[proofplan]
Assume that an odd continuous map $g: S^n \to S^{n-1}$ exists. Include $S^{n-1}$ as the equator in $S^n$ and form the composite $F: S^n \to S^n$. The composite is an odd self-map of $S^n$, so its mod-$2$ degree is $1$ by the odd self-map theorem. But its image lies in the equator, so the induced map on top-dimensional mod-$2$ homology factors through $H_n(S^{n-1};\mathbb{F}_2)=0$, forcing its mod-$2$ degree to be $0$, a contradiction.
[/proofplan]
[step:Embed the target sphere as the equator of $S^n$]
Let $n \geq 1$ be fixed, as in the theorem statement. Assume, toward a contradiction, that there exists a continuous map $g: S^n \to S^{n-1}$ such that $g(-x) = -g(x)$ for every $x \in S^n$.
Define the equatorial inclusion
\begin{align*}
i: S^{n-1} &\to S^n
\end{align*}
by
\begin{align*}
i(y_1,\dots,y_n) = (y_1,\dots,y_n,0).
\end{align*}
This map is continuous and satisfies $i(-y) = -i(y)$ for every $y \in S^{n-1}$.
Define the composite map
\begin{align*}
F: S^n &\to S^n
\end{align*}
by
\begin{align*}
F(x) = (i \circ g)(x).
\end{align*}
Since both $g$ and $i$ are continuous, $F$ is continuous.
[/step]
[step:Show that the composite is an odd self-map of $S^n$]
For every $x \in S^n$, using the oddness of $g$ and the equivariance of the equatorial inclusion $i$, we compute
\begin{align*}
F(-x) = i(g(-x)).
\end{align*}
Since $g(-x) = -g(x)$, this becomes
\begin{align*}
F(-x) = i(-g(x)).
\end{align*}
Since $i(-y) = -i(y)$ for every $y \in S^{n-1}$, we obtain
\begin{align*}
F(-x) = -i(g(x)) = -F(x).
\end{align*}
Thus $F: S^n \to S^n$ is an odd continuous self-map.
By the odd self-map theorem for spheres, every odd continuous map $S^n \to S^n$ has odd degree, equivalently mod-$2$ degree equal to $1$; citing a result not yet in the wiki: Odd Self-Map Theorem for Spheres. Therefore
\begin{align*}
\deg_2(F) = 1.
\end{align*}
[guided]
The point of composing with the equatorial inclusion is to turn the assumed map $g: S^n \to S^{n-1}$ into a self-map of $S^n$, because degree is defined for maps between closed oriented manifolds of the same dimension, and mod-$2$ degree is especially convenient for spheres.
We define
\begin{align*}
i: S^{n-1} &\to S^n
\end{align*}
by
\begin{align*}
i(y_1,\dots,y_n) = (y_1,\dots,y_n,0).
\end{align*}
This is the standard inclusion of $S^{n-1}$ as the equator of $S^n$. It preserves antipodes, since
\begin{align*}
i(-y_1,\dots,-y_n) = (-y_1,\dots,-y_n,0) = -i(y_1,\dots,y_n).
\end{align*}
Now define
\begin{align*}
F: S^n &\to S^n
\end{align*}
by
\begin{align*}
F(x) = i(g(x)).
\end{align*}
The map $F$ is continuous because it is the composition of the continuous maps $g$ and $i$. We verify its oddness directly. For $x \in S^n$,
\begin{align*}
F(-x) = i(g(-x)).
\end{align*}
Since $g$ is odd, $g(-x) = -g(x)$. Hence
\begin{align*}
F(-x) = i(-g(x)).
\end{align*}
Since $i$ also preserves antipodes,
\begin{align*}
F(-x) = -i(g(x)) = -F(x).
\end{align*}
Thus $F$ is an odd continuous self-map of $S^n$.
We now use the odd self-map theorem for spheres: every continuous map $h: S^n \to S^n$ satisfying $h(-x) = -h(x)$ has mod-$2$ degree equal to $1$; citing a result not yet in the wiki: Odd Self-Map Theorem for Spheres. The map $F$ satisfies exactly these hypotheses, so
\begin{align*}
\deg_2(F) = 1.
\end{align*}
[/guided]
[/step]
[step:Compute the same mod-$2$ degree from the equatorial factorization]
Let $H_k(X;\mathbb{F}_2)$ denote singular homology with coefficients in the field $\mathbb{F}_2$. The map induced by $F$ on top-dimensional homology is
\begin{align*}
F_*: H_n(S^n;\mathbb{F}_2) \to H_n(S^n;\mathbb{F}_2).
\end{align*}
Since $F = i \circ g$, functoriality of singular homology gives
\begin{align*}
F_* = i_* \circ g_*,
\end{align*}
where
\begin{align*}
g_*: H_n(S^n;\mathbb{F}_2) \to H_n(S^{n-1};\mathbb{F}_2)
\end{align*}
and
\begin{align*}
i_*: H_n(S^{n-1};\mathbb{F}_2) \to H_n(S^n;\mathbb{F}_2)
\end{align*}
are the induced homomorphisms.
Since $n \geq 1$, the top nonzero homology of $S^{n-1}$ occurs in degree $n-1$, and therefore
\begin{align*}
H_n(S^{n-1};\mathbb{F}_2) = 0.
\end{align*}
Thus $g_*$ has zero codomain, so $F_* = i_* \circ g_*$ is the zero homomorphism. By the definition of mod-$2$ degree, $F_*$ is multiplication by $\deg_2(F)$ on the one-dimensional [vector space](/page/Vector%20Space) $H_n(S^n;\mathbb{F}_2) \cong \mathbb{F}_2$. Since $F_* = 0$, we obtain
\begin{align*}
\deg_2(F) = 0.
\end{align*}
[/step]
[step:Derive the contradiction]
The previous two steps give
\begin{align*}
\deg_2(F) = 1
\end{align*}
and
\begin{align*}
\deg_2(F) = 0
\end{align*}
in $\mathbb{F}_2$. This is impossible because $0 \neq 1$ in $\mathbb{F}_2$. Therefore the assumed odd continuous map $g: S^n \to S^{n-1}$ cannot exist.
[/step]
