[proofplan] We must verify two claims about the element $x \in C_n(K)$, where $K$ is the simplicial complex on $S^n$ with vertex set $V_K = \{\pm \mathbf{e}_0, \ldots, \pm \mathbf{e}_n\}$ and top simplices $\{(\varepsilon_0 \mathbf{e}_0, \ldots, \varepsilon_n \mathbf{e}_n) : \boldsymbol{\varepsilon} \in \{\pm 1\}^{n+1}\}$. First, we compute $d x$ using the simplicial boundary formula and show that the $2^{n+1}$ summands cancel in pairs determined by flipping a single sign $\varepsilon_i$, so $dx = 0$ and $[x] \in H_n(K)$. Second, since $H_n(K) = H_n(S^n) \cong \mathbb{Z}$ by [Homology of Spheres](/theorems/1945), every nonzero class is an integer multiple of a generator; we show $[x]$ is primitive (its coefficients are not all divisible by any integer $> 1$ in the chain basis), hence a generator. The only candidate primitive class is one whose simplicial-chain coefficients are all $\pm 1$, which is exactly the form of $x$. [/proofplan]

[step:Set up the simplicial complex and the chain $x$] Let $K$ be the simplicial complex with vertex set \begin{align*} V_K = \{\pm \mathbf{e}_0, \pm \mathbf{e}_1, \ldots, \pm \mathbf{e}_n\} \subset \mathbb{R}^{n+1}. \end{align*} The top-dimensional simplices are indexed by sign vectors $\boldsymbol{\varepsilon} = (\varepsilon_0, \ldots, \varepsilon_n) \in \{\pm 1\}^{n+1}$: for each $\boldsymbol{\varepsilon}$, the simplex $\sigma_{\boldsymbol{\varepsilon}} = (\varepsilon_0 \mathbf{e}_0, \ldots, \varepsilon_n \mathbf{e}_n)$ is an oriented $n$-simplex with vertices $\varepsilon_i \mathbf{e}_i$. The geometric realisation $|K|$ is the boundary of the cross-polytope $\{x \in \mathbb{R}^{n+1} : \sum |x_i| \leq 1\}$, and the radial projection $|K| \to S^n$, $y \mapsto y/|y|$, is a homeomorphism, so $H_*(K) \cong H_*(S^n)$. The element under consideration is \begin{align*} x = \sum_{\boldsymbol{\varepsilon} \in \{\pm 1\}^{n+1}} \operatorname{sgn}(\boldsymbol{\varepsilon}) \cdot \sigma_{\boldsymbol{\varepsilon}} \in C_n(K), \end{align*} where $\operatorname{sgn}(\boldsymbol{\varepsilon}) := \varepsilon_0 \varepsilon_1 \cdots \varepsilon_n \in \{\pm 1\}$ is the product of signs. [/step]

[step:Compute the boundary $dx$ via the standard simplicial boundary formula]The simplicial boundary operator acts on an oriented $n$-simplex by \begin{align*} d(v_0, v_1, \ldots, v_n) = \sum_{i=0}^{n} (-1)^i (v_0, \ldots, \widehat{v_i}, \ldots, v_n), \end{align*} where $\widehat{v_i}$ means $v_i$ is omitted. Applying this to $\sigma_{\boldsymbol{\varepsilon}} = (\varepsilon_0 \mathbf{e}_0, \ldots, \varepsilon_n \mathbf{e}_n)$, \begin{align*} d \sigma_{\boldsymbol{\varepsilon}} = \sum_{i=0}^{n} (-1)^i \tau_i(\boldsymbol{\varepsilon}), \end{align*} where \begin{align*} \tau_i(\boldsymbol{\varepsilon}) := (\varepsilon_0 \mathbf{e}_0, \ldots, \widehat{\varepsilon_i \mathbf{e}_i}, \ldots, \varepsilon_n \mathbf{e}_n) \end{align*} is the oriented $(n-1)$-simplex whose vertex set is $\{\varepsilon_j \mathbf{e}_j : j \neq i\}$. Crucially, $\tau_i(\boldsymbol{\varepsilon})$ depends only on the coordinates $\varepsilon_j$ for $j \neq i$ — it is *independent* of $\varepsilon_i$. Expanding $dx$ by linearity, \begin{align*} dx = \sum_{\boldsymbol{\varepsilon}} \operatorname{sgn}(\boldsymbol{\varepsilon}) \sum_{i=0}^{n} (-1)^i \tau_i(\boldsymbol{\varepsilon}) = \sum_{i=0}^{n} (-1)^i \sum_{\boldsymbol{\varepsilon}} \operatorname{sgn}(\boldsymbol{\varepsilon}) \tau_i(\boldsymbol{\varepsilon}). \end{align*} For each fixed $i$, we partition the sum over $\boldsymbol{\varepsilon} \in \{\pm 1\}^{n+1}$ into pairs $(\boldsymbol{\varepsilon}, \boldsymbol{\varepsilon}')$ where $\boldsymbol{\varepsilon}'$ is obtained from $\boldsymbol{\varepsilon}$ by flipping the $i$-th sign: $\varepsilon_j' = \varepsilon_j$ for $j \neq i$ and $\varepsilon_i' = -\varepsilon_i$. Within each such pair: - $\tau_i(\boldsymbol{\varepsilon}) = \tau_i(\boldsymbol{\varepsilon}')$, since $\tau_i$ does not depend on $\varepsilon_i$. - $\operatorname{sgn}(\boldsymbol{\varepsilon}') = -\operatorname{sgn}(\boldsymbol{\varepsilon})$, since flipping $\varepsilon_i$ flips the sign of the product. So \begin{align*} \operatorname{sgn}(\boldsymbol{\varepsilon}) \tau_i(\boldsymbol{\varepsilon}) + \operatorname{sgn}(\boldsymbol{\varepsilon}') \tau_i(\boldsymbol{\varepsilon}') = \operatorname{sgn}(\boldsymbol{\varepsilon})\, \tau_i(\boldsymbol{\varepsilon}) - \operatorname{sgn}(\boldsymbol{\varepsilon})\, \tau_i(\boldsymbol{\varepsilon}) = 0. \end{align*} Since the $2^{n+1}$ sign vectors partition into $2^n$ disjoint such pairs (indexed by the values of $\varepsilon_j$ for $j \neq i$), the inner sum over $\boldsymbol{\varepsilon}$ vanishes for each $i$. Therefore $dx = 0$ and $[x] \in H_n(K)$.[/step]

