ by $\delta_L/4$ if needed (or directly taking $\varepsilon = \delta_L/4$) ensures the diameter bound holds. To keep the constant clean, fix \begin{align*} \varepsilon := \varepsilon(L) := \delta_L / 4, \end{align*} so that any set of diameter $< 4\varepsilon = \delta_L$ lies inside some $\operatorname{St}_L(w)$. [guided] We need a quantitative constant $\varepsilon = \varepsilon(L)$ that encodes "how close maps into $|L|$ must be in order that both maps fall inside a common open star of $L$ pointwise". This is exactly the setting of the [Lebesgue Number Lemma](/theorems/???): the cover $\{\operatorname{St}_L(w) : w \in V_L\}$ is open (open stars are open in $|L|$), and it is a cover (every $x \in |L|$ lies in some open simplex, hence in $\operatorname{St}_L(w)$ for each vertex $w$ of the carrier of $x$). Compactness of $|L|$ is the key hypothesis of the Lebesgue Number Lemma. A finite simplicial complex has finite-union-of-compact-simplices as its polyhedron, hence $|L|$ is compact. We want $\varepsilon$ to satisfy: if $f, g: |K| \to |L|$ have $\|f(x) - g(x)\| < \varepsilon$ for all $x$, and some additional fineness condition on $f$ holds, then a single vertex $w \in V_L$ can be chosen with both $f(x), g(x) \in \operatorname{St}_L(w)$. The two image points $f(x)$ and $g(x)$ together span a set of diameter $< \varepsilon + (\text{fineness of } f)$; if both terms are controlled, the two points lie in a small-diameter set, hence in a common star. Taking $\varepsilon = \delta_L/4$ gives a clean factor of safety. [/guided] [/step] [step:Assume $\|f - g\|_\infty < \varepsilon$ and apply the Lebesgue Number Lemma to $f$] Assume $f, g: |K| \to |L|$ satisfy $\|f(x) - g(x)\| < \varepsilon$ for all $x \in |K|$. Consider the open cover of $|K|$ by preimages of open $\varepsilon$-balls: \begin{align*} \mathcal{U} := \bigl\{ f^{-1}(B(y, \varepsilon) \cap |L|) : y \in |L| \bigr\}. \end{align*} Each member is open in $|K|$ because $f$ is continuous and $B(y, \varepsilon) \cap |L|$ is open in $|L|$. The collection covers $|K|$ because every $x \in |K|$ lies in $f^{-1}(B(f(x), \varepsilon) \cap |L|)$. Since $K$ is a finite simplicial complex, $|K| \subseteq \mathbb{R}^{N'}$ is compact. Apply the [Lebesgue Number Lemma](/theorems/???) to $\mathcal{U}$: there exists $\delta > 0$ such that every subset of $|K|$ of diameter less than $\delta$ is contained in some $f^{-1}(B(y, \varepsilon) \cap |L|)$, equivalently, $f$ maps every such subset into some open $\varepsilon$-ball in $|L|$. [/step] [step:Subdivide $K$ so that every closed star has diameter less than $\delta$] By [Mesh Shrinks under Subdivision](/theorems/1918), the mesh $\mu(K^{(r)}) \to 0$ as $r \to \infty$. Choose $r \geq 0$ large enough that \begin{align*} \mu(K^{(r)}) < \delta / 2. \end{align*} For any vertex $v \in V_{K^{(r)}}$, the closed star $\overline{\operatorname{St}}_{K^{(r)}}(v) = \bigcup \{\sigma \in K^{(r)} : v \in \sigma\}$ has diameter at most $2 \mu(K^{(r)}) < \delta$ (any two points lie in simplices sharing $v$, and each such simplex has diameter at most $\mu(K^{(r)})$, so the two points are within $\mu(K^{(r)}) + \mu(K^{(r)})$ of $v$). In particular the open star $\operatorname{St}_{K^{(r)}}(v) \subseteq \overline{\operatorname{St}}_{K^{(r)}}(v)$ has diameter less than $\delta$. By Step 2, $f$ maps $\operatorname{St}_{K^{(r)}}(v)$ into some open ball $B(y_v, \varepsilon) \cap |L|$ for a chosen point $y_v \in |L|$. [/step] [step:Show that both $f$ and $g$ map each open star into a common open star of $L$] Fix $v \in V_{K^{(r)}}$. By Step 3, $f(\operatorname{St}_{K^{(r)}}(v)) \subseteq B(y_v, \varepsilon)$. For any $x \in \operatorname{St}_{K^{(r)}}(v)$, $\|f(x) - y_v\| < \varepsilon$, and by hypothesis $\|f(x) - g(x)\| < \varepsilon$, so \begin{align*} \|g(x) - y_v\| \leq \|g(x) - f(x)\| + \|f(x) - y_v\| < \varepsilon + \varepsilon = 2\varepsilon. \end{align*} Therefore both $f(x)$ and $g(x)$ lie in the ball $B(y_v, 2\varepsilon) \cap |L|$, and in fact the set $f(\operatorname{St}_{K^{(r)}}(v)) \cup g(\operatorname{St}_{K^{(r)}}(v))$ has diameter at most $4\varepsilon = \delta_L$. By the choice of $\varepsilon = \delta_L / 4$ in Step 1 and the defining property of $\delta_L$ from the [Lebesgue Number Lemma](/theorems/???), every set of diameter less than $\delta_L$ in $|L|$ is contained in $\operatorname{St}_L(w)$ for some vertex $w \in V_L$. We apply this to the set $f(\operatorname{St}_{K^{(r)}}(v)) \cup g(\operatorname{St}_{K^{(r)}}(v))$: there exists $w_v \in V_L$ with \begin{align*} f(\operatorname{St}_{K^{(r)}}(v)) \cup g(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(w_v). \end{align*} [guided] The core of the argument is to exhibit a single vertex $w_v \in V_L$ that works simultaneously for $f$ and $g$ on the star of $v$. The separate star-condition \begin{align*} f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(w_v) \quad \text{and} \quad g(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(w_v) \end{align*} is what turns the map $v \mapsto w_v$ into a simplicial approximation of both $f$ and $g$ simultaneously. Why do we need the two-step control ($\varepsilon$ for the $f$-$g$ gap, $\delta$ for the $f$-fineness)? Because the star condition is a statement about $f(\operatorname{St}(v))$ being small (diameter-wise) in $|L|$: shrinking $\operatorname{St}(v)$ in $|K|$ shrinks $f(\operatorname{St}(v))$ because $f$ is continuous. But $g$'s fineness on the star is not controlled by $K$-subdivision alone — it is controlled by closeness to $f$ via the $\|f - g\|_\infty < \varepsilon$ hypothesis. So we use subdivision to make $f$ fine, and the hypothesis to make $g$ track $f$. The specific constant $\varepsilon = \delta_L / 4$ ensures: the $f$-image of a star is in a $\varepsilon$-ball (diameter $< 2\varepsilon$), the $g$-image is in a $2\varepsilon$-ball around the same center (since each $g(x)$ is within $\varepsilon$ of some $f(x) \in B(y_v, \varepsilon)$), so the union has diameter $< 4\varepsilon = \delta_L$, which falls inside a star by the Lebesgue property. This gives the required common target vertex $w_v$. [/guided] [/step] [step:Define the common simplicial approximation $s: K^{(r)} \to L$] Define the vertex map \begin{align*} s: V_{K^{(r)}} &\to V_L \\ v &\mapsto w_v, \end{align*} where $w_v$ is any vertex of $L$ satisfying Step 4 (choosing one for each $v$). By construction, for every vertex $v \in V_{K^{(r)}}$, \begin{align*} f(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(s(v)) \quad \text{and} \quad g(\operatorname{St}_{K^{(r)}}(v)) \subseteq \operatorname{St}_L(s(v)). \end{align*} The first inclusion shows $s$ is a simplicial approximation to $f$; the second shows $s$ is a simplicial approximation to $g$. That $s$ extends to a simplicial map $K^{(r)} \to L$ (rather than just a vertex map) follows automatically from the star condition: for any simplex $\sigma = [v_0, \dots, v_k] \in K^{(r)}$, the intersection $\bigcap_{i=0}^k \operatorname{St}_{K^{(r)}}(v_i)$ is non-empty (it contains the open simplex $\operatorname{int}(\sigma)$), so its $f$-image is non-empty and contained in $\bigcap_{i=0}^k \operatorname{St}_L(s(v_i))$; a non-empty intersection of open stars in $L$ forces $[s(v_0), \dots, s(v_k)]$ to span a simplex of $L$. [/step] [step:Conclude that $f_* = g_*$] By the construction in [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939), \begin{align*} f_* = s_* \circ \nu_{K,r}^{-1} = g_*, \end{align*} since $s$ is a simplicial approximation to both $f$ and $g$ at level $r$ and the definition of $f_*$ (resp. $g_*$) is independent of the approximation chosen. This holds for every $n$, completing the proof. [/step]