I'm searching for a proof for a version of Borsuk theorem, which says:
Any continuous antipodal map $f:S^n\to S^n$(antipodal means $f(x)=-f(-x)$) is not nullhomotopic
I'm eager for a "direct" proof, which means we can see that by simple argument without many tools from homology or homotopy theory, or using other versions of Borsuk-Ulam theorem. For instance, directly use the definition of homotopy to get the result. Here's the motivation of the question:
The proof for $S^1$ is easy, first define $g:[0,1]\to S^1$ that $g(x)=e^{2\pi i x}$. We consider antipodal points $0$ and $1/2$, they are mapped to antipodal points so that the winding $f\circ g:[0,1/2]\to S^1$ is $n+1/2,n\in\mathbb{Z}$. Since the mapping is antipodal, by simple argument we know that the winding number of image of $[1/2,1]$ remains $n+1/2$(by definition of winding number). So the total winding number is $2n+1$, which is not nullhomotopic.
But when winding number for $f:S^n\to S^n$ is replaced by the Brouwer degree(since we are dealing with $S^n$ it's not necessary to use the index to define it), I'm not quite sure how to prove that an antipodal map is not nullhomotopic. I tried to prove that $[f]$ is not zero component in $\pi_n(S^n)$. This attempt failed. Applying the similar argument in $S^1$ is hard because the addition of two degree can not be simply done. So I wonder is there any other way to add the degree, or directly prove from definition of homotopy that the map is not nullhomotopic? Thanks for your attention!
Here you must know a very few basics results from degree theory. I would imply you know the definition of the degree of a map $f: S^n \to S^n$; by this definition, a nullhomotopic map has degree $0$.
I need a few basic results to be clear.
$\operatorname{deg}f = -1$ if $f$ is a reflection of $S^n$, fixing the points in a subsphere $S^{n-1}$ and interchanging the two complementary hemispheres.
For we can give $S^n$ a $\Delta$-complex structure with these two hemispheres as its two $n$-simplices $\Delta_1^n$ and $\Delta_2^n$, and the $n$-chain $\Delta_1^n - \Delta_2^n$ represents a generator of $H_n(S^n)$. So the reflection interchanging $\Delta_1^n$ and $\Delta_2^n$ sens this generator to its negative.
So we have:
A little generalization shows that:
If $f: S^n \to S^n$ has no fixed points then $\operatorname{deg}f = (-1)^{n+1}$. In particular, the antipodal map has degree $(-1)^{n+1}$.
In fact, if $f(x) \neq x$, then the line segment from $f(x)$ to $-x$, defined by $t \mapsto (1-t)f(x) - tx$ for $t \in [0,1]$, does not pass through the origin. hence if $f$ has no fixed points, the formula $f_t(x) = \dfrac{(1-t) f(x) - tx}{|(1-t) f(x) - tx|}$ defines an homotopy from $f$ to the antipodal map. (What I have hidden here is that two homotopic maps have the same degree, but this is really elementary from the definition.)
What I shall prove is that if $f: S^n \to S^n$ is odd (that's the term for what you call antipodal) then $\operatorname{deg}f$ is odd.
Proof
I'll go by induction on $n$. If $n=1$, as you writed in your answer, the result is (almost) immediate. Let's assume the thesis is true for every $n < N \in \mathbb N$ and let's consider the pair $\left( S^n, A \right)$ where $A$ is the equator "line", a copy of $S^{n-1}$. Now there's an induced map
$ \tilde{f} : \mathbb{R}P^n \to \mathbb{R}P^n $
By cellular approximation, this may be assumed to take the hyperplane at infinity (the $n−1$-cell of the standard cell structure on $\mathbb{R}P^n$) to itself. Since whether a map lifts to a covering depends only on its homotopy class, $f$ is homotopic to an odd map taking $A$ to itself. We may assume that $f$ is such a map.
The map $f$ gives us a morphism of the long exact sequences:
Clearly, the map $f|A$ is odd, so by the induction hypothesis, $f|A$ has odd degree. Note that a map has odd degree if and only if $f^*: H_n(S^n;\mathbb{Z}_2) \to H_n(S^n;\mathbb{Z}_2)$ is an isomorphism. Thus $f^*: H_n(A;\mathbb{Z}_2) \to H_n(A;\mathbb{Z}_2)$ is an isomorphism. By the commutativity of the diagram, the map $f^*: H_n(S^n, A;\mathbb{Z}_2) \to H_n(S^n,A;\mathbb{Z}_2)$ is not trivial.
I claim it is an isomorphism. $H_n(S^n,A;\mathbb{Z}_2)$ is generated by cycles $[R+]$ and $[R−]$ which are the fundamental classes of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of $A$, $[A]\in H_{n−1}(A;\mathbb{Z}_2)$.
By the commutativity of the diagram, $\partial(f^*([R\pm]))=f^*(\partial([R\pm]))=f^*([A])=[A]$. Thus $f^*([R+])=[R\pm]$ and $f^*([R−])=[R\mp]$ since $f$ commutes with the antipodal map. Thus $f^*$ is an isomorphism on $H_n(S^n,A;\mathbb{Z}_2)$.
Since $H_n(A,\mathbb{Z}_2)=0$, by the exactness of the sequence $i:H_n(S^n;\mathbb{Z}_2) \to H_n(S^n,A;\mathbb{Z}_2)$ is injective, and so by the commutativity of the diagram (or equivalently by the 5-lemma) $f^*:H_n(S^n;\mathbb{Z}_2) \to H_n(S^n;\mathbb{Z}_2)$ is an isomorphism.
Thus $f$ has odd degree.