[proofplan] We verify the three properties of barycentric subdivision separately. For the chain map property (1), we use the definition of $\rho^X$ via the cone construction and verify naturality by checking that $f_\# \circ \rho^X = \rho^Y \circ f_\#$ on generators and that $\rho^X$ commutes with the boundary. For the chain homotopy to the identity (2), we construct an explicit chain homotopy $T_n: C_n(X) \to C_{n+1}(X)$ using iterated coning and verify the chain homotopy relation $d \circ T + T \circ d = \operatorname{id} - \rho^X$. For the diameter bound (3), we prove by induction on the number of subdivisions that the maximum diameter of simplices in $(\rho^{\Delta^n})^k(\iota_n)$ is at most $\left(\frac{n}{n+1}\right)^k \operatorname{diam}(\Delta^n)$. [/proofplan]

[step:Verify that $\rho^X$ commutes with the boundary operator]Recall the definition of the barycentric subdivision operator. For a singular $n$-simplex $\sigma: \Delta^n \to X$, the subdivision $\rho^X_n(\sigma)$ is defined inductively: $\rho^X_0 = \operatorname{id}$ on $0$-chains, and for $n \geq 1$, \begin{align*} \rho^X_n(\sigma) = \sigma_\#\bigl(\rho^{\Delta^n}_n(\iota_n)\bigr), \end{align*} where $\iota_n: \Delta^n \to \Delta^n$ is the identity simplex and $\rho^{\Delta^n}_n(\iota_n)$ is defined by the cone formula \begin{align*} \rho^{\Delta^n}_n(\iota_n) = b_n * \rho^{\Delta^n}_{n-1}(d_n(\iota_n)), \end{align*} with $b_n$ the barycentre of $\Delta^n$ and $*$ denoting the cone operator. We verify $d_n \circ \rho^X_n = \rho^X_{n-1} \circ d_n$ by induction on $n$. The base case $n = 0$ is immediate since both sides act as the identity on $0$-chains. For the inductive step, we work on the model simplex $\Delta^n$ with the identity $\iota_n$. The cone formula and the relation $d(b * c) = c - b * d(c)$ (valid for any chain $c$ with $\varepsilon(c) = 0$, where $\varepsilon$ is the augmentation) give \begin{align*} d_n\bigl(\rho^{\Delta^n}_n(\iota_n)\bigr) &= d_n\bigl(b_n * \rho^{\Delta^n}_{n-1}(d_n(\iota_n))\bigr) \\ &= \rho^{\Delta^n}_{n-1}(d_n(\iota_n)) - b_n * d_{n-1}\bigl(\rho^{\Delta^n}_{n-1}(d_n(\iota_n))\bigr). \end{align*} By the inductive hypothesis, $d_{n-1} \circ \rho^{\Delta^n}_{n-1} = \rho^{\Delta^n}_{n-2} \circ d_{n-1}$, so \begin{align*} d_n\bigl(\rho^{\Delta^n}_n(\iota_n)\bigr) = \rho^{\Delta^n}_{n-1}(d_n(\iota_n)) - b_n * \rho^{\Delta^n}_{n-2}(d_{n-1}(d_n(\iota_n))). \end{align*} Since $d_{n-1} \circ d_n = 0$ by the [Boundary Squared is Zero](/theorems/2232) theorem, the second term vanishes, and we obtain \begin{align*} d_n\bigl(\rho^{\Delta^n}_n(\iota_n)\bigr) = \rho^{\Delta^n}_{n-1}(d_n(\iota_n)). \end{align*} For a general singular simplex $\sigma: \Delta^n \to X$, we apply $\sigma_\#$ to both sides and use the naturality $\sigma_\# \circ d = d \circ \sigma_\#$, concluding $d_n(\rho^X_n(\sigma)) = \rho^X_{n-1}(d_n(\sigma))$.[/step]

[guided]The chain map property says that subdividing and then taking the boundary is the same as taking the boundary and then subdividing. The proof uses the inductive definition of $\rho$ via coning. The cone formula states that for a point $b$ and an $(n-1)$-chain $c$ with $\varepsilon(c) = 0$ (where $\varepsilon: C_0 \to \mathbb{Z}$ is the augmentation), the cone satisfies $d(b * c) = c - b * d(c)$. Why does this hold? The cone $b * c$ is the chain obtained by "filling in" from $b$ to $c$. Its boundary consists of $c$ itself (the "base" of the cone) minus the cone over the boundary of $c$ (the "sides" of the cone). The condition $\varepsilon(c) = 0$ ensures the formula works correctly in degree $0$. Starting from $\rho^{\Delta^n}_n(\iota_n) = b_n * \rho^{\Delta^n}_{n-1}(d_n(\iota_n))$, we apply $d_n$: \begin{align*} d_n\bigl(b_n * \rho^{\Delta^n}_{n-1}(d_n(\iota_n))\bigr) = \rho^{\Delta^n}_{n-1}(d_n(\iota_n)) - b_n * d_{n-1}\bigl(\rho^{\Delta^n}_{n-1}(d_n(\iota_n))\bigr). \end{align*} The augmentation condition $\varepsilon(\rho^{\Delta^n}_{n-1}(d_n(\iota_n))) = 0$ is satisfied because $\rho$ preserves augmentation and $\varepsilon(d_n(\iota_n)) = 0$ (the alternating sum of the $n+1$ vertices of $\Delta^n$ has augmentation zero for $n \geq 1$). By the inductive hypothesis, $d_{n-1} \circ \rho_{n-1} = \rho_{n-2} \circ d_{n-1}$, so the second term becomes $b_n * \rho_{n-2}(d_{n-1}(d_n(\iota_n)))$. Since $d_{n-1} \circ d_n = 0$, this vanishes entirely. The result is $d_n(\rho_n(\iota_n)) = \rho_{n-1}(d_n(\iota_n))$. To pass from the model simplex to a general $\sigma: \Delta^n \to X$, observe that $\rho^X_n(\sigma) = \sigma_\#(\rho^{\Delta^n}_n(\iota_n))$, and the pushforward $\sigma_\#$ commutes with $d$ by functoriality: \begin{align*} d(\rho^X(\sigma)) = d(\sigma_\#(\rho(\iota_n))) = \sigma_\#(d(\rho(\iota_n))) = \sigma_\#(\rho(d(\iota_n))) = \rho^X(d(\sigma)). \end{align*}[/guided]

[step:Verify naturality: $f_\# \circ \rho^X = \rho^Y \circ f_\#$ for any continuous $f: X \to Y$] For any singular $n$-simplex $\sigma: \Delta^n \to X$, we compute both sides on $\sigma$. On the left: \begin{align*} f_\#(\rho^X_n(\sigma)) = f_\#(\sigma_\#(\rho^{\Delta^n}_n(\iota_n))) = (f \circ \sigma)_\#(\rho^{\Delta^n}_n(\iota_n)), \end{align*} using the functoriality $(f \circ \sigma)_\# = f_\# \circ \sigma_\#$. On the right: \begin{align*} \rho^Y_n(f_\#(\sigma)) = \rho^Y_n(f \circ \sigma) = (f \circ \sigma)_\#(\rho^{\Delta^n}_n(\iota_n)), \end{align*} by the definition of $\rho^Y$ applied to the singular simplex $f \circ \sigma: \Delta^n \to Y$. The two expressions are identical, so $f_\# \circ \rho^X = \rho^Y \circ f_\#$. [/step]

[step:Construct a chain homotopy $T$ between $\rho^X$ and $\operatorname{id}_{C_\bullet(X)}$]We construct homomorphisms $T_n: C_n(X) \to C_{n+1}(X)$ satisfying \begin{align*} d_{n+1} \circ T_n + T_{n-1} \circ d_n = \operatorname{id}_{C_n(X)} - \rho^X_n. \end{align*} Define $T_n$ on a singular simplex $\sigma: \Delta^n \to X$ by \begin{align*} T_n(\sigma) = \sigma_\#(T_n^{\Delta^n}(\iota_n)), \end{align*} where $T_n^{\Delta^n}(\iota_n)$ is defined inductively: $T_0 = 0$ on $0$-chains, and for $n \geq 1$, \begin{align*} T_n^{\Delta^n}(\iota_n) = b_n * \bigl(\iota_n - \rho^{\Delta^n}_n(\iota_n) - T_{n-1}^{\Delta^n}(d_n(\iota_n))\bigr). \end{align*} This is the "acyclic models" construction: on the model simplex $\Delta^n$, which is contractible and hence acyclic, the chain $\iota_n - \rho_n(\iota_n) - T_{n-1}(d_n(\iota_n))$ is a cycle (by induction), and the cone from $b_n$ provides a filling. We verify the chain homotopy relation. On the model simplex: \begin{align*} d(T_n(\iota_n)) &= d\Bigl(b_n * \bigl(\iota_n - \rho_n(\iota_n) - T_{n-1}(d_n(\iota_n))\bigr)\Bigr) \\ &= \iota_n - \rho_n(\iota_n) - T_{n-1}(d_n(\iota_n)) - b_n * d\bigl(\iota_n - \rho_n(\iota_n) - T_{n-1}(d_n(\iota_n))\bigr). \end{align*} By the inductive hypothesis, $d \circ T_{n-1} + T_{n-2} \circ d = \operatorname{id} - \rho_{n-1}$ on $(n-1)$-chains, and using the chain map property $d \circ \rho_n = \rho_{n-1} \circ d$, the argument of the cone in the second term vanishes. Therefore \begin{align*} d(T_n(\iota_n)) + T_{n-1}(d_n(\iota_n)) = \iota_n - \rho_n(\iota_n). \end{align*} For a general $\sigma$, we apply $\sigma_\#$ and use its commutativity with $d$ and the naturality of $T$ (which follows from the same argument as naturality of $\rho$).[/step]

