[proofplan] Two chain maps that are chain homotopic induce the same map on homology, so it suffices to construct a chain homotopy between the chain maps $f_\sharp, g_\sharp: C_\bullet(K) \to C_\bullet(L)$ induced by $f$ and $g$. On a basis $n$-simplex $[a_0, \ldots, a_n]$ of $C_n(K)$ we define $h_n[a_0, \ldots, a_n]$ as the alternating sum of the $(n+1)$-simplices $[f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n)]$ obtained by interpolating between the $f$-image and the $g$-image one vertex at a time. The contiguity of $f$ and $g$ guarantees that each vertex list lies inside a single simplex of $L$, so the formula lands in $C_{n+1}(L)$. A direct computation of $d_{n+1}^L h_n + h_{n-1} d_n^K$, organised as a telescoping sum, produces exactly $g_\sharp - f_\sharp$, which is the chain-homotopy identity we need. [/proofplan]

[step:Reduce to producing a chain homotopy between $f_\sharp$ and $g_\sharp$] Let $C_\bullet(K)$ and $C_\bullet(L)$ denote the oriented simplicial chain complexes of $K$ and $L$, with differentials $d_\bullet^K$ and $d_\bullet^L$. A simplicial map $\varphi: K \to L$ induces a chain map $\varphi_\sharp: C_\bullet(K) \to C_\bullet(L)$ which on a basis simplex $[a_0, \ldots, a_n]$ is given by \begin{align*} \varphi_\sharp[a_0, \ldots, a_n] = \begin{cases} [\varphi(a_0), \ldots, \varphi(a_n)] & \text{if } \varphi(a_0), \ldots, \varphi(a_n) \text{ are distinct,} \\ 0 & \text{otherwise,} \end{cases} \end{align*} where the oriented simplex $[v_0, \ldots, v_n]$ is understood modulo even permutations of its vertices. We recall that a chain homotopy between two chain maps $\varphi_\sharp, \psi_\sharp: C_\bullet(K) \to C_\bullet(L)$ is a sequence of homomorphisms \begin{align*} h_n: C_n(K) &\to C_{n+1}(L) \end{align*} such that \begin{align*} d_{n+1}^L \circ h_n + h_{n-1} \circ d_n^K = \psi_\sharp - \varphi_\sharp \qquad (n \ge 0), \end{align*} with the convention $h_{-1} = 0$. By the [Chain Homotopy Induces Equal Maps on Homology](/theorems/???), the existence of such an $h_\bullet$ implies $\varphi_* = \psi_*$ on every $H_n$. Applying this with $\varphi = f$ and $\psi = g$ reduces the theorem to the construction of a chain homotopy $h_\bullet$ from $f_\sharp$ to $g_\sharp$. [/step]

[step:Define the prism operator $h_n$ on basis simplices]Fix $n \ge 0$ and a basis $n$-simplex $[a_0, \ldots, a_n]$ of $C_n(K)$, where $\{a_0, \ldots, a_n\}$ is the vertex set of a simplex of $K$. For each $i = 0, \ldots, n$, form the ordered vertex list \begin{align*} V_i := (f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n)) \in V_L^{n+2}, \end{align*} where $V_L$ is the vertex set of $L$. Define \begin{align*} [V_i] := \begin{cases} [f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n)] \in C_{n+1}(L) & \text{if the entries of } V_i \text{ are all distinct,} \\ 0 & \text{otherwise,} \end{cases} \end{align*} and set \begin{align*} h_n[a_0, \ldots, a_n] := \sum_{i=0}^{n} (-1)^i [V_i]. \end{align*} Extend $h_n$ linearly to all of $C_n(K)$. We must verify that $[V_i]$ is a legitimate chain in $C_{n+1}(L)$, i.e., that the vertex set of $V_i$ spans a simplex of $L$. The vertices of $V_i$ are \begin{align*} \{f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n)\} \subseteq \{f(a_0), \ldots, f(a_n)\} \cup \{g(a_0), \ldots, g(a_n)\}. \end{align*} Because $f$ and $g$ are contiguous, for every simplex $\sigma = \{a_0, \ldots, a_n\}$ of $K$ the union $f(\sigma) \cup g(\sigma)$ is contained in a single simplex $\tau$ of $L$ (this is the definition of contiguity). Hence the vertex set of $V_i$ is contained in $\tau$, which means the set spans a face of $\tau$, hence a simplex of $L$. Therefore $[V_i] \in C_{n+1}(L)$, and $h_n$ is a well-defined homomorphism $C_n(K) \to C_{n+1}(L)$.[/step]

