[proofplan] We compare the chain complex of $K = \partial \Delta^n$ with that of $L = \Delta^n$ (all faces). In every degree $k < n$ both complexes have identical chain groups and boundary maps, since every simplex of dimension $< n$ is a proper face of $\Delta^n$ and therefore lies in $K$. The only difference is in degree $n$: $C_n(L) = \mathbb{Z}$ (generated by the fundamental simplex) while $C_n(K) = 0$. This single discrepancy determines the whole answer: for $k < n-1$ the identical boundary data forces $H_k(K) = H_k(L) = 0$ by [Homology of the Standard $n$-Simplex](/theorems/1928); in degree $n-1$ the difference eliminates the image of $d_n^L$ from the denominator, and the known fact $H_{n-1}(L) = 0$ identifies this image with $\ker d_{n-1}^L$, so $H_{n-1}(K) = \ker d_{n-1}^L = \operatorname{im} d_n^L \cong \mathbb{Z}$ once we check $d_n^L$ is injective. [/proofplan] [step:Relate the chain complexes of $K$ and $L$ in each degree] Denote $L = \Delta^n$ with all its faces and $K = \partial \Delta^n$, the subcomplex consisting of all proper faces of $\Delta^n$. The simplices of $L$ of dimension $k \leq n-1$ are exactly the proper faces of $\Delta^n$ of dimension $k$, and these are precisely the simplices of $K$ of dimension $k$. Hence \begin{align*} C_k(K) &= C_k(L) \qquad \text{for } 0 \leq k \leq n-1. \end{align*} In dimension $n$, $L$ has exactly one $n$-simplex — the fundamental simplex $[v_0, \dots, v_n]$ — so $C_n(L) \cong \mathbb{Z}$, while $K$ has no $n$-simplices (the unique $n$-simplex of $\Delta^n$ is not a proper face of itself), so $C_n(K) = 0$. In dimension $k > n$, both chain groups vanish. Furthermore, the boundary maps \begin{align*} d_k^K: C_k(K) \to C_{k-1}(K) \end{align*} and $d_k^L: C_k(L) \to C_{k-1}(L)$ coincide for $0 \leq k \leq n-1$, because they agree as $\mathbb{Z}$-linear maps on the identical generating sets (the simplicial boundary formula depends only on the simplex and its faces, all of which are proper faces of $\Delta^n$ and thus common to both complexes). [/step] [step:Conclude $H_k(K) = 0$ for $0 < k < n-1$] Let $0 < k < n-1$, so that $k, k+1 \leq n-1$. By the previous step, $C_k(K) = C_k(L)$, $C_{k+1}(K) = C_{k+1}(L)$, and the boundary maps $d_k^K = d_k^L$ and $d_{k+1}^K = d_{k+1}^L$ are identical. Hence \begin{align*} H_k(K) = \frac{\ker d_k^K}{\operatorname{im} d_{k+1}^K} = \frac{\ker d_k^L}{\operatorname{im} d_{k+1}^L} = H_k(L). \end{align*} By [Homology of the Standard $n$-Simplex](/theorems/1928), $H_k(L) = 0$ for $k > 0$. Therefore $H_k(K) = 0$ for $0 < k < n-1$. The case $k = 0$ follows in the same way when $n \geq 2$: the complexes in degrees $0$ and $1$ coincide, so $H_0(K) = H_0(L) = \mathbb{Z}$. [guided] The key structural observation is that truncating $L$ by removing the top $n$-simplex only affects $H_{n-1}$ and above. In degrees strictly below $n-1$, everything that goes into computing homology — the $k$-chains, the $(k+1)$-chains, and all boundary maps between them — is untouched by the truncation, because both $C_k$ and $C_{k+1}$ involve simplices of dimension at most $n-1$, all of which are proper faces of $\Delta^n$ and therefore present in both $K$ and $L$. Why does the top simplex not affect $H_k$ for $k < n-1$? The formula $H_k = \ker d_k / \operatorname{im} d_{k+1}$ depends on $C_k, C_{k+1}, d_k, d_{k+1}$. For $k < n-1$, we have $k+1 \leq n-1$, so the chain group $C_{k+1}$ is built from proper faces alone — the fundamental $n$-simplex does not appear. Hence the computation is identical in $K$ and in $L$, and we can import the answer from the homology of the full simplex. [/guided] [/step] [step:Compute $H_{n-1}(K)$ via the difference at the top] Now consider $k = n-1$ for $n \geq 2$. We have $C_{n-1}(K) = C_{n-1}(L)$ with the same boundary map $d_{n-1}^K = d_{n-1}^L$, so \begin{align*} \ker d_{n-1}^K = \ker d_{n-1}^L. \end{align*} The image of the incoming boundary differs: $\operatorname{im} d_n^K = 0$ because $C_n(K) = 0$, whereas $\operatorname{im} d_n^L$ is a subgroup of $C_{n-1}(L)$ that we must compute. Hence \begin{align*} H_{n-1}(K) = \frac{\ker d_{n-1}^K}{\operatorname{im} d_n^K} = \frac{\ker d_{n-1}^L}{0} = \ker d_{n-1}^L. \end{align*} We now identify $\ker d_{n-1}^L$ using the already-known homology of $L$. By [Homology of the Standard $n$-Simplex](/theorems/1928), $H_{n-1}(L) = 0$ (since $n - 1 \geq 1$). Hence \begin{align*} \frac{\ker d_{n-1}^L}{\operatorname{im} d_n^L} = 0, \qquad \text{so} \qquad \ker d_{n-1}^L = \operatorname{im} d_n^L. \end{align*} Combining, \begin{align*} H_{n-1}(K) = \operatorname{im} d_n^L. \end{align*} [/step] [step:Verify that $d_n^L: C_n(L) \to C_{n-1}(L)$ is injective] We claim $d_n^L$ is injective, which will give $\operatorname{im} d_n^L \cong C_n(L) \cong \mathbb{Z}$. The chain group $C_n(L)$ is the free abelian group on the single generator $\sigma = [v_0, v_1, \dots, v_n]$, so $C_n(L) \cong \mathbb{Z}$. The boundary is \begin{align*} d_n^L \sigma = \sum_{i=0}^{n} (-1)^i [v_0, \dots, \widehat{v_i}, \dots, v_n]. \end{align*} The simplices $\tau_i := [v_0, \dots, \widehat{v_i}, \dots, v_n]$ for $i = 0, 1, \dots, n$ are $n+1$ distinct $(n-1)$-simplices of $L$ (distinct because they have distinct vertex sets — each $\tau_i$ omits a different vertex). Hence they are linearly independent elements of the free abelian group $C_{n-1}(L)$. An integer multiple $m\sigma \in C_n(L)$ has boundary \begin{align*} d_n^L(m\sigma) = m \sum_{i=0}^{n} (-1)^i \tau_i. \end{align*} If $d_n^L(m\sigma) = 0$, linear independence of $\{\tau_i\}$ forces $m(-1)^i = 0$ for each $i$, hence $m = 0$. Therefore $\ker d_n^L = 0$ and $d_n^L$ is injective. Consequently, $\operatorname{im} d_n^L \cong C_n(L) \cong \mathbb{Z}$, and combining with the previous step, \begin{align*} H_{n-1}(K) \cong \mathbb{Z}. \end{align*} [guided] We need $d_n^L$ to be injective so that its image is free abelian of rank one, matching $C_n(L)$. The argument is direct: $C_n(L)$ has a single generator $\sigma$, and $d_n^L \sigma$ is a formal $\mathbb{Z}$-linear combination of $n+1$ different $(n-1)$-simplices with alternating signs. Because the $(n-1)$-simplices are distinct elements of the free abelian group $C_{n-1}(L)$ (they have different vertex sets), they are linearly independent, and a non-zero integer multiple of their alternating sum is non-zero. Another way to see injectivity: if $d_n^L$ had a non-trivial kernel, then since $C_n(L) \cong \mathbb{Z}$ the kernel would be all of $C_n(L)$, meaning $d_n^L = 0$. But the coefficient of (say) $\tau_0 = [v_1, \dots, v_n]$ in $d_n^L \sigma$ is $+1 \neq 0$, so $d_n^L \neq 0$. Hence the kernel is trivial. This is the step where the geometry of the boundary of a simplex enters: the boundary of the fundamental $n$-cycle is the alternating sum of its $(n-1)$-faces, which is itself a non-trivial cycle (and in fact generates $H_{n-1}$ of the sphere, as we have just shown). [/guided] [/step] [step:Assemble the three cases into the full statement] Combining Steps 2 and 4, for $n \geq 2$: \begin{align*} H_k(K) &= \begin{cases} \mathbb{Z} & k = 0, \\ 0 & 0 < k < n-1, \\ \mathbb{Z} & k = n-1. \end{cases} \end{align*} In dimensions $k \geq n$ both $C_k(K)$ and $C_{k+1}(K)$ are trivial, so $H_k(K) = 0$ there as well (this case is implicit in the statement, which ranges over dimensions where $H_k$ may be non-zero). This establishes the claimed homology of the triangulated sphere $S^{n-1}$ and completes the proof. [/step]