[guided]The computation looks formidable but the cancellation mechanism is elegant. Let us unpack it. **The key observation.** When we compute the $i$-th face $\tau_i(\boldsymbol{\varepsilon})$ of the simplex $\sigma_{\boldsymbol{\varepsilon}}$, we delete the $i$-th vertex $\varepsilon_i \mathbf{e}_i$. The remaining vertices $\{\varepsilon_j \mathbf{e}_j : j \neq i\}$ do not involve $\varepsilon_i$ at all. So the face $\tau_i(\boldsymbol{\varepsilon})$ is *independent of the sign $\varepsilon_i$*. This creates the pairing mechanism. Consider the involution on $\{\pm 1\}^{n+1}$ that flips only the $i$-th coordinate: $\boldsymbol{\varepsilon} \mapsto \boldsymbol{\varepsilon}'$. Why does it produce a sign cancellation? *On the face.* $\tau_i(\boldsymbol{\varepsilon}) = \tau_i(\boldsymbol{\varepsilon}')$, as just observed. *On the sign.* $\operatorname{sgn}(\boldsymbol{\varepsilon}') = \varepsilon_0 \cdots (-\varepsilon_i) \cdots \varepsilon_n = -\operatorname{sgn}(\boldsymbol{\varepsilon})$. So when we sum the contributions to $(dx)_i$ (the $i$-th piece of the boundary, contributed by deleting the $i$-th vertex) over $\boldsymbol{\varepsilon}$ and $\boldsymbol{\varepsilon}'$, the face is the same but the signs differ, and the pair cancels. **Why pair up this way?** We are looking for an involution on $\{\pm 1\}^{n+1}$ that preserves $\tau_i$ (to produce a common face) but flips $\operatorname{sgn}$ (to produce cancellation). The involution "flip $\varepsilon_i$" does exactly this — and it is the *only* such involution we need, because for each $i$ we run the argument with a different pairing. **The partition.** For fixed $i$, the group $\mathbb{Z}/2$ acts on $\{\pm 1\}^{n+1}$ by flipping $\varepsilon_i$, and the orbits are exactly the $2^n$ pairs. No sign vector is fixed by this involution (since flipping a $\pm 1$ coordinate never returns the same value), so the orbits are all pairs. **Why $dx = 0$ globally.** For each $i \in \{0, 1, \ldots, n\}$, the inner sum over $\boldsymbol{\varepsilon}$ is zero. Thus the outer sum over $i$ is also zero, giving $dx = 0$. This is a characteristically "symmetry-based" computation: the structure of the chain $x$ encodes a sign-cancellation that makes it automatically closed.[/guided]

