Equivalences of the Borsuk-Ulam Theorem I have the following problem:
Problem: Show that the following statements are equivalent:

*

*If $f:\mathbb{S}^{n}\longrightarrow\mathbb{S}^{m}$ is a continuous function such that $f(-x)=-f(x)$, then $n\leq m$.


*If $f:\mathbb{S}^{n}\longrightarrow\mathbb{R}^{n}$ is a continuous function, then there exists $x\in\mathbb{S}^{n}$ such that $f(x)=f(-x)$. (Borsuk-Ulam Theorem)


*If $\mathbb{S}^{n}$ is covered by $n+1$ closed sets $A_{1},...,A_{n+1}$ then at least one of the $A_{i}$ contains an antipodal pair of points. (Lusternik-Schnirelman Theorem).


*If $f:\mathbb{S}^{n}\longrightarrow\mathbb{S}^{n}$ is a continuous function such that $f(-x)=-f(x)$, then $f$ has odd degree.


*If $f:\mathbb{S}^{n}\longrightarrow\mathbb{R}^{n}$ is a continuous function such that $f(-x)=-f(x)$, then there exists $x_{0}\in \mathbb{S}^{n}$ such that $f(x_{0})=0$.
The implications $(1)\Rightarrow (2)\Rightarrow (3)$, $(1)\Rightarrow (5)$ and $(2)\Leftrightarrow (5)$ are not complicated, but I don't know how to finish testing the remaining implications. Although from the form of the statements, I think that $(3)$ implies $(1)$ would be perhaps the most viable option.
Any hint will help me. Thanks!
 A: This answer is incomplete, because I can’t see how to easily deduce (4) from the other statements – it looks to me like a significant strengthening of the theorem, but I may easily be looking at it from the wrong angle. Anyway, let’s see all the other implications:
$(4) \Rightarrow (1)$: by contradiction. If $f: S^n \rightarrow S^m$ is continuous and odd with $m<n$, its composition with the natural embedding $S^m \rightarrow S^n$ is still continuous and odd and nonsurjective, hence of degree zero, a contradiction.
$(1) \Rightarrow (2)$: by contradiction. If $f$ does not take the same value on two antipodal points, let $g(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|} \in S^{n-1}$, then $g: S^n \rightarrow S^{n-1}$ is odd continuous, a contradiction.
$(2) \Rightarrow (3)$: consider $f(x)=(d(x,A_i))_{1 \leq i \leq n}$.
$(3) \Rightarrow (1)$: by contradiction, we can assume $m<n$. Let $g$ be the composition of $f$ with the inclusion $S^m \subset S^{n}$. Then $g$ is odd and $S^n \rightarrow S^n$.
For $0 \leq i \leq n$, let $A_i$ be the set of $x \in S^n$ such that the $i$-th coordinate of $g(x)$ is not less than any other one. Then the $A_i$ are closed subsets covering $S^n$ so one of them contains two antipodal points: it means that for some $x \in S^n$ and some $0 \leq i \leq n$, the $i$-th coordinate of $g(x)$ is the largest and the smallest, so that $g(x)$ has equal coordinates and is thus not in the image of $S^m$ in $S^n$: we get our desired contradiction.
$(2) \Rightarrow (5)$: if $f: S^n \rightarrow \mathbb{R}^n$ is odd, then there is (by (2)) some $x \in S^n$ such that $f(x)=f(-x)$. As $f(x)=-f(-x)$, $f(x)=0$.
$(5) \Rightarrow (2)$: let $f: S^n \rightarrow \mathbb{R}^n$, then $x \longmapsto f(x)-f(-x)$ is odd so vanishes at some $x \in S^n$, then $f(x)=f(-x)$.
