[proofplan] The theorem has three assertions, which we prove in turn. For (1), we show that the vertex map $g$ sends simplices to simplices by fixing a simplex $\sigma \in K$, choosing an interior point $x \in \mathring{\sigma}$, and using the approximation condition to force $f(x)$ into the intersection $\bigcap_i \operatorname{St}_L(g(a_i))$; a point of this intersection must sit in the interior of a simplex $\tau \in L$ whose vertices include every $g(a_i)$, so $\{g(a_0), \dots, g(a_n)\}$ spans a face of $\tau$. For (2), we build the straight-line homotopy $H(x,t) = (1-t) f(x) + t |g|(x)$ in $|L|$; convexity of each simplex shows $H$ lands in $|L|$. For (3), the approximation condition combined with the hypothesis that $f$ is simplicial on $M$ pins $g$ to $f$ on the vertices of $M$, which forces the straight-line homotopy to be stationary on $|M|$. [/proofplan]

[step:Show that $g$ sends each simplex of $K$ to a simplex of $L$]Let $\sigma = \langle a_0, \ldots, a_n \rangle \in K$. By the [Unique Interior Simplex](/theorems/1914) property, the interior $\mathring{\sigma}$ is non-empty (it is the open $n$-simplex of barycentric coordinates strictly positive on each $a_i$), so we may fix a point $x \in \mathring{\sigma}$. Since $x \in \mathring{\sigma}$ and $a_i$ is a vertex of $\sigma$, the simplex $\sigma$ itself witnesses $x \in \operatorname{St}_K(a_i)$ for every $i \in \{0, \ldots, n\}$. Applying the simplicial approximation condition $f(\operatorname{St}_K(a_i)) \subseteq \operatorname{St}_L(g(a_i))$ at each vertex, we obtain \begin{align*} f(x) \in \bigcap_{i=0}^n \operatorname{St}_L(g(a_i)). \end{align*} By the [Unique Interior Simplex](/theorems/1914) theorem applied in $L$, $f(x)$ lies in the interior of a unique simplex $\tau \in L$. The condition $f(x) \in \operatorname{St}_L(g(a_i))$ means $f(x) \in \mathring{\tau'}$ for some simplex $\tau' \in L$ with $g(a_i)$ as a vertex; by uniqueness of the interior simplex, $\tau' = \tau$, so $g(a_i)$ is a vertex of $\tau$ for every $i$. Hence $\{g(a_0), \ldots, g(a_n)\}$ is a (possibly with repetitions) subset of the vertex set of $\tau$, so it spans a face of $\tau$. This verifies the simpliciality condition: $g$ sends the vertex set of $\sigma$ to the vertex set of a simplex of $L$. Since $\sigma$ was arbitrary, $g: K \to L$ is a simplicial map.[/step]