[step:Identify $[x]$ as a generator of $H_n(S^n) \cong \mathbb{Z}$] Under the homeomorphism $|K| \cong S^n$ (radial projection from the cross-polytope boundary), simplicial homology gives $H_n(K) \cong H_n(S^n) \cong \mathbb{Z}$ by [Homology of Spheres](/theorems/1945) (applied with parameter $n$ where the statement reads $H_{n-1}(S^{n-1}) = \mathbb{Z}$, but with the dimension indices shifted to match our ambient $S^n$). Fix a generator $y \in H_n(K)$. Then $[x] = k \cdot y$ for some $k \in \mathbb{Z}$; we must show $k = \pm 1$. Choose a representative cycle $\tilde y \in Z_n(K)$ for $y$ in the simplicial basis $\{\sigma_{\boldsymbol{\varepsilon}}\}_{\boldsymbol{\varepsilon} \in \{\pm 1\}^{n+1}}$ of $C_n(K)$ (these are all the $n$-simplices of $K$): write \begin{align*} \tilde y = \sum_{\boldsymbol{\varepsilon}} c_{\boldsymbol{\varepsilon}}\, \sigma_{\boldsymbol{\varepsilon}}, \qquad c_{\boldsymbol{\varepsilon}} \in \mathbb{Z}. \end{align*} Since $[x] = k \cdot [\tilde y]$ in $H_n(K)$, we have $x - k \tilde y \in B_n(K) = \operatorname{Range}(d_{n+1})$. But $C_{n+1}(K) = 0$ (the simplicial complex $K$ has no simplices of dimension above $n$, since the top-dimensional simplices of $|K| = \partial(\text{cross-polytope})$ are exactly the $n$-dimensional $\sigma_{\boldsymbol{\varepsilon}}$). Hence $B_n(K) = 0$, and so $x = k \tilde y$ as chains: \begin{align*} \operatorname{sgn}(\boldsymbol{\varepsilon}) = k\, c_{\boldsymbol{\varepsilon}} \quad \text{for every } \boldsymbol{\varepsilon} \in \{\pm 1\}^{n+1}. \end{align*} Since $\operatorname{sgn}(\boldsymbol{\varepsilon}) = \pm 1$ and $c_{\boldsymbol{\varepsilon}} \in \mathbb{Z}$, the equation $\pm 1 = k \cdot c_{\boldsymbol{\varepsilon}}$ forces $k \in \{\pm 1\}$ (and $c_{\boldsymbol{\varepsilon}} = \pm 1$ for every $\boldsymbol{\varepsilon}$, of sign determined by $\operatorname{sgn}(\boldsymbol{\varepsilon})$). Therefore $[x] = \pm y$, which is a generator of $H_n(K) \cong \mathbb{Z}$. This completes the proof.[/step]

[guided]**Why is there nothing to quotient by at the chain level?** The simplicial complex $K$ has $n$-dimensional simplices as its top cells — there are no $(n+1)$-simplices. Hence the boundary operator $d_{n+1}: C_{n+1}(K) \to C_n(K)$ has trivial domain, so $B_n(K) = \operatorname{Range}(d_{n+1}) = 0$. Consequently $H_n(K) = Z_n(K) / B_n(K) = Z_n(K)$: the $n$-cycles *are* the $n$-th homology, with no quotienting. This is a huge simplification. The usual difficulty in showing an element is a generator is that you do not know what lies in $B_n$; here $B_n = 0$, so you can work directly at the chain level. **The primitive-coefficient argument.** We know $H_n(K) \cong \mathbb{Z}$, so pick any generator $y \in H_n(K)$. Lift it to a cycle $\tilde y \in Z_n(K)$ with integer coefficients $c_{\boldsymbol{\varepsilon}}$ in the simplex basis. Since $[x] \in H_n(K) = \mathbb{Z} \cdot [\tilde y]$, we have $[x] = k [\tilde y]$ for some $k \in \mathbb{Z}$, and then $x - k \tilde y \in B_n(K) = 0$, so $x = k \tilde y$ as chains. Reading off coefficients in the $\sigma_{\boldsymbol{\varepsilon}}$ basis, $\operatorname{sgn}(\boldsymbol{\varepsilon}) = k c_{\boldsymbol{\varepsilon}}$. Since the LHS is $\pm 1$, $k$ must divide $1$ in $\mathbb{Z}$, so $k \in \{\pm 1\}$. **The upshot.** Any chain whose coefficients are all $\pm 1$ and which is a cycle must be a generator of $H_n \cong \mathbb{Z}$ (as long as $B_n$ is trivial, which holds here). The chain $x$ fits this description, so $x$ generates $H_n(K) \cong H_n(S^n) \cong \mathbb{Z}$. **Sanity check.** The element $-x$ is also a generator (it is the other generator of $\mathbb{Z}$). The sign ambiguity corresponds to the choice of orientation of $S^n$ — reflecting any single coordinate $\varepsilon_i \mapsto -\varepsilon_i$ sends $x$ to $-x$, consistent with the fact that such reflections reverse the orientation of $S^n$.[/guided]