There is no equivariant map $f:S^2 \to S^1$ To fix some notation, let $n \geq 2$ and let $p:S^n \to P^n$ be the canonical double cover. Let $\gamma:I \to S^n$ be a lift of a representative of a nontrivial element in $\pi_1(P_n) \cong \mathbb{Z}/2;$ hence $\gamma(0) = -\gamma(1)$ and $\gamma' := p \circ \gamma \not\simeq *.$ (For instance, let $\gamma$ be a great circle from the north pole of $S^n$ to the south pole.) Assuming there exists an equivariant map $f:S^n \to S^{n-1}$ (i.e., such that $f(-x) = -f(x)$), let $f':P^n \to P^{n-1}$ be the induced map. Show that $f' \circ \gamma' \not\simeq *.$ Conclude that there exists no such $f$ when $n = 2.$
I came up with a proof below and am wondering if it is the most straightforward way to proceed. Your input is greatly appreciated!
 A: By definition, $f' \circ \gamma' = f' \circ (p \circ \gamma) = (f' \circ p) \circ \gamma = (p \circ f) \circ \gamma = p \circ (f \circ \gamma).$ Thus the path $f \circ \gamma$ in $S^{n-1}$ lifts the loop $f' \circ \gamma'$ in $P^{n-1}.$ Note that $(f \circ \gamma)(0) = f(\gamma(0)) = f(-\gamma(1)) = -f(\gamma(1)) = -(f \circ \gamma)(1),$ so that $f \circ \gamma$ isn't a loop. Hence $f' \circ \gamma'$ cannot be null-homotopic (for then it would lift to null-homotopic loop, see e.g. Bredon, chapter 3, corollary 3.6).
Let us now show that $f' \circ \gamma' \simeq *$ when $n = 2.$ Identify $\pi_1(P^2)$ with $\mathbb{Z}/2$ and $\pi_1(P^1),\pi_1(S^1)$ with $\mathbb{Z};$ then we see geometrically that $p_*:\pi_1(S^1) \to \pi_1(P^1)$ is multiplication by $2,$ hence injective (although more generally a covering map always induces a monomorphism), and also it must be the case that $f'_*:\mathbb{Z}/2 \to \mathbb{Z}$ is the zero map (since $0 = f'_*(0) = f'_*(2) = 2f'_*(1)$). It follows that
$$ 0 = f'_* \circ \gamma'_* = (f' \circ \gamma')_* = (p \circ f \circ \gamma)_* = p_* \circ (f \circ \gamma)_* $$
and so we must have $(f \circ \gamma)_* = 0$ by injectivity of $p_*.$ But this says that $f \circ \gamma:S^1 \to S^1$ has degree $0$ and hence is null-homotopic. Finally we compose this null-homotopy $S^1 \times I \to S^1$ with $p:S^1 \to P^1$ to conclude that $f' \circ \gamma':S^1 \to P^1$ is also null-homotopic.