[guided]We must check that the vertex map $g: V_K \to V_L$ satisfies the simpliciality condition: for every simplex $\sigma = \langle a_0, \ldots, a_n \rangle \in K$, the image set $\{g(a_0), \ldots, g(a_n)\}$ spans a simplex of $L$. The strategy is geometric. We pick a single point $x$ in the interior of $\sigma$ and use the approximation condition to land $f(x)$ in a simplex $\tau \in L$ whose vertices must include every $g(a_i)$. Why the interior? Because a point in the *interior* of $\sigma$ simultaneously lies in the open star of *every* vertex of $\sigma$, which is exactly what we need to apply the approximation condition at every $a_i$. The interior $\mathring{\sigma}$ is non-empty: by the [Unique Interior Simplex](/theorems/1914) theorem, the open simplices $\mathring{\sigma}$ partition $|K|$, and in particular each $\sigma$ has a non-empty interior (the set of strictly positive barycentric coordinates). Fix $x \in \mathring{\sigma}$. Now verify $x \in \operatorname{St}_K(a_i)$ for every $i$. Recall $\operatorname{St}_K(a_i) = \bigcup_{\sigma \ni a_i} \mathring{\sigma}$. The simplex $\sigma$ itself has $a_i$ as a vertex and $x \in \mathring{\sigma}$, so $x \in \operatorname{St}_K(a_i)$ by the defining union. The hypothesis of the theorem — that $g$ is a simplicial approximation — gives $f(\operatorname{St}_K(a_i)) \subseteq \operatorname{St}_L(g(a_i))$ for every vertex $a_i \in V_K$. Applied to our $x$: \begin{align*} f(x) \in \bigcap_{i=0}^n \operatorname{St}_L(g(a_i)). \end{align*} What does it mean for a point to lie in $\bigcap_i \operatorname{St}_L(g(a_i))$? By the [Unique Interior Simplex](/theorems/1914) theorem applied in $L$, every point of $|L|$ — in particular $f(x)$ — lies in the interior $\mathring{\tau}$ of a *unique* simplex $\tau \in L$. The condition $f(x) \in \operatorname{St}_L(g(a_i))$ unpacks as: there exists a simplex $\tau_i \in L$ with $g(a_i)$ as a vertex and $f(x) \in \mathring{\tau_i}$. By uniqueness of the interior simplex, every $\tau_i$ equals $\tau$; so $g(a_i)$ is a vertex of $\tau$ for every $i$. Therefore $\{g(a_0), \ldots, g(a_n)\}$ is a subset (with possible repetitions) of the vertex set of $\tau$, and hence spans a face of $\tau$, which is itself a simplex of $L$. By the definition of simplicial map, $g$ is simplicial.[/guided]

[step:Define the straight-line homotopy $H$ and verify it lands in $|L|$]Define the map \begin{align*} H: |K| \times [0, 1] &\to |L| \\ (x, t) &\mapsto (1 - t) f(x) + t\, |g|(x), \end{align*} where the convex combination is formed in the ambient Euclidean space in which $|L|$ sits. Since $f$ and $|g|$ are continuous (the latter by the [Geometric Realization of Simplicial Maps](/theorems/1913)) and the convex combination is continuous in $(x, t)$, the map $H$ is continuous as a map into Euclidean space. We verify the codomain: $H(x, t) \in |L|$ for all $(x, t) \in |K| \times [0, 1]$. Fix $x \in |K|$. By the [Unique Interior Simplex](/theorems/1914) theorem applied in $L$, $f(x)$ lies in the interior of a unique simplex $\tau \in L$. We claim $|g|(x) \in \tau$ as well. Indeed, by the [Unique Interior Simplex](/theorems/1914) theorem applied in $K$, $x$ lies in the interior of a unique simplex $\sigma = \langle a_0, \ldots, a_n \rangle \in K$. Write $x = \sum_{i=0}^n t_i a_i$ with $t_i > 0$ and $\sum t_i = 1$. By the [Geometric Realization of Simplicial Maps](/theorems/1913), \begin{align*} |g|(x) = \sum_{i=0}^n t_i\, g(a_i). \end{align*} From Step 1 applied to the simplex $\sigma$, every $g(a_i)$ is a vertex of $\tau$. Hence $|g|(x)$ is a convex combination of vertices of $\tau$ and therefore lies in $\tau$ (the convex hull of its vertices). We now have $f(x), |g|(x) \in \tau$. Since $\tau$ is a simplex in Euclidean space, it is a convex set. The convex combination \begin{align*} H(x, t) = (1 - t) f(x) + t\, |g|(x) \end{align*} therefore lies in $\tau \subseteq |L|$ for every $t \in [0, 1]$. At the endpoints, $H(x, 0) = f(x)$ and $H(x, 1) = |g|(x)$, so $H$ is a homotopy from $f$ to $|g|$. This proves (2).[/step]

