[proofplan]
The identity map on the antipodal sphere gives the immediate upper bound $\operatorname{ind}_{\mathbb{Z}/2}(S^n) \leq n$. To compute the cohomological index, we identify the orbit space $S^n/(\mathbb{Z}/2)$ with $\mathbb{R}P^n$ and use the standard mod-$2$ [cohomology ring](/theorems/2271) computation for real projective space. The non-vanishing of the degree-$n$ power and vanishing of the degree-$(n+1)$ power give $\operatorname{coind}_{\mathbb{Z}/2}(S^n)=n$, and the general cohomological lower bound then forces $\operatorname{ind}_{\mathbb{Z}/2}(S^n) \geq n$.
[/proofplan]
[step:Use the identity map to bound the equivariant index from above]
Let $\tau_n: S^n \to S^n$ denote the antipodal action map defined by $\tau_n(x)=-x$. Define
\begin{align*}
\operatorname{id}_{S^n}: S^n \to S^n,\quad x \mapsto x.
\end{align*}
This map is $\mathbb{Z}/2$-equivariant because, for every $x \in S^n$,
\begin{align*}
\operatorname{id}_{S^n}(\tau_n(x)) = -x = \tau_n(\operatorname{id}_{S^n}(x)).
\end{align*}
By the definition of $\operatorname{ind}_{\mathbb{Z}/2}(X)$ as the least integer $m$ for which there exists a $\mathbb{Z}/2$-equivariant map $X \to S^m$, this equivariant identity map gives
\begin{align*}
\operatorname{ind}_{\mathbb{Z}/2}(S^n) \leq n.
\end{align*}
[/step]
[step:Compute the cohomological index from the quotient $\mathbb{R}P^n$]
Let
\begin{align*}
q_n: S^n \to S^n/(\mathbb{Z}/2)
\end{align*}
be the quotient map for the antipodal action. The orbit space $S^n/(\mathbb{Z}/2)$ is naturally homeomorphic to $\mathbb{R}P^n$, since each orbit is the unordered pair $\{x,-x\}$ and therefore determines the line $\mathbb{R}x \subset \mathbb{R}^{n+1}$. Let $\mathbb{F}_2$ denote the field with two elements, and let $H^*(\mathbb{R}P^n;\mathbb{F}_2)$ denote the graded singular cohomology ring of $\mathbb{R}P^n$ with coefficients in $\mathbb{F}_2$.
First suppose $n=0$. Then $S^0/(\mathbb{Z}/2)$ is a single point, so $H^0(S^0/(\mathbb{Z}/2);\mathbb{F}_2)\cong \mathbb{F}_2$ and $H^1(S^0/(\mathbb{Z}/2);\mathbb{F}_2)=0$. If $w_0 \in H^1(S^0/(\mathbb{Z}/2);\mathbb{F}_2)$ denotes the canonical degree-$1$ cohomology class used in the definition of the mod-$2$ cohomological index, then $w_0=0$, while its zeroth power is the unit:
\begin{align*}
w_0^0 = 1 \in H^0(S^0/(\mathbb{Z}/2);\mathbb{F}_2).
\end{align*}
Thus $w_0^0 \neq 0$ and $w_0^1=0$, so the largest integer $k$ for which $w_0^k\neq 0$ is $0$. Hence
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(S^0)=0.
\end{align*}
Now suppose $n \geq 1$. By the standard mod-$2$ cohomology computation for real projective space,
\begin{align*}
H^*(\mathbb{R}P^n;\mathbb{F}_2) \cong \mathbb{F}_2[a]/(a^{n+1}),
\end{align*}
where $a \in H^1(\mathbb{R}P^n;\mathbb{F}_2)$ is the degree-$1$ generator. Hence
\begin{align*}
a^n \neq 0
\end{align*}
and
\begin{align*}
a^{n+1}=0.
\end{align*}
By the definition of the mod-$2$ cohomological index as the largest integer $k$ for which the $k$th power of the degree-$1$ class on the quotient is nonzero, it follows that
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(S^n)=n.