[guided]The chain homotopy $T$ is constructed by the method of acyclic models. The idea is that on the model simplex $\Delta^n$, which is contractible, every cycle is a boundary. We define $T$ on $\Delta^n$ by solving the chain homotopy equation using the cone operator, then extend to arbitrary spaces via pushforward. Why does the inductive construction work? At each stage, we need $\iota_n - \rho_n(\iota_n) - T_{n-1}(d(\iota_n))$ to be a cycle so that we can fill it. Computing its boundary using the inductive hypothesis $d \circ T_{n-1} + T_{n-2} \circ d = \operatorname{id} - \rho_{n-1}$: \begin{align*} d(\iota_n - \rho_n(\iota_n) - T_{n-1}(d(\iota_n))) &= d(\iota_n) - d(\rho_n(\iota_n)) - d(T_{n-1}(d(\iota_n))) \\ &= d(\iota_n) - \rho_{n-1}(d(\iota_n)) - \bigl[d(\iota_n) - \rho_{n-1}(d(\iota_n)) - T_{n-2}(d(d(\iota_n)))\bigr] \\ &= T_{n-2}(d_{n-1}(d_n(\iota_n))) = 0, \end{align*} where the last equality uses $d \circ d = 0$. So the chain we need to fill is indeed a cycle, and the cone from $b_n$ provides the required $(n+1)$-chain. Once $T$ is defined on model simplices, the extension to general simplices $\sigma: \Delta^n \to X$ via $T_n(\sigma) = \sigma_\#(T_n(\iota_n))$ is natural by the same functoriality argument used for $\rho$. The chain homotopy relation on $X$ follows from the relation on $\Delta^n$ by pushing forward.[/guided]

