[proofplan]
We pass from the odd map $f: S^n \to S^n$ to the induced quotient map $\bar f: \mathbb{R}P^n \to \mathbb{R}P^n$. Oddness identifies the pullback of the antipodal double cover along $\bar f$ with the original antipodal double cover, so the first Stiefel-Whitney class of that cover is fixed by $\bar f^*$. Since the mod two [cohomology ring](/theorems/2271) of $\mathbb{R}P^n$ is generated by this degree-one class, $\bar f$ has mod two degree $1$. Finally, counting preimages in antipodal pairs shows that the mod two degree of $f$ agrees with the mod two degree of $\bar f$.
[/proofplan]
[step:Descend the odd map to projective space]
Let $q: S^n \to \mathbb{R}P^n$ denote the antipodal quotient map, defined by $q(x) = [x]$, where $[x] = \{x,-x\}$. Since $f(-x) = -f(x)$ for every $x \in S^n$, the formula
\begin{align*}
\bar f: \mathbb{R}P^n \to \mathbb{R}P^n
\end{align*}
given by $\bar f([x]) = [f(x)]$ is well-defined. Indeed, if $[x] = [x']$, then $x' = x$ or $x' = -x$, and therefore $[f(x')] = [f(x)]$ in both cases. The map $\bar f$ is continuous because it is the unique continuous map satisfying
\begin{align*}
\bar f \circ q = q \circ f,
\end{align*}
with $q$ a quotient map.
[/step]
[step:Identify the pulled-back double cover with the antipodal cover]
Let $P_{\bar f}$ be the pullback space
\begin{align*}
P_{\bar f} := \{([x],y) \in \mathbb{R}P^n \times S^n : \bar f([x]) = q(y)\}.
\end{align*}
Let $\pi_{\bar f}: P_{\bar f} \to \mathbb{R}P^n$ be the projection map $\pi_{\bar f}([x],y) = [x]$. This is the pullback of the double cover $q: S^n \to \mathbb{R}P^n$ along $\bar f$.
Define
\begin{align*}
\Phi: S^n &\to P_{\bar f}
\end{align*}
by
\begin{align*}
\Phi(x) = ([x], f(x)).
\end{align*}
This is well-defined because
\begin{align*}
q(f(x)) = [f(x)] = \bar f([x]).
\end{align*}
The map $\Phi$ is continuous as a product of continuous maps.
We claim that $\Phi$ is an isomorphism of double covers over $\mathbb{R}P^n$. First,
\begin{align*}
\pi_{\bar f}(\Phi(x)) = \pi_{\bar f}([x],f(x)) = [x] = q(x),
\end{align*}
so $\Phi$ lies over the identity map on $\mathbb{R}P^n$. Second, for any $([x],y) \in P_{\bar f}$, the equality $\bar f([x]) = q(y)$ means $[f(x)] = [y]$, so $y = f(x)$ or $y = -f(x) = f(-x)$. Thus $([x],y)$ equals either $\Phi(x)$ or $\Phi(-x)$. Since $x$ and $-x$ are precisely the two points over $[x]$, this proves that $\Phi$ is bijective on every fiber. This fiberwise bijection is a covering isomorphism: for any evenly covered neighbourhood $V \subset \mathbb{R}P^n$, both covers restrict over $V$ to two disjoint sheets mapped homeomorphically onto $V$, and a fiber-preserving continuous bijection sends each connected sheet into one sheet over $V$ and restricts there to the inverse-composed homeomorphism onto $V$. Thus $P_{\bar f} \to \mathbb{R}P^n$ is isomorphic to the antipodal cover $q: S^n \to \mathbb{R}P^n$.
[guided]
The point of this step is to translate oddness into a statement about double covers. The antipodal quotient map
\begin{align*}
q: S^n \to \mathbb{R}P^n
\end{align*}
sends a point $x$ to the unordered pair $[x] = \{x,-x\}$. Pulling this cover back along $\bar f$ produces the space
\begin{align*}
P_{\bar f} := \{([x],y) \in \mathbb{R}P^n \times S^n : \bar f([x]) = q(y)\},
\end{align*}
with projection
\begin{align*}
\pi_{\bar f}: P_{\bar f} \to \mathbb{R}P^n
\end{align*}
defined by $\pi_{\bar f}([x],y) = [x]$.
We now construct the comparison map explicitly:
\begin{align*}
\Phi: S^n &\to P_{\bar f}
\end{align*}
by
\begin{align*}
\Phi(x) = ([x], f(x)).
\end{align*}
Why does this land in $P_{\bar f}$? By definition of $\bar f$,
\begin{align*}
\bar f([x]) = [f(x)] = q(f(x)),
\end{align*}
so the pair $([x],f(x))$ satisfies the pullback condition. Also,
\begin{align*}
\pi_{\bar f}(\Phi(x)) = [x] = q(x),
\end{align*}
so $\Phi$ respects the projections to $\mathbb{R}P^n$.
It remains to see that $\Phi$ gives the same double cover, not merely a map between the two total spaces. Fix $[x] \in \mathbb{R}P^n$. The fiber of $q$ over $[x]$ is $\{x,-x\}$. The fiber of $\pi_{\bar f}$ over $[x]$ consists of all pairs $([x],y)$ such that $q(y) = \bar f([x]) = [f(x)]$. Thus $y$ must be either $f(x)$ or $-f(x)$. Since $f$ is odd, $-f(x)=f(-x)$. Therefore the two points in this fiber are
\begin{align*}
([x],f(x)) = \Phi(x)
\end{align*}
and
\begin{align*}
([x],-f(x)) = ([x],f(-x)) = \Phi(-x).
\end{align*}
So $\Phi$ is bijective on every fiber and commutes with the covering projections. To see that this is a covering isomorphism, take an evenly covered neighbourhood $V \subset \mathbb{R}P^n$. Over $V$, both total spaces are the disjoint union of two sheets, each mapped homeomorphically onto $V$. Because $\Phi$ preserves the base point and is fiberwise bijective, it sends each sheet of $q^{-1}(V)$ into one sheet of $\pi_{\bar f}^{-1}(V)$ and restricts to the homeomorphism obtained by composing the projection to $V$ with the inverse sheet parametrisation. Hence it identifies the pullback cover $\pi_{\bar f}:P_{\bar f}\to \mathbb{R}P^n$ with the original antipodal cover $q:S^n\to \mathbb{R}P^n$.
[/guided]
[/step]
[step:Compute the induced map on the first cohomology generator]
Let $t \in H^1(\mathbb{R}P^n;\mathbb{F}_2)$ denote the first Stiefel-Whitney class of the antipodal double cover $q: S^n \to \mathbb{R}P^n$. By naturality of the first Stiefel-Whitney class for pullback double covers, the first Stiefel-Whitney class of $\pi_{\bar f}:P_{\bar f}\to \mathbb{R}P^n$ is $\bar f^*(t)$. By the covering isomorphism constructed above, $\pi_{\bar f}$ is isomorphic to $q$, so its first Stiefel-Whitney class is also $t$. Therefore
\begin{align*}
\bar f^*(t) = t.
\end{align*}
Here $t$ is the nonzero generator of $H^1(\mathbb{R}P^n;\mathbb{F}_2) \cong \mathbb{F}_2$.
[/step]
[step:Use the cohomology ring of projective space to get top mod two degree]
The mod two cohomology ring of real projective space is
\begin{align*}
H^*(\mathbb{R}P^n;\mathbb{F}_2) \cong \mathbb{F}_2[t]/(t^{n+1}),
\end{align*}
where $t$ has degree $1$. Since $\bar f^*(t)=t$ and $\bar f^*$ is a graded ring homomorphism, we have
\begin{align*}
\bar f^*(t^n) = \bar f^*(t)^n = t^n.
\end{align*}
The class $t^n$ is the generator of $H^n(\mathbb{R}P^n;\mathbb{F}_2) \cong \mathbb{F}_2$. Therefore $\bar f$ acts as the identity on top-dimensional mod two cohomology, so
\begin{align*}
\deg_2(\bar f)=1.
\end{align*}
[/step]
[step:Approximate by a smooth odd map and compare degrees by regular values]
We first reduce the degree comparison to the smooth case. Regard $S^n$ as the unit sphere in $\mathbb{R}^{n+1}$. Since $f:S^n\to S^n$ is continuous and $S^n$ is compact, an equivariant smoothing argument gives a smooth odd map
\begin{align*}
g:S^n \to S^n
\end{align*}
that is uniformly close enough to $f$ so that $g(x)\neq -f(x)$ for every $x\in S^n$. The straight-line normalization
\begin{align*}
H:S^n\times[0,1]\to S^n,\qquad H(x,s)=\frac{(1-s)f(x)+sg(x)}{|(1-s)f(x)+sg(x)|}
\end{align*}
is then a well-defined homotopy from $f$ to $g$. Because both $f$ and $g$ are odd, $H(-x,s)=-H(x,s)$ for every $(x,s)\in S^n\times[0,1]$. Hence $H$ also descends to a homotopy between $\bar f$ and the quotient map $\bar g:\mathbb{R}P^n\to\mathbb{R}P^n$ induced by $g$. Homotopy invariance of mod two degree gives
\begin{align*}
\deg_2(f)=\deg_2(g)
\end{align*}
and
\begin{align*}
\deg_2(\bar f)=\deg_2(\bar g).
\end{align*}
Now $g$ and $\bar g$ are smooth maps between compact smooth $n$-manifolds. By Sard's theorem, choose $y\in S^n$ such that $y$ and $-y$ are regular values of $g$; then $[y]\in\mathbb{R}P^n$ is a regular value of $\bar g$, because the quotient map $q:S^n\to\mathbb{R}P^n$ is a local diffeomorphism and $q\circ g=\bar g\circ q$. The mod two degree of $g$ is computed by the finite regular fiber
\begin{align*}
\deg_2(g)\equiv |g^{-1}(y)|\pmod 2,
\end{align*}
and the mod two degree of $\bar g$ is computed by
\begin{align*}
\deg_2(\bar g)\equiv |\bar g^{-1}([y])|\pmod 2.
\end{align*}
For each projective point $[x]\in\bar g^{-1}([y])$, the equality $[g(x)]=[y]$ means $g(x)=y$ or $g(x)=-y$. Since $g$ is odd, exactly one of $x$ and $-x$ lies in $g^{-1}(y)$. Therefore the map sending $[x]$ to the unique representative $x'\in[x]$ with $g(x')=y$ is a bijection
\begin{align*}
\bar g^{-1}([y])\to g^{-1}(y).
\end{align*}
Thus $\deg_2(g)=\deg_2(\bar g)$. Combining this equality with the homotopy-invariance equalities and with $\deg_2(\bar f)=1$ from the preceding step gives
\begin{align*}
\deg_2(f)=\deg_2(g)=\deg_2(\bar g)=\deg_2(\bar f)=1.
\end{align*}
This proves the theorem.
[/step]