[guided]We want to build a chain homotopy $h_n: C_n(K) \to C_{n+1}(L)$. Geometrically, the standard model is the **prism** $|\sigma| \times [0,1]$ over a simplex $|\sigma|$: its top face $|\sigma| \times \{1\}$ represents $g_\sharp(\sigma)$, its bottom face $|\sigma| \times \{0\}$ represents $f_\sharp(\sigma)$, and the prism can be triangulated by the $(n+1)$-simplices \begin{align*} [(a_0, 0), \ldots, (a_i, 0), (a_i, 1), \ldots, (a_n, 1)] \quad (i = 0, \ldots, n). \end{align*} Sending the bottom copy of $a_j$ to $f(a_j)$ and the top copy to $g(a_j)$ gives the formula for $V_i$ above. The sign $(-1)^i$ is forced by the orientation of the prism, so that the algebraic boundary matches the expected geometric boundary. Before we can even call $[V_i]$ a chain of $L$, we need its vertex set to span a simplex of $L$. This is exactly where **contiguity** is used: by definition, $f$ and $g$ are contiguous if for every simplex $\sigma = \{a_0, \ldots, a_n\}$ of $K$, there exists a simplex $\tau$ of $L$ containing both $f(\sigma)$ and $g(\sigma)$. The vertex set \begin{align*} \{f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n)\} \subseteq f(\sigma) \cup g(\sigma) \subseteq \tau \end{align*} is thus a subset of the vertex set of $\tau$, and any subset of the vertex set of a simplex of $L$ spans a face of that simplex, hence a simplex of $L$. The convention $[V_i] = 0$ when entries repeat absorbs the **degenerate** cases in which $V_i$ is not a genuine $(n+1)$-simplex (some vertex appears twice because $f(a_i) = g(a_i)$, or more generally because of collisions inside $\tau$); this is the standard convention for the oriented simplex basis and is essential for the boundary computation to telescope cleanly in the next step.[/guided]

[step:Compute $d_{n+1}^L h_n$ by expanding each $[V_i]$] We compute $d_{n+1}^L h_n[a_0, \ldots, a_n]$. Recall that for a non-degenerate oriented $(n+1)$-simplex $[w_0, \ldots, w_{n+1}]$, \begin{align*} d_{n+1}^L[w_0, \ldots, w_{n+1}] = \sum_{k=0}^{n+1} (-1)^k [w_0, \ldots, \hat{w}_k, \ldots, w_{n+1}], \end{align*} where the hat denotes omission. Applied to $[V_i]$ with vertex list $(f(a_0), \ldots, f(a_i), g(a_i), \ldots, g(a_n))$, the omission at position $k$ yields three types of terms: **(A) Omit a vertex on the $f$-side strictly before index $i$** — i.e., $k = j$ with $0 \le j < i$. The result is \begin{align*} (-1)^j [f(a_0), \ldots, \widehat{f(a_j)}, \ldots, f(a_i), g(a_i), \ldots, g(a_n)]. \end{align*} **(B) Omit a vertex on the $g$-side strictly after index $i$** — i.e., $k = j+1$ with $i < j \le n$, so the omitted vertex is $g(a_j)$ at position $j+1$ in the vertex list of $[V_i]$. The result is \begin{align*} (-1)^{j+1} [f(a_0), \ldots, f(a_i), g(a_i), \ldots, \widehat{g(a_j)}, \ldots, g(a_n)]. \end{align*} **(C) Omit one of the two "middle" vertices $f(a_i)$ or $g(a_i)$** — i.e., $k = i$ or $k = i+1$. These produce \begin{align*} (-1)^i [f(a_0), \ldots, \widehat{f(a_i)}, g(a_i), \ldots, g(a_n)] &= (-1)^i [f(a_0), \ldots, f(a_{i-1}), g(a_i), g(a_{i+1}), \ldots, g(a_n)], \\ (-1)^{i+1} [f(a_0), \ldots, f(a_i), \widehat{g(a_i)}, g(a_{i+1}), \ldots, g(a_n)] &= (-1)^{i+1} [f(a_0), \ldots, f(a_i), g(a_{i+1}), \ldots, g(a_n)]. \end{align*} Multiplying by the outer sign $(-1)^i$ from the definition of $h_n$ and summing over $i$, the total contribution to $d_{n+1}^L h_n[a_0, \ldots, a_n]$ from type (A) and type (B) is \begin{align*} \mathrm{(A)}+\mathrm{(B)} &= \sum_{i=0}^{n} \sum_{0 \le j < i} (-1)^{i+j} [f(a_0), \ldots, \widehat{f(a_j)}, \ldots, f(a_i), g(a_i), \ldots, g(a_n)] \\ &\quad + \sum_{i=0}^{n} \sum_{i < j \le n} (-1)^{i+j+1} [f(a_0), \ldots, f(a_i), g(a_i), \ldots, \widehat{g(a_j)}, \ldots, g(a_n)]. \end{align*} The type (C) contribution is \begin{align*} \mathrm{(C)} = \sum_{i=0}^{n} \Big\{ [f(a_0), \ldots, f(a_{i-1}), g(a_i), \ldots, g(a_n)] - [f(a_0), \ldots, f(a_i), g(a_{i+1}), \ldots, g(a_n)] \Big\}, \end{align*} where we have used $(-1)^i \cdot (-1)^i = 1$ and $(-1)^i \cdot (-1)^{i+1} = -1$. [/step]

