I am trying to prove that there is no antipode-preserving map $f:S^{n+1}\rightarrow S^n$. Here is what I have:
Suppose there is an antipode-preserving map $f:S^{n+1}\rightarrow S^n$. If we restrict $f$ the the "equator" $S^n$ then we have an antipode-preserving map $S^n\rightarrow S^n$, so $f$ must be nullhomotopic. But the restriction of $f$ extends continuously to the hemisphere $D^{n+1}$, so the degree of $f$ must be zero which is a contradicton.
I am not at all satisfied with this proof. I would like a proof that does not assume that there is no antipode-preserving map $S^n\rightarrow S^n$ that is nullhomotopic. Any suggestions are appreciated.
Best Answer
If we consider $S^{n}$ as a subset of $\mathbb R^{n+1}$, then we get a continuous map $S^{n+1}\to\mathbb R^{n+1}$. By the Borsuk-Ulam Theorem, such a map must map two antipodal points in $S^{n+1}$ to the same point. In particular, these two antipodal points are not mapped to antipodal points in $S^n$.
Alternatively, prove it directly - if $f\colon S^{n+1}\to S^n$ preserves antipodes, then it induces a map $f'\colon \mathbb RP^{n+1}\to\mathbb RP^n$. Now, the cohomology ring of $\mathbb RP^m$ with $\mathbb Z_2$ coefficients is $\mathbb Z_2[\alpha]/(\alpha^{m+1})$, where $\alpha$ is the generator of $H^1(\mathbb RP^m\;;\;\mathbb Z_2)=\mathbb Z_2$. So $f$ induces a map $$ f^*\;\colon\;\mathbb Z_2[\alpha]/(\alpha^{n+1})\to\mathbb Z_2[\beta]/(\beta^{n+2}) $$ Since $f^*$ maps $H^1(\mathbb RP^{n})$ to $H^1(\mathbb RP^{n+1})$, it must map $\alpha$ either to $0$ or to $\beta$. But the second case is impossible: then we have $$ 0=f^*(0)=f^*(\alpha^{n+1})=f^*(\alpha)^{n+1}=\beta^{n+1}\ne 0 $$ Therefore, $f^*(\alpha)=0$.
By the isomorphism given by the universal coefficient theorem, we see that the induced map on homology groups - $$ f_*\;\colon\;H_1(\mathbb RP^{n+1})\to H_1(\mathbb RP^n) $$ - is trivial also, and it follows that $f$ induces a trivial map on fundamental groups $\pi_1(\mathbb RP^{n+1})\to\mathbb RP^n)$ (since $\pi_1(\mathbb RP^m=\mathbb Z_2$).
It follows immediately that $f'$ satisfies the lifting criterion for the covering $S^n\to\mathbb RP^n$, so we get a lifting $\overline{f}\colon\mathbb RP^{n+1}\to S^n$.
Now compose $\overline f$ with the projection from $S^{n+1}$ to $\mathbb RP^{n+1}$ to get a map $$ \overline{f}\circ\pi_{n+1}\;\colon \;S^{n+1}\xrightarrow{\pi}\mathbb RP^{n+1}\xrightarrow{\overline{f}}S^n $$ We now have two maps $S^{n+1}\to S^n$ - our original map $f$ and $\overline{f}\circ\pi_{n+1}$, and we can readily see that they are both lifts of the map $$ \pi_n\circ f\;\colon\;S^{n+1}\xrightarrow{f}S^n\xrightarrow{\pi_n}\mathbb RP^n $$ The unique lifting property tells us that two such liftings are equal if they agree at a point.
Fix some point $x_0\in S^{n+1}$. Then $\pi_{n+1}(x_0)=\pi_{n+1}(-x_0)$, so $\overline f\pi_{n+1}(x_0)=\overline f\pi_{n+1}(-x_0)$. But $f(x_0)=-f(-x_0)$, so these maps are certainly not equal. But now, by the definition of a lifting, we must have either $f(x_0)=\overline f\pi_{n+1}(x_0)$ or $f(-x_0)=\overline f\pi_{n+1}(-x_0)$ - so these two liftings agree at a point. This is a contradiction.