[proofplan] The argument is a three-line consequence of two substantial inputs: the identification of a concrete generator of $H_n(S^n)$ (the antipodal-symmetric cycle $x$ from [Generator of $H_n(S^n)$ in the Antipodal Triangulation](/theorems/1948)), and the homotopy invariance of the induced homology map ([Homotopic Maps Induce Equal Maps on Homology](/theorems/1944)). We compute $a_*$ on the generator $x$ by directly applying the simplicial chain map of $a$, obtaining $a_*[x] = (-1)^{n+1}[x]$. For even $n$, this equals $-[x]$, which differs from $\operatorname{id}_*[x] = [x]$, so $a$ cannot be homotopic to the identity. [/proofplan] [step:Set up the antipodal-symmetric triangulation and the simplicial chain map of $a$] Let $K$ be the simplicial complex on $S^n$ with vertex set \begin{align*} V_K = \{\pm e_0, \pm e_1, \ldots, \pm e_n\} \subset \mathbb{R}^{n+1}, \end{align*} whose top-dimensional simplices are $\sigma_{\varepsilon} = (\varepsilon_0 e_0, \ldots, \varepsilon_n e_n)$ for $\varepsilon \in \{\pm 1\}^{n+1}$. The antipodal map \begin{align*} a: S^n &\to S^n \\ v &\mapsto -v \end{align*} sends $e_i \mapsto -e_i$, hence permutes the vertex set $V_K$ (it is an involution exchanging $e_i$ with $-e_i$) and sends each top simplex $\sigma_{\varepsilon}$ to another top simplex. So $a$ is a simplicial map with respect to $K$, and the induced chain map $a_\#: C_n(K) \to C_n(K)$ is defined on the basis by \begin{align*} a_\#(\sigma_{\varepsilon}) = a_\# (\varepsilon_0 e_0, \ldots, \varepsilon_n e_n) = (-\varepsilon_0 e_0, \ldots, -\varepsilon_n e_n) = \sigma_{-\varepsilon}, \end{align*} where $-\varepsilon = (-\varepsilon_0, \ldots, -\varepsilon_n)$. [/step] [step:Apply $a_\#$ to the generator $x$ and compute the coefficient] By [Generator of $H_n(S^n)$ in the Antipodal Triangulation](/theorems/1948), the chain \begin{align*} x = \sum_{\varepsilon \in \{\pm 1\}^{n+1}} \operatorname{sgn}(\varepsilon)\, \sigma_{\varepsilon}, \qquad \operatorname{sgn}(\varepsilon) := \varepsilon_0 \cdots \varepsilon_n, \end{align*} is a cycle and $[x]$ generates $H_n(K) \cong H_n(S^n) \cong \mathbb{Z}$. Applying $a_\#$ and using the formula from the previous step, \begin{align*} a_\#(x) = \sum_{\varepsilon} \operatorname{sgn}(\varepsilon) \sigma_{-\varepsilon}. \end{align*} Reindex the sum by $\delta := -\varepsilon$ (so $\varepsilon = -\delta$, and as $\varepsilon$ ranges over $\{\pm 1\}^{n+1}$ so does $\delta$): \begin{align*} a_\#(x) = \sum_{\delta} \operatorname{sgn}(-\delta)\, \sigma_{\delta}. \end{align*} The sign computation is \begin{align*} \operatorname{sgn}(-\delta) = (-\delta_0)(-\delta_1) \cdots (-\delta_n) = (-1)^{n+1}\, \delta_0 \delta_1 \cdots \delta_n = (-1)^{n+1}\, \operatorname{sgn}(\delta). \end{align*} Therefore \begin{align*} a_\#(x) = (-1)^{n+1} \sum_{\delta} \operatorname{sgn}(\delta)\, \sigma_{\delta} = (-1)^{n+1} x. \end{align*} Passing to homology classes via the induced map $a_*: H_n(K) \to H_n(K)$ (see [Simplicial Maps Induce Chain Maps](/theorems/1926) and the fact that chain maps descend to homology), we obtain \begin{align*} a_*[x] = [a_\#(x)] = (-1)^{n+1} [x]. \end{align*} [guided] The computation is essentially the tracking of a single sign. Let us walk through it. **The action on top simplices.** The antipodal map permutes vertices by $e_i \leftrightarrow -e_i$. On a top simplex $\sigma_{\varepsilon}$, it sends each vertex $\varepsilon_i e_i$ to $-\varepsilon_i e_i$, producing the oriented simplex $(-\varepsilon_0 e_0, \ldots, -\varepsilon_n e_n) = \sigma_{-\varepsilon}$. Crucially, the *orientation* of the simplex is determined by the ordering of vertices — we have *not* permuted the order, only replaced each vertex with its antipode. So $a_\#(\sigma_{\varepsilon}) = \sigma_{-\varepsilon}$ with a "+" sign. **Reindexing.** In $a_\#(x) = \sum_{\varepsilon} \operatorname{sgn}(\varepsilon) \sigma_{-\varepsilon}$, we want to collect coefficients of each $\sigma_{\delta}$ separately. Substitute $\delta = -\varepsilon$; the sum runs over the same set $\{\pm 1\}^{n+1}$ (just reindexed), and the coefficient of $\sigma_{\delta}$ is $\operatorname{sgn}(-\delta)$. **The sign flip.** Each of the $n+1$ coordinates gets negated: $-\delta_i = (-1)\delta_i$. The product has $n+1$ factors of $-1$ pulled out: \begin{align*} \operatorname{sgn}(-\delta) = \prod_{i=0}^{n} (-\delta_i) = (-1)^{n+1} \prod_{i=0}^{n} \delta_i = (-1)^{n+1} \operatorname{sgn}(\delta). \end{align*} **Result.** Factoring out $(-1)^{n+1}$ from every coefficient, $a_\#(x) = (-1)^{n+1} x$. Passing to homology (chain maps induce maps on homology by [Simplicial Maps Induce Chain Maps](/theorems/1926) and the standard "chain maps descend to homology" construction; see also [Chain Maps Induce Maps on Homology](/theorems/1922)) gives $a_*[x] = (-1)^{n+1} [x]$. **Why is the dimension $n+1$ in the exponent, not $n$?** There are $n+1$ vertices in an $n$-simplex (a triangle has 3 vertices, a tetrahedron has 4, etc.). Each contributes a sign flip, so the total is $(-1)^{n+1}$. [/guided] [/step] [step:Conclude that $a \not\simeq \operatorname{id}$ when $n$ is even] Since $n$ is even, $n + 1$ is odd, so $(-1)^{n+1} = -1$. By the previous step, \begin{align*} a_*[x] = -[x] \end{align*} in $H_n(S^n) \cong \mathbb{Z}$. On the other hand, the identity map $\operatorname{id}: S^n \to S^n$ satisfies $(\operatorname{id})_* = \operatorname{id}_{H_n(S^n)}$ by functoriality (see [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939)), so \begin{align*} (\operatorname{id})_*[x] = [x]. \end{align*} Since $[x] \neq 0$ (it is a generator of $\mathbb{Z}$, in particular nonzero), $-[x] \neq [x]$ in $\mathbb{Z}$, so $a_* \neq (\operatorname{id})_*$ on $H_n(S^n)$. Suppose, for contradiction, that $a \simeq \operatorname{id}$. By [Homotopic Maps Induce Equal Maps on Homology](/theorems/1944), we would have $a_* = (\operatorname{id})_*$ on $H_k(S^n)$ for every $k$ — contradicting the inequality we just established. Therefore $a$ is not homotopic to $\operatorname{id}$ when $n$ is even, completing the proof. [/step]