[step:Telescope the (C) terms to obtain $g_\sharp - f_\sharp$]The sum $\mathrm{(C)}$ is a telescoping sum. Introducing the notation \begin{align*} T_i := [f(a_0), \ldots, f(a_{i-1}), g(a_i), g(a_{i+1}), \ldots, g(a_n)] \qquad (i = 0, \ldots, n+1), \end{align*} with the conventions $T_0 = [g(a_0), \ldots, g(a_n)]$ and $T_{n+1} = [f(a_0), \ldots, f(a_n)]$, the two parentheses inside $\mathrm{(C)}$ are precisely $T_i$ and $T_{i+1}$: \begin{align*} \mathrm{(C)} = \sum_{i=0}^{n} (T_i - T_{i+1}) = T_0 - T_{n+1} = [g(a_0), \ldots, g(a_n)] - [f(a_0), \ldots, f(a_n)]. \end{align*} By the definition of $f_\sharp$ and $g_\sharp$ — where vertex repetitions on either side force the oriented simplex to be $0$ — this is exactly \begin{align*} \mathrm{(C)} = g_\sharp[a_0, \ldots, a_n] - f_\sharp[a_0, \ldots, a_n]. \end{align*}[/step]

[guided]The core of the chain-homotopy identity is the identification of the "middle" contribution (C) with $g_\sharp - f_\sharp$. Why does this telescope? For each $i$, the first bracket omits the "bottom" vertex at index $i$ of the prism simplex $V_i$ — which erases the $f$-image of $a_i$ — and the second bracket omits the "top" vertex at index $i+1$ of $V_i$ — which erases the $g$-image of $a_i$. After erasure, the first bracket has vertex list \begin{align*} (f(a_0), \ldots, f(a_{i-1}), g(a_i), g(a_{i+1}), \ldots, g(a_n)), \end{align*} which is precisely the vertex list after the $f$-to-$g$ "transition" has moved from position $i$ to position $i-1$. Similarly the second bracket's vertex list is the list after the transition has moved from position $i$ to position $i+1$. These two families of vertex lists, as $i$ varies, form a single chain of adjacent transitions from $(g(a_0), \ldots, g(a_n))$ down to $(f(a_0), \ldots, f(a_n))$. The telescoping then delivers the two endpoints. We must check two endpoint conventions. At $i = 0$ the first bracket reads \begin{align*} [f(a_0), \ldots, \widehat{f(a_0)}, g(a_0), \ldots, g(a_n)] = [g(a_0), g(a_1), \ldots, g(a_n)], \end{align*} which is $g_\sharp[a_0, \ldots, a_n]$. At $i = n$ the second bracket reads \begin{align*} [f(a_0), \ldots, f(a_n), \widehat{g(a_n)}] = [f(a_0), \ldots, f(a_n)], \end{align*} which is $f_\sharp[a_0, \ldots, a_n]$. No other endpoint terms survive. The convention that an oriented simplex with a repeated vertex equals $0$ plays no role here in the non-degenerate case, but it ensures the formula remains correct even when some $f(a_i)$ coincides with $f(a_{i-1})$ or similar: any repetition in $V_i$ kills the whole prism term, and the contributions from neighbouring $T_i$ and $T_{i+1}$ cancel term-by-term under the same convention. The reader can verify this by tracking any single repetition through the expansion; we will not need it for the clean (non-degenerate) case.[/guided]