[guided]For assertion (2) we must produce a continuous homotopy $H: |K| \times [0, 1] \to |L|$ with $H(\cdot, 0) = f$ and $H(\cdot, 1) = |g|$. The natural candidate in a linear ambient space is the straight-line homotopy — for each $x$, run along the segment from $f(x)$ to $|g|(x)$ at unit speed. The content of the argument is not that $H$ is continuous (that is automatic from continuity of $f$, $|g|$, and bilinearity) but that the straight line *stays inside* $|L|$. A priori, $|L|$ is only a finite union of simplices, not convex; a segment with endpoints in $|L|$ can leave $|L|$ through a gap between simplices. The key observation — and this is where simpliciality of $g$ as a simplicial approximation does the work — is that for each individual $x$, both endpoints $f(x)$ and $|g|(x)$ lie in a *single* simplex $\tau \in L$, not just in $|L|$ at large. Once we know this, convexity of the simplex $\tau$ finishes the job. Let us verify this. Fix $x \in |K|$ and apply the [Unique Interior Simplex](/theorems/1914) theorem in $L$: the point $f(x)$ sits in the interior of a unique simplex $\tau \in L$. We show $|g|(x) \in \tau$. Apply the [Unique Interior Simplex](/theorems/1914) theorem in $K$: $x$ sits in the interior of a unique simplex $\sigma = \langle a_0, \ldots, a_n \rangle \in K$. Expand $x$ in barycentric coordinates as $x = \sum t_i a_i$ with $t_i > 0$ and $\sum t_i = 1$. By the [Geometric Realization of Simplicial Maps](/theorems/1913) formula for $|g|$, \begin{align*} |g|(x) = \sum_{i=0}^n t_i\, g(a_i). \end{align*} From Step 1 we know every $g(a_i)$ is a vertex of $\tau$ (that was exactly the content of simpliciality applied to $\sigma$: the vertices of $g(\sigma)$ all lie in $\tau$). Hence $|g|(x)$ is a convex combination of vertices of $\tau$, and therefore lies in $\tau$. Now both $f(x), |g|(x) \in \tau$, and $\tau$ — as an affine $n$-simplex in Euclidean space — is convex. The straight-line homotopy \begin{align*} H(x, t) = (1 - t) f(x) + t\, |g|(x) \end{align*} therefore stays in $\tau \subseteq |L|$ for all $t \in [0, 1]$. Continuity of $H$ as a map into $|L|$ now follows from continuity into the ambient Euclidean space together with the fact that $|L|$ carries the subspace topology. At $t = 0$ and $t = 1$ we recover $f$ and $|g|$, so $H$ is a homotopy $f \simeq |g|$.[/guided]

[step:Pin $g$ to $f$ on vertices of $M$ under the relative hypothesis]Suppose $f$ is simplicial on the subcomplex $M \subseteq K$, i.e., its restriction $f\big|_{|M|}: |M| \to |L|$ equals $|h|$ for some simplicial map $h: M \to L$. For each vertex $v \in V_M$, the simpliciality of $h$ forces $f(v) = h(v) \in V_L$. By the simplicial approximation condition applied to $v$, \begin{align*} f(\operatorname{St}_K(v)) \subseteq \operatorname{St}_L(g(v)). \end{align*} Since $v \in \operatorname{St}_K(v)$ (the simplex $\langle v \rangle$ has $v$ in its interior $= \{v\}$), we obtain $f(v) \in \operatorname{St}_L(g(v))$. The open star $\operatorname{St}_L(g(v)) = \bigcup_{g(v) \in \tau' \in L} \mathring{\tau'}$ contains exactly one vertex of $L$, namely $g(v)$ itself: indeed, if $w \in V_L$ satisfies $w \in \operatorname{St}_L(g(v))$, then $w = \{w\} \in \mathring{\langle w \rangle}$ forces $\langle w \rangle$ to be a simplex of $L$ with $g(v)$ as a vertex, which happens only when $w = g(v)$ (since a $0$-simplex has exactly one vertex). Combining, $f(v) \in V_L$ and $f(v) \in \operatorname{St}_L(g(v))$ force $f(v) = g(v)$. Hence $g\big|_{V_M} = f\big|_{V_M}$.[/step]