[step:Prove the diameter bound $\left(\frac{n}{n+1}\right)^k \operatorname{diam}(\Delta^n)$ for iterated subdivision] We first establish the base case $k = 1$: each simplex appearing in $\rho^{\Delta^n}(\iota_n)$ has diameter at most $\frac{n}{n+1} \operatorname{diam}(\Delta^n)$. [claim:Each simplex in the barycentric subdivision has diameter at most $\frac{n}{n+1}$ times the original] Let $\tau$ be an $n$-simplex in the barycentric subdivision of $\Delta^n$. The vertices of $\tau$ are barycentres $b_{F_0}, b_{F_1}, \ldots, b_{F_n}$ of a chain of faces $F_0 \subset F_1 \subset \cdots \subset F_n$ of $\Delta^n$, where $\dim(F_i) = i$. The diameter of $\tau$ is \begin{align*} \operatorname{diam}(\tau) = \max_{0 \le i < j \le n} |b_{F_j} - b_{F_i}|. \end{align*} Since $F_i \subset F_j$ and $b_{F_i} \in F_i \subset F_j$, we have $|b_{F_j} - b_{F_i}| \le \operatorname{diam}(F_j) \le \operatorname{diam}(\Delta^n)$. We sharpen this: the barycentre $b_{F_j}$ of a face $F_j$ with vertices $v_0, \ldots, v_j$ is $b_{F_j} = \frac{1}{j+1}(v_0 + \cdots + v_j)$. For any vertex $v_\ell$ of $F_j$, \begin{align*} |b_{F_j} - v_\ell| = \left|\frac{1}{j+1} \sum_{m=0}^{j} (v_m - v_\ell)\right| \le \frac{1}{j+1} \sum_{m \neq \ell} |v_m - v_\ell| \le \frac{j}{j+1} \operatorname{diam}(F_j). \end{align*} Since $b_{F_i}$ is a convex combination of vertices of $F_j$ (as $F_i \subset F_j$), and the distance from a barycentre to any point in a simplex is at most the maximum distance from the barycentre to a vertex, we obtain $|b_{F_j} - b_{F_i}| \le \frac{j}{j+1} \operatorname{diam}(F_j) \le \frac{n}{n+1} \operatorname{diam}(\Delta^n)$. [/claim] [proof] We prove the sharper bound that for any two points $p, q$ in a simplex $\sigma$ of dimension $n$, if $q$ is the barycentre of a face $F$ of $\sigma$ containing the point $p$, then $|p - q| \le \frac{n}{n+1} \operatorname{diam}(\sigma)$. Let $F$ have vertices $w_0, \ldots, w_m$ with $m \le n$, and write $p = \sum_{i=0}^{m} \alpha_i w_i$ with $\alpha_i \ge 0$ and $\sum \alpha_i = 1$. The barycentre is $q = \frac{1}{m+1} \sum_{i=0}^{m} w_i$. Then \begin{align*} |p - q| = \left|\sum_{i=0}^{m} \left(\alpha_i - \frac{1}{m+1}\right) w_i\right|. \end{align*} Writing $w_i = w_0 + (w_i - w_0)$ and noting that $\sum (\alpha_i - \frac{1}{m+1}) = 0$, we get \begin{align*} |p - q| = \left|\sum_{i=1}^{m} \left(\alpha_i - \frac{1}{m+1}\right)(w_i - w_0)\right| \le \sum_{i=1}^{m} \left|\alpha_i - \frac{1}{m+1}\right| \cdot |w_i - w_0|. \end{align*} Since $|w_i - w_0| \le \operatorname{diam}(F) \le \operatorname{diam}(\sigma)$ and $\sum_{i=1}^{m} |\alpha_i - \frac{1}{m+1}| \le \frac{m}{m+1} \le \frac{n}{n+1}$ (the maximum of the total variation $\sum |\alpha_i - \frac{1}{m+1}|$ over the simplex is achieved at a vertex, where it equals $\frac{m}{m+1}$), we conclude $|p - q| \le \frac{n}{n+1} \operatorname{diam}(\sigma)$. In the barycentric subdivision, any simplex $\tau$ has vertices $b_{F_0}, \ldots, b_{F_n}$ with $F_0 \subset \cdots \subset F_n$. Each pair $(b_{F_i}, b_{F_j})$ with $i < j$ satisfies $b_{F_i} \in F_i \subset F_j$ and $b_{F_j}$ is the barycentre of $F_j$, so $|b_{F_i} - b_{F_j}| \le \frac{n}{n+1} \operatorname{diam}(\Delta^n)$ by the bound above. [/proof] The general case follows by induction on $k$. After one application of $\rho$, every simplex in $\rho(\iota_n)$ has diameter at most $\frac{n}{n+1} \operatorname{diam}(\Delta^n)$. After $k$ applications, each simplex in $\rho^k(\iota_n)$ is obtained by subdividing a simplex from $\rho^{k-1}(\iota_n)$, and the subdivision ratio $\frac{n}{n+1}$ applies to each step. Therefore the maximum diameter is at most \begin{align*} \left(\frac{n}{n+1}\right)^k \operatorname{diam}(\Delta^n). \end{align*} Since $\frac{n}{n+1} < 1$, this tends to $0$ as $k \to \infty$, which is the crucial property used in the proof of excision. [/step]

[step:Assemble the three conclusions] Combining the three steps: 1. The operators $\rho^X_n$ commute with $d_n$ and with pushforward $f_\#$, so $\rho^X_\bullet: C_\bullet(X) \to C_\bullet(X)$ is a natural chain map. 2. The chain homotopy $T$ satisfies $d \circ T + T \circ d = \operatorname{id} - \rho^X$, so $\rho^X$ is chain homotopic to the identity. By the theorem that [Chain Homotopic Maps Induce Equal Maps on Homology](/theorems/1923), $\rho^X$ induces the identity on homology: $(\rho^X)_* = \operatorname{id}_{H_n(X)}$ for all $n$. 3. The diameter of each simplex in $(\rho^{\Delta^n})^k(\iota_n)$ is bounded by $\left(\frac{n}{n+1}\right)^k \operatorname{diam}(\Delta^n)$, which tends to $0$ as $k \to \infty$. [/step]