[step:Compute $h_{n-1} d_n^K$ and match it against the $\mathrm{(A)} + \mathrm{(B)}$ terms] Using $d_n^K[a_0, \ldots, a_n] = \sum_{j=0}^n (-1)^j [a_0, \ldots, \hat{a}_j, \ldots, a_n]$ and the definition of $h_{n-1}$ on the $(n-1)$-simplex $[a_0, \ldots, \hat{a}_j, \ldots, a_n]$, \begin{align*} h_{n-1} d_n^K[a_0, \ldots, a_n] &= \sum_{j=0}^{n} (-1)^j h_{n-1}[a_0, \ldots, \hat{a}_j, \ldots, a_n] \\ &= \sum_{j=0}^{n} (-1)^j \left( \sum_{\substack{0 \le i \le n \\ i \ne j}} (-1)^{\iota(i,j)} W_{i,j} \right), \end{align*} where $\iota(i,j) = i$ if $i < j$ and $\iota(i,j) = i - 1$ if $i > j$ (the index shift accounts for the removal of $a_j$), and \begin{align*} W_{i,j} := \begin{cases} [f(a_0), \ldots, \widehat{f(a_j)}, \ldots, f(a_i), g(a_i), \ldots, g(a_n)] & \text{if } i > j, \\ [f(a_0), \ldots, f(a_i), g(a_i), \ldots, \widehat{g(a_j)}, \ldots, g(a_n)] & \text{if } i < j. \end{cases} \end{align*} Collecting the signs: \begin{itemize} \item For $i > j$ (omission on the $f$-side): the sign is $(-1)^j (-1)^{i-1} = -(-1)^{i+j}$, and the simplex is $[f(a_0), \ldots, \widehat{f(a_j)}, \ldots, f(a_i), g(a_i), \ldots, g(a_n)]$. \item For $i < j$ (omission on the $g$-side): the sign is $(-1)^j (-1)^i = (-1)^{i+j}$, and the simplex is $[f(a_0), \ldots, f(a_i), g(a_i), \ldots, \widehat{g(a_j)}, \ldots, g(a_n)]$. \end{itemize} Thus \begin{align*} h_{n-1} d_n^K[a_0, \ldots, a_n] &= -\sum_{0 \le j < i \le n} (-1)^{i+j} [f(a_0), \ldots, \widehat{f(a_j)}, \ldots, f(a_i), g(a_i), \ldots, g(a_n)] \\ &\quad + \sum_{0 \le i < j \le n} (-1)^{i+j} [f(a_0), \ldots, f(a_i), g(a_i), \ldots, \widehat{g(a_j)}, \ldots, g(a_n)]. \end{align*} Comparing with the expressions for $\mathrm{(A)}$ and $\mathrm{(B)}$ in Step 3, we see \begin{align*} \mathrm{(A)} + h_{n-1} d_n^K[a_0, \ldots, a_n]\big|_{i>j \text{ terms}} &= 0, \\ \mathrm{(B)} + h_{n-1} d_n^K[a_0, \ldots, a_n]\big|_{i<j \text{ terms}} &= 0, \end{align*} since the signs $(-1)^{i+j}$ and $-(-1)^{i+j}$ (resp.\ $(-1)^{i+j+1}$ and $(-1)^{i+j}$) are opposite. Therefore $\mathrm{(A)} + \mathrm{(B)} + h_{n-1} d_n^K[a_0, \ldots, a_n] = 0$. [/step]

[step:Combine to obtain the chain-homotopy identity, and conclude] Summing the contributions from Steps 3-5: \begin{align*} d_{n+1}^L h_n[a_0, \ldots, a_n] + h_{n-1} d_n^K[a_0, \ldots, a_n] &= \big( \mathrm{(A)} + \mathrm{(B)} + \mathrm{(C)} \big) + h_{n-1} d_n^K[a_0, \ldots, a_n] \\ &= \mathrm{(C)} \\ &= g_\sharp[a_0, \ldots, a_n] - f_\sharp[a_0, \ldots, a_n]. \end{align*} Since every basis $n$-simplex satisfies this identity and every map involved is a homomorphism of abelian groups, the identity extends by linearity to all of $C_n(K)$: \begin{align*} d_{n+1}^L \circ h_n + h_{n-1} \circ d_n^K = g_\sharp - f_\sharp \qquad (n \ge 0), \end{align*} with $h_{-1} = 0$. Thus $h_\bullet$ is a chain homotopy from $f_\sharp$ to $g_\sharp$. By the reduction in Step 1, $f_* = g_*: H_n(K) \to H_n(L)$ for all $n \ge 0$, which completes the proof. [/step]