[guided]In the relative setting we are given that $f$ is already simplicial on a subcomplex $M \subseteq K$, and we want to show that (i) $g$ agrees with $f$ on $V_M$ and (ii) the homotopy $H$ from Step 2 is stationary on $|M|$. The strategy is: if $f$ is simplicial on $M$, then $f$ sends each vertex $v \in V_M$ to a *vertex* of $L$. The approximation condition puts $f(v) \in \operatorname{St}_L(g(v))$. But open stars of vertices are tight — they contain exactly one vertex of $L$ each — so $f(v)$ must be $g(v)$. Let $v \in V_M$. Simpliciality of $f\big|_{|M|} = |h|$ means $f(v) = h(v) \in V_L$. The approximation condition applied at $v$ gives $f(\operatorname{St}_K(v)) \subseteq \operatorname{St}_L(g(v))$; applying this to the point $v$ itself — and $v$ *is* in $\operatorname{St}_K(v)$, because $v \in \mathring{\langle v \rangle}$ and $\langle v \rangle \in K$ — we obtain \begin{align*} f(v) \in \operatorname{St}_L(g(v)). \end{align*} Now the key lemma: the only vertex of $L$ inside $\operatorname{St}_L(g(v))$ is $g(v)$ itself. Why? A vertex $w \in V_L$ is its own interior, i.e., $w \in \mathring{\langle w \rangle}$. For $w$ to lie in $\operatorname{St}_L(g(v)) = \bigcup_{g(v) \in \tau' \in L} \mathring{\tau'}$, the interior $\mathring{\langle w \rangle}$ must meet some $\mathring{\tau'}$ with $g(v) \in \tau'$. By the [Unique Interior Simplex](/theorems/1914) theorem, the open simplices of $L$ partition $|L|$, so $\mathring{\langle w \rangle} = \mathring{\tau'}$, forcing $\langle w \rangle = \tau'$. But $\tau'$ has $g(v)$ as a vertex, so $w = g(v)$. Therefore $f(v) = g(v)$. This holds for every $v \in V_M$, so $g\big|_{V_M} = f\big|_{V_M}$.[/guided]

[step:Deduce that the homotopy $H$ is constant on $|M|$] We show the homotopy $H: |K| \times [0,1] \to |L|$ from Step 2 satisfies $H(x, t) = f(x)$ for all $x \in |M|$ and $t \in [0, 1]$. Let $x \in |M|$. By the [Unique Interior Simplex](/theorems/1914) theorem applied within the subcomplex $M$, $x$ lies in the interior of a unique simplex $\sigma = \langle a_0, \ldots, a_n \rangle \in M$, with $x = \sum t_i a_i$, $t_i > 0$, $\sum t_i = 1$. Since $f$ is simplicial on $|M|$, by the [Geometric Realization of Simplicial Maps](/theorems/1913) formula, \begin{align*} f(x) = \sum_{i=0}^n t_i\, f(a_i) = \sum_{i=0}^n t_i\, g(a_i) = |g|(x), \end{align*} where the middle equality uses $f(a_i) = g(a_i)$ from Step 3 (since each $a_i \in V_M$) and the final equality uses the [Geometric Realization of Simplicial Maps](/theorems/1913) formula for $|g|$. Substituting into the definition of $H$: \begin{align*} H(x, t) = (1 - t) f(x) + t\, |g|(x) = (1 - t) f(x) + t\, f(x) = f(x). \end{align*} Thus $H\big|_{|M| \times [0,1]}$ is the stationary homotopy $(x, t) \mapsto f(x)$, which proves the relative statement in (3). Combining Steps 1-4, the three assertions of the theorem are established, and the proof is complete. [/step]