[proofplan] The cases $n=0$ and $n=1$ are handled separately by elementary topology. For $n\geq 2$, we argue by contradiction: if $f$ separates every antipodal pair, then the difference $f(x)-f(-x)$ is a nowhere-zero odd map into $\mathbb R^n$, and normalizing it produces an odd continuous map $S^n\to S^{n-1}$. The obstruction is that no odd map can lower the sphere dimension by one; this follows from the mod-$2$ cohomology rings of real projective spaces. The contradiction forces an antipodal coincidence for $f$. [/proofplan] [step:Handle the zero-dimensional sphere separately] If $n=0$, then $\mathbb R^0$ is the one-point [vector space](/page/Vector%20Space). Hence every map $f:S^0\to \mathbb R^0$ is constant, so for either point $x\in S^0$ we have $f(x)=f(-x)$. For the rest of the proof, assume $n\geq 1$. [/step] [step:Handle the circle case by connectedness] Assume $n=1$. Suppose, for contradiction, that $f(x)\ne f(-x)$ for every $x\in S^1$. Define $h:S^1\to \mathbb R$ by $h(x)=f(x)-f(-x)$. The map $h$ is continuous and nowhere zero, so the normalized map $g:S^1\to S^0$ that sends $x$ to the sign of the nonzero real number $h(x)$ is continuous. Since $S^1$ is connected and $S^0=\{-1,1\}$ is discrete, the map $g$ is constant. But $h(-x)=-h(x)$, so $g(-x)=-g(x)$, contradicting constancy. Therefore the conclusion holds when $n=1$. For the rest of the proof, assume $n\geq 2$. [/step] [step:Reduce a separated antipodal pair map to an odd sphere map] Assume, for contradiction, that $f(x)\ne f(-x)$ for every $x\in S^n$. Define $h:S^n\to \mathbb R^n$ by $h(x)=f(x)-f(-x)$. The map $h$ is continuous because it is built from $f$, the antipodal map $x\mapsto -x$, and subtraction in $\mathbb R^n$. The assumption gives $h(x)\ne 0$ for every $x\in S^n$, so the normalized map $g:S^n\to S^{n-1}$ defined by $g(x)=h(x)/\|h(x)\|$ is continuous. Since $h(-x)=f(-x)-f(x)=-h(x)$, we have $g(-x)=-g(x)$. Hence $g$ is an odd continuous map from $S^n$ to $S^{n-1}$. [/step] [step:Exclude odd maps from $S^n$ to $S^{n-1}$ by projective cohomology] We show that an odd continuous map $g:S^n\to S^{n-1}$ cannot exist for $n\geq 2$. Suppose such a map exists. Because $g(-x)=-g(x)$, it descends to a continuous map $\bar g:\mathbb RP^n\to \mathbb RP^{n-1}$ between orbit spaces of the antipodal actions. For each integer $m\geq 0$, write $\mathbb RP^m$ for the quotient of $S^m$ by the antipodal relation $x\sim -x$. Write $\mathbb F_2$ for the field with two elements, and write $H^*(-;\mathbb F_2)$ for singular cohomology with coefficients in that field. We use the standard mod-$2$ cohomology calculation for real projective spaces: for $m\geq 1$, the ring $H^*(\mathbb RP^m;\mathbb F_2)$ is the truncated [polynomial ring](/page/Polynomial%20Ring) generated by the class $t\in H^1(\mathbb RP^m;\mathbb F_2)$ that classifies the antipodal double cover $S^m\to \mathbb RP^m$, with nonzero powers through degree $m$ and zero powers above degree $m$. Let $a\in H^1(\mathbb RP^{n-1};\mathbb F_2)$ and $b\in H^1(\mathbb RP^n;\mathbb F_2)$ denote these standard generators. The hypothesis $n\geq 2$ ensures that $\mathbb RP^{n-1}$ has such a degree-one generator. Naturality of the classifying class for double covers gives $\bar g^*(a)=b$. The same cohomology calculation gives $a^n=0$ in $H^*(\mathbb RP^{n-1};\mathbb F_2)$ and $b^n\ne 0$ in $H^*(\mathbb RP^n;\mathbb F_2)$. Since $\bar g^*$ is a ring homomorphism, applying it to the relation $a^n=0$ forces $b^n=0$. This contradicts $b^n\ne 0$. Therefore no odd continuous map $S^n\to S^{n-1}$ exists for $n\geq 2$. [guided] The point of passing to projective space is to turn the oddness condition into an ordinary map between quotient spaces. Since $g(-x)=-g(x)$, the two antipodal points $x$ and $-x$ in $S^n$ have images that represent the same point of $\mathbb RP^{n-1}$. Therefore $g$ defines a continuous map $\bar g:\mathbb RP^n\to \mathbb RP^{n-1}$. For each integer $m\geq 0$, the notation $\mathbb RP^m$ means the quotient of $S^m$ by the antipodal relation $x\sim -x$. The notation $\mathbb F_2$ means the field with two elements, and $H^*(-;\mathbb F_2)$ means singular cohomology with coefficients in $\mathbb F_2$. Now use the standard mod-$2$ cohomology calculation for real projective spaces. For every $m\geq 1$, the ring $H^*(\mathbb RP^m;\mathbb F_2)$ is a truncated polynomial ring generated by one degree-one class $t$, and $t$ classifies the antipodal double cover $S^m\to \mathbb RP^m$. The powers of $t$ are nonzero through degree $m$ and zero above degree $m$. Since we are in the case $n\geq 2$, both $\mathbb RP^{n-1}$ and $\mathbb RP^n$ fall under this calculation. Let $a\in H^1(\mathbb RP^{n-1};\mathbb F_2)$ and $b\in H^1(\mathbb RP^n;\mathbb F_2)$ be the corresponding degree-one classes. The commutativity forced by $g(-x)=-g(x)$ says that pulling back the double cover $S^{n-1}\to \mathbb RP^{n-1}$ along $\bar g$ gives the double cover $S^n\to \mathbb RP^n$. By naturality of the classifying cohomology class, $\bar g^*(a)=b$. The ring calculation also says $a^n=0$ in the smaller projective space and $b^n\ne 0$ in the larger one. Since cohomology pullback preserves products and sends zero to zero, applying $\bar g^*$ to the relation $a^n=0$ forces $b^n=0$. This contradicts $b^n\ne 0$, so the assumed odd map $S^n\to S^{n-1}$ cannot exist. [/guided] [/step] [step:Conclude that the original map has an antipodal coincidence] For $n=0$ and $n=1$, the conclusion was proved in the first two steps. For $n\geq 2$, the reduction step constructed an odd continuous map $g:S^n\to S^{n-1}$ from the assumption that $f(x)\ne f(-x)$ for every $x\in S^n$. The obstruction step proves that no such odd map exists. Therefore the assumption is false, and there exists $x\in S^n$ such that $f(x)=f(-x)$. [/step]