[proofplan]
The second and third statements are the same assertion written with and without the word "odd." To compare the first statement with the odd-map formulation, we prove both contrapositives. If the antipodal coincidence statement fails, then the difference $f(x)-f(-x)$ never vanishes, so normalizing it produces an odd map into $S^{n-1}$. Conversely, an odd map into $S^{n-1} \subset \mathbb{R}^n$ is itself a continuous map into $\mathbb{R}^n$ that separates every antipodal pair.
[/proofplan]
[step:Identify the odd-map formulation with its defining condition]
By definition, a map $g:S^n \to S^{n-1}$ is odd precisely when
\begin{align*}
g(-x)=-g(x) \quad \text{for every } x \in S^n.
\end{align*}
Therefore statement 2 and statement 3 are logically identical.
[/step]
[step:Construct an odd sphere-valued map from a failure of antipodal coincidence]
Assume statement 1 fails. Then there exists a continuous map $f:S^n \to \mathbb{R}^n$ such that
\begin{align*}
f(x)\neq f(-x) \quad \text{for every } x \in S^n.
\end{align*}
Define the punctured Euclidean space $\mathbb{R}^n_0 := \mathbb{R}^n \setminus \{0\}$. Define
\begin{align*}
F:S^n \to \mathbb{R}^n_0, \quad x \mapsto f(x)-f(-x).
\end{align*}
The map $F$ is continuous because it is obtained from $f$, the continuous antipodal map $x \mapsto -x$ on $S^n$, and subtraction in $\mathbb{R}^n$. Its codomain is $\mathbb{R}^n_0$ because $f(x)\neq f(-x)$ for every $x \in S^n$.
Define the normalization map
\begin{align*}
N:\mathbb{R}^n_0 \to S^{n-1}, \quad y \mapsto \frac{y}{|y|}.
\end{align*}
Since $|y|>0$ on $\mathbb{R}^n_0$, the map $N$ is continuous. Now define
\begin{align*}
u:S^n \to S^{n-1}, \quad x \mapsto N(F(x))=\frac{f(x)-f(-x)}{|f(x)-f(-x)|}.
\end{align*}
The map $u$ is continuous as a [composition of continuous maps](/theorems/4960). For every $x \in S^n$,
\begin{align*}
F(-x)=f(-x)-f(x)=-F(x).
\end{align*}
Hence
\begin{align*}
u(-x)=\frac{F(-x)}{|F(-x)|}=\frac{-F(x)}{|F(x)|}=-u(x).
\end{align*}
Thus $u:S^n \to S^{n-1}$ is an odd continuous map. Therefore, if statement 1 fails, then statement 2 fails.
[guided]
We prove the contrapositive: instead of assuming an odd map does not exist and trying directly to force an antipodal coincidence, we assume that the antipodal coincidence conclusion fails and build an odd map.
So suppose there is a continuous map $f:S^n \to \mathbb{R}^n$ such that no antipodal pair has the same image:
\begin{align*}
f(x)\neq f(-x) \quad \text{for every } x \in S^n.
\end{align*}
The natural object to measure the failure of equality is the difference between the two values. Define
\begin{align*}
F:S^n \to \mathbb{R}^n_0, \quad x \mapsto f(x)-f(-x),
\end{align*}
where $\mathbb{R}^n_0 := \mathbb{R}^n \setminus \{0\}$. This is well-defined as a map into $\mathbb{R}^n_0$ because the assumption $f(x)\neq f(-x)$ says exactly that $f(x)-f(-x)\neq 0$ for every $x \in S^n$.
The map $F$ is continuous: the antipodal map $S^n \to S^n$, $x \mapsto -x$, is continuous, so $x \mapsto f(-x)$ is continuous, and subtraction in $\mathbb{R}^n$ preserves continuity. Now we convert a nonzero vector in $\mathbb{R}^n$ into a unit vector by normalization. Define
\begin{align*}
N:\mathbb{R}^n_0 \to S^{n-1}, \quad y \mapsto \frac{y}{|y|}.
\end{align*}
This is continuous because the denominator $|y|$ is never zero on $\mathbb{R}^n_0$. Therefore the composite
\begin{align*}
u:S^n \to S^{n-1}, \quad x \mapsto N(F(x))=\frac{f(x)-f(-x)}{|f(x)-f(-x)|}
\end{align*}
is continuous.
It remains to check oddness. For every $x \in S^n$,
\begin{align*}
F(-x)=f(-x)-f(x)=-F(x).
\end{align*}
Since Euclidean norm is unchanged by multiplication by $-1$, we have $|F(-x)|=|F(x)|$. Hence
\begin{align*}
u(-x)=\frac{F(-x)}{|F(-x)|}=\frac{-F(x)}{|F(x)|}=-u(x).
\end{align*}
Thus $u$ is an odd continuous map $S^n \to S^{n-1}$. This proves that failure of statement 1 implies failure of statement 2.
[/guided]
[/step]
[step:Use an odd sphere-valued map to violate antipodal coincidence]
Assume statement 2 fails. Then there exists an odd continuous map $g:S^n \to S^{n-1}$. Let
\begin{align*}
\iota:S^{n-1} \to \mathbb{R}^n, \quad y \mapsto y
\end{align*}
be the inclusion map. Define
\begin{align*}
f:S^n \to \mathbb{R}^n, \quad x \mapsto \iota(g(x)).
\end{align*}
Then $f$ is continuous. For every $x \in S^n$, oddness gives $g(-x)=-g(x)$. Since $g(x)\in S^{n-1}$, we have $|g(x)|=1$, so $g(x)\neq 0$. Therefore $g(x)\neq -g(x)$, because equality would imply $2g(x)=0$ in $\mathbb{R}^n$. Thus
\begin{align*}
f(x)=g(x)\neq -g(x)=g(-x)=f(-x)
\end{align*}
for every $x \in S^n$. Hence statement 1 fails. Therefore, if statement 2 fails, then statement 1 fails.
[/step]
[step:Conclude the equivalence of all three statements]
The previous two steps prove that statement 1 fails if and only if statement 2 fails. Hence statement 1 and statement 2 are equivalent. Since statement 2 and statement 3 are identical by the definition of oddness, all three statements are equivalent.
[/step]
