# A possible proof of the Borsuk Ulam theorem without “Homology-Cohomology”

Assume that $n>1$.

The configuration space of $S^n$ is defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$

We have two questions:

1.Is there a continuous function $f:M_n \to S^{n-1}$ with $f(y,x)=-f(x,y)$, for all $x,y \in$S^{n}$? 2.Is there a continuos function$h: M_n \to \mathbb{R}^n$such that$h(x,y)=-h(y,x) $and$h(x,-x)\neq 0$for all$x,y \in S^n$If the answer to either of these two questions is "affirmative ", then we can provide an alternative proof for the Borsuk Ulam theorem, inductively. Because an equivalent formulation of the Borsuk Ulam theorem is that: There is no an odd continuous function$g:S^{n+1}\to S^n$Assuming that the answer to either of the above two questions is affirmative, we give a proof for the above equivalent formulation of the Borsuk Ulam theorem. The proof is as follows: Assume that$g:S^{n+1}\to S^n$is an odd continuous function. then$f(g(x),g(-x))$( or$h(g(x),g(-x))$) is an odd continuous function from$S^{n+1}$to$S^{n-1}$( or to$\mathbb{R}^n \setminus \{0\}$). This obviously gives a contradiction by induction. Because this situation leads to existence of an odd continuous function from$S^n$to$S^{n-1}$. Now we apply the induction argument. • @NoahSchweber Yes We want$n>1$as i wrote in the first line of the question. – Ali Taghavi Aug 14 '17 at 6:28 ## 1 Answer Suppose such a function$M_n \to S^{n-1}$existed. Consider the composition$S^n \to M_n \to S^{n-1}$, where the first map sends$x$to the pair$(x,-x)$. That composition is an odