[proofplan] The statement has two parts: that any choice of $a: V_{K'} \to V_K$ sending $\hat{\sigma}$ to a vertex of $\sigma$ extends to a simplicial map and defines a simplicial approximation to $\operatorname{id}_{|K|}$, and conversely that every simplicial approximation has this form. The proof proceeds in three stages. First, we check that $a$ is well-defined as a vertex map and extends to a simplicial map $a: K' \to K$ by verifying that the vertices of any simplex of $K'$ — a flag $\sigma_0 < \sigma_1 < \cdots < \sigma_p$ of simplices of $K$ — are sent into the vertex set of the largest $\sigma_p$. Second, we verify the simplicial-approximation condition: for any $x \in |K'| = |K|$, $a$ carries $x$ into the closed carrier of $\operatorname{id}(x) = x$ in $K$. Third, we establish the converse by showing that the approximation condition applied to the vertex $\hat{\sigma}$ of $K'$ forces $a(\hat{\sigma})$ to be a vertex of $\sigma$. [/proofplan] [step:Set up the vertex data and the candidate simplicial extension] Let $K$ be a simplicial complex with vertex set $V_K$ and let $K'$ denote its [barycentric subdivision](/theorems/???), whose vertex set $V_{K'}$ is in bijection with the set of simplices of $K$ via $\sigma \leftrightarrow \hat{\sigma}$, where $\hat{\sigma}$ is the barycenter of $|\sigma|$. The simplices of $K'$ are precisely the finite totally ordered chains (flags) of simplices of $K$ under proper inclusion: \begin{align*} \{\hat{\sigma}_0, \hat{\sigma}_1, \ldots, \hat{\sigma}_p\} \in K' \iff \sigma_0 \subsetneq \sigma_1 \subsetneq \cdots \subsetneq \sigma_p \text{ in } K. \end{align*} On the level of underlying spaces, $|K'| = |K|$ (as topological subspaces of the same Euclidean embedding), and each open simplex $\mathring{\rho}$ of $K'$ indexed by a flag $\sigma_0 \subsetneq \cdots \subsetneq \sigma_p$ is the subset of $|\sigma_p|$ consisting of points whose barycentric coordinates relative to the vertices of $\sigma_p$ induce, under the flag, a strictly decreasing pattern (the precise formula is not needed below — only the inclusion $|\rho| \subseteq |\sigma_p|$). Let $a: V_{K'} \to V_K$ be any map such that for every simplex $\sigma \in K$, $a(\hat{\sigma})$ is a vertex of $\sigma$. [/step] [step:Verify that $a$ extends to a simplicial map $a: K' \to K$] We must check that for every simplex $\{\hat{\sigma}_0, \ldots, \hat{\sigma}_p\} \in K'$, the image vertex set $\{a(\hat{\sigma}_0), \ldots, a(\hat{\sigma}_p)\}$ spans a simplex of $K$. The simplex $\{\hat{\sigma}_0, \ldots, \hat{\sigma}_p\} \in K'$ corresponds to a flag $\sigma_0 \subsetneq \sigma_1 \subsetneq \cdots \subsetneq \sigma_p$ in $K$. By the hypothesis on $a$, each $a(\hat{\sigma}_i)$ is a vertex of $\sigma_i$, and because $\sigma_i \subseteq \sigma_p$ for every $i \le p$, we have \begin{align*} \{a(\hat{\sigma}_0), a(\hat{\sigma}_1), \ldots, a(\hat{\sigma}_p)\} \subseteq \{\text{vertices of } \sigma_p\}. \end{align*} Therefore the image vertex set spans a face of $\sigma_p$, which is itself a simplex of $K$ by the face-closure axiom of simplicial complexes. This holds for every simplex of $K'$, so the map $a: V_{K'} \to V_K$ extends (uniquely) to a simplicial map $a: K' \to K$, which we denote by the same letter. [/step] [step:Show that $a: K' \to K$ is a simplicial approximation to the identity] We must verify: for every $x \in |K'| = |K|$, if $\tau_x \in K$ is the unique simplex of $K$ with $x \in \mathring{\tau_x}$, then the image $|a|(x) \in |K|$ (where $|a|$ is the geometric realisation of $a$) lies in $\overline{\tau_x}$. Fix $x \in |K'|$. Let $\rho$ be the unique simplex of $K'$ with $x \in \mathring{\rho}$, corresponding to some flag $\sigma_0 \subsetneq \sigma_1 \subsetneq \cdots \subsetneq \sigma_p$ in $K$. Then $|\rho| \subseteq |\sigma_p|$, so $x \in |\sigma_p|$. Let $\tau_x$ be the unique simplex of $K$ with $x \in \mathring{\tau_x}$; since the open simplices of $K$ partition $|K|$, and $x \in |\sigma_p|$ is equivalent to $x \in \bigcup_{\sigma \subseteq \sigma_p} \mathring{\sigma}$, we have $\tau_x \subseteq \sigma_p$ (i.e., $\tau_x$ is a face of $\sigma_p$). We claim, in fact, that $\tau_x = \sigma_p$. To see this, observe more sharply: by the description of open simplices of $K'$, every point of $\mathring{\rho}$ has strictly positive barycentric coordinate at each vertex of $\sigma_p$ (this is the content of the flag-based coordinate analysis for barycentric subdivisions). Hence $x$ has strictly positive barycentric coordinate at every vertex of $\sigma_p$, which by the defining property of $\mathring{\tau_x}$ forces $\tau_x = \sigma_p$. Now $|a|(x)$ is a convex combination of the vertices $a(\hat{\sigma}_0), \ldots, a(\hat{\sigma}_p)$, each of which is a vertex of $\sigma_p = \tau_x$ (by Step 2). Consequently \begin{align*} |a|(x) \in \operatorname{conv}\{a(\hat{\sigma}_0), \ldots, a(\hat{\sigma}_p)\} \subseteq \overline{\sigma_p} = \overline{\tau_x}. \end{align*} This is the simplicial-approximation condition for $a: K' \to K$ with respect to $\operatorname{id}: |K'| \to |K|$. [guided] The simplicial-approximation condition compares the value $|a|(x)$ to the carrier of $\operatorname{id}(x) = x$ in the target complex $K$. We must show $|a|(x)$ lies in the closed carrier $\overline{\tau_x}$. The argument has two pieces. First, we locate the open simplex of $K'$ containing $x$: call it $\mathring{\rho}$, with $\rho$ indexed by a flag $\sigma_0 \subsetneq \cdots \subsetneq \sigma_p$. The top simplex of the flag, $\sigma_p$, contains $|\rho|$ and hence $x$. This already shows $\tau_x$ is a face of $\sigma_p$ — but we need the sharper claim $\tau_x = \sigma_p$. Why is $\tau_x$ the **top** simplex of the flag and not some proper face? The geometry of barycentric subdivision is built so that every point of $\mathring{\rho}$ mixes contributions from **all** vertices of $\sigma_p$: explicitly, if $\rho$ has flag $\sigma_0 \subsetneq \cdots \subsetneq \sigma_p$ and $x = \sum_{i=0}^p t_i \hat{\sigma}_i$ with $t_i > 0$ and $\sum t_i = 1$, then expanding each $\hat{\sigma}_i$ in the barycentric coordinates of $\sigma_p$ gives a coefficient of $x$ at every vertex of $\sigma_p$ that is a strictly positive convex combination of strictly positive numbers — hence strictly positive. So $x$ has strictly positive barycentric coordinates at **every** vertex of $\sigma_p$, which means $\tau_x = \sigma_p$. Second, once we know $\tau_x = \sigma_p$, we observe that the image vertices $a(\hat{\sigma}_0), \ldots, a(\hat{\sigma}_p)$ all lie in the vertex set of $\sigma_p$ (by construction of $a$ and the nesting $\sigma_i \subseteq \sigma_p$), so any convex combination of them lies in $\overline{\sigma_p}$. The affine-linear nature of $|a|$ on the closed simplex $|\rho|$ makes $|a|(x)$ exactly such a convex combination. [/guided] [/step] [step:Prove the converse: every simplicial approximation to the identity sends $\hat{\sigma}$ to a vertex of $\sigma$] Let $b: K' \to K$ be any simplicial approximation to $\operatorname{id}: |K'| \to |K|$. Fix a simplex $\sigma \in K$ and consider the vertex $\hat{\sigma} \in V_{K'}$. As a point of $|K'| = |K|$, the barycenter $\hat{\sigma}$ lies in $\mathring{\sigma}$: its barycentric coordinates with respect to the vertices of $\sigma$ are all equal to $1/(\dim \sigma + 1) > 0$, and its coordinates with respect to any other vertex of $K$ are zero. By the simplicial-approximation property applied to the point $x = \hat{\sigma}$, we have $\tau_x = \sigma$ (since $\hat{\sigma} \in \mathring{\sigma}$), and hence \begin{align*} b(\hat{\sigma}) \in \overline{\sigma}. \end{align*} But $b(\hat{\sigma})$ is a vertex of $K$ (because $b$ is simplicial), and the vertices of $\overline{\sigma}$ are exactly the vertices of $\sigma$. Therefore $b(\hat{\sigma})$ is a vertex of $\sigma$, which is the condition required of the vertex map in the statement. This shows that the vertex map of $b$ has the form described in the theorem. [/step] [step:Conclude the equivalence and close the proof] Step 2 shows that any vertex map $a: V_{K'} \to V_K$ satisfying $a(\hat{\sigma}) \in \sigma$ for every $\sigma \in K$ extends to a simplicial map $a: K' \to K$. Step 3 shows that this extension is a simplicial approximation to the identity. Step 4 shows that every simplicial approximation $b: K' \to K$ to the identity has a vertex map satisfying the same condition, hence arises in this way. Combining these, the simplicial approximations to the identity $|K'| = |K|$ are in bijection with the functions $a: V_{K'} \to V_K$ sending each $\hat{\sigma}$ to some vertex of $\sigma$. This completes the proof. [/step]