Direct proof of version of Borsuk theorem 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!
 A: 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:

*

*The antipodal map $-\operatorname{Id} : S^n \to S^n$, $x \mapsto -x$, has degree $(-1)^{n+1}$ since it is the composition of $n+1$ reflections, each changing the sign of one coordinate in $\mathbb{R}^{n+1}$.

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.
