[proofplan] Contiguity of two simplicial maps $f, g: K \to L$ is a simplex-by-simplex condition: for every simplex $\sigma$ of $K$, the union $f(\sigma) \cup g(\sigma)$ must be contained in a single simplex of $L$. We fix a simplex $\sigma \in K$ and locate, via the open-simplex decomposition of $|L|$, a single simplex $\tau \in L$ whose open interior $\mathring{\tau}$ contains $F(s)$ for every $s$ in the interior $\mathring{\sigma}$. The defining property of a simplicial approximation then places the vertex images $f(\sigma), g(\sigma)$ inside the closed simplex $\tau$, giving the required joint containment. [/proofplan] [step:Fix a simplex of $K$ and locate a simplex of $L$ whose interior meets $F(\mathring{\sigma})$] Fix a simplex $\sigma \in K$ and pick an arbitrary point $s \in \mathring{\sigma}$, where $\mathring{\sigma}$ denotes the interior of $|\sigma|$ in the affine subspace it spans (equivalently, the points of $|\sigma|$ whose barycentric coordinates with respect to the vertices of $\sigma$ are all strictly positive). The open simplices $\{\mathring{\tau} : \tau \in L\}$ partition $|L|$, so there is a unique simplex $\tau = \tau(s) \in L$ with \begin{align*} F(s) \in \mathring{\tau}. \end{align*} [/step] [step:Apply the definition of simplicial approximation to place $f(\sigma)$ and $g(\sigma)$ inside $\tau$] Recall the definition of simplicial approximation: a simplicial map $\varphi: K \to L$ is a [simplicial approximation](/theorems/???) to $F: |K| \to |L|$ if for every point $x \in |K|$, the image $\varphi(x) \in |L|$ lies in the closed simplex $\overline{\tau_x}$, where $\tau_x \in L$ is the unique simplex of $L$ with $F(x) \in \mathring{\tau_x}$. Equivalently (and this is the form we use), for every simplex $\sigma \in K$ and every $s \in \mathring{\sigma}$, every vertex of $\sigma$ is sent by $\varphi$ to a vertex of $\tau(s)$; in particular $\varphi(\sigma) \subseteq \{\text{vertices of } \tau(s)\}$. Applying this to $f$ and to $g$ with the same choice of $s \in \mathring{\sigma}$ and the same simplex $\tau = \tau(s) \in L$ constructed in Step 1, we conclude \begin{align*} f(\sigma) \subseteq \{\text{vertices of } \tau\}, \qquad g(\sigma) \subseteq \{\text{vertices of } \tau\}. \end{align*} Hence \begin{align*} f(\sigma) \cup g(\sigma) \subseteq \{\text{vertices of } \tau\}. \end{align*} [guided] The definition of simplicial approximation is a *pointwise* condition: for every $x \in |K|$, the value $\varphi(x)$ lies in the closed simplex of $L$ containing $F(x)$ in its interior. We need to package this pointwise data at the level of simplices. The standard packaging is: > For any simplex $\sigma \in K$ and any point $s$ in its open interior $\mathring{\sigma}$, there exists a simplex $\tau = \tau(s) \in L$ with $F(s) \in \mathring{\tau}$, and every vertex of $\sigma$ is mapped by $\varphi$ to a vertex of $\tau$. Let us see why this follows from the pointwise statement. Take a vertex $a$ of $\sigma$. The carrier of $a$ is $\{a\}$, so $F(a) \in \mathring{\tau_a}$ for a unique $\tau_a \in L$, and $\varphi(a) \in \overline{\tau_a}$ — but $\varphi(a)$ is a vertex of $L$, and the vertices of $\overline{\tau_a}$ are exactly the vertices of $\tau_a$. We still need $\tau_a$ to be a face of $\tau$. This is where the geometry of the open-simplex decomposition enters: the simplex $\tau$ containing $F(s)$ in its interior is characterised by \begin{align*} \overline{\tau} = \bigcap_{\epsilon > 0} \overline{F(B(s,\epsilon) \cap \mathring{\sigma})}, \end{align*} and any point $F(a)$ lying in the closure of $F(\mathring{\sigma})$ (which holds because $a \in \overline{\mathring{\sigma}}$ and $F$ is continuous) has its open-simplex $\mathring{\tau_a}$ satisfying $\overline{\tau_a} \subseteq \overline{\tau}$, i.e., $\tau_a$ is a face of $\tau$. In particular the vertices of $\tau_a$ are vertices of $\tau$, so $\varphi(a)$ is a vertex of $\tau$. This is the packaging we cite above. With the packaging in hand, applying it to both $f$ and $g$ — with the **same** test point $s \in \mathring{\sigma}$ and hence the **same** simplex $\tau = \tau(s)$ — gives $f(\sigma), g(\sigma) \subseteq \{\text{vertices of } \tau\}$ simultaneously. The key point is that the simplex $\tau$ is determined entirely by $F$ and $s$; it does not depend on which approximation we are considering. [/guided] [/step] [step:Conclude that $f$ and $g$ are contiguous] The vertex set $f(\sigma) \cup g(\sigma)$ is contained in the vertex set of the simplex $\tau$ of $L$, so it spans a face of $\tau$, which is itself a simplex of $L$ (by the face-closure axiom for simplicial complexes). Since the choice of $\sigma \in K$ was arbitrary, the contiguity condition \begin{align*} \text{for every simplex } \sigma \in K, \ f(\sigma) \cup g(\sigma) \text{ is contained in a simplex of } L \end{align*} is satisfied. Therefore $f$ and $g$ are contiguous, completing the proof. [/step]