\end{align*}
[guided]
The cohomological index is designed to measure how far the quotient remembers powers of its canonical degree-$1$ class. For the antipodal sphere, the quotient map
\begin{align*}
q_n: S^n \to S^n/(\mathbb{Z}/2)
\end{align*}
identifies each point $x \in S^n$ with its antipode $-x$. Thus an orbit is exactly the pair $\{x,-x\}$, and this pair determines the one-dimensional linear subspace $\mathbb{R}x \subset \mathbb{R}^{n+1}$. This gives the natural identification
\begin{align*}
S^n/(\mathbb{Z}/2) \cong \mathbb{R}P^n.
\end{align*}
Let $\mathbb{F}_2$ denote the field with two elements, and let $H^*(\mathbb{R}P^n;\mathbb{F}_2)$ denote the graded singular cohomology ring of $\mathbb{R}P^n$ with coefficients in $\mathbb{F}_2$.
There is a small endpoint case to separate. If $n=0$, then $S^0$ consists of two antipodal points and the quotient $S^0/(\mathbb{Z}/2)$ is a single point. Therefore $H^0(S^0/(\mathbb{Z}/2);\mathbb{F}_2)\cong \mathbb{F}_2$ and $H^1(S^0/(\mathbb{Z}/2);\mathbb{F}_2)=0$. If $w_0 \in H^1(S^0/(\mathbb{Z}/2);\mathbb{F}_2)$ denotes the canonical degree-$1$ class used in the definition of cohomological index, then $w_0=0$. The convention for powers in a unital graded ring gives
\begin{align*}
w_0^0 = 1 \in H^0(S^0/(\mathbb{Z}/2);\mathbb{F}_2),
\end{align*}
so $w_0^0\neq 0$, while $w_0^1=0$. Thus the largest integer $k$ such that $w_0^k\neq 0$ is $0$, and hence
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(S^0)=0.
\end{align*}
Now assume $n\geq 1$. We use the standard computation of the mod-$2$ cohomology ring of real projective space:
\begin{align*}
H^*(\mathbb{R}P^n;\mathbb{F}_2) \cong \mathbb{F}_2[a]/(a^{n+1}),
\end{align*}
where $a \in H^1(\mathbb{R}P^n;\mathbb{F}_2)$ is the degree-$1$ generator. The quotient by the ideal $(a^{n+1})$ means precisely that powers of $a$ up to degree $n$ survive, while the next power vanishes. Therefore
\begin{align*}
a^n \neq 0
\end{align*}
and
\begin{align*}
a^{n+1}=0.
\end{align*}
By definition, the mod-$2$ cohomological index of a free $\mathbb{Z}/2$-space is the largest integer $k$ such that the $k$th power of the canonical degree-$1$ class on the orbit space is nonzero. For $S^n$ with $n\geq 1$, the largest such integer is exactly $n$. Hence
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(S^n)=n.
\end{align*}
[/guided]
[/step]
[step:Apply the cohomological lower bound to force the reverse inequality]
The cohomological lower bound for the equivariant index states that, for every free $\mathbb{Z}/2$-space $X$, the same mod-$2$ cohomological index used in the previous step satisfies
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(X) \leq \operatorname{ind}_{\mathbb{Z}/2}(X).
\end{align*}
The antipodal action on $S^n$ is free because $x=-x$ would imply $x=0$, while $0 \notin S^n$. Therefore the bound applies to $X=S^n$, and the previous step gives
\begin{align*}
n = \operatorname{coind}_{\mathbb{Z}/2}(S^n) \leq \operatorname{ind}_{\mathbb{Z}/2}(S^n).
\end{align*}
[/step]
[step:Combine the two inequalities]
The first step proved
\begin{align*}
\operatorname{ind}_{\mathbb{Z}/2}(S^n) \leq n.
\end{align*}
The cohomological lower bound proved
\begin{align*}
n \leq \operatorname{ind}_{\mathbb{Z}/2}(S^n).
\end{align*}
Combining these inequalities gives
\begin{align*}
\operatorname{ind}_{\mathbb{Z}/2}(S^n)=n.
\end{align*}
The second step already established
\begin{align*}
\operatorname{coind}_{\mathbb{Z}/2}(S^n)=n.
\end{align*}
Thus both claimed equalities hold.
[/step]
