There is no antipode-preserving map $f:S^{n+1}\rightarrow S^n$ 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.    
 A: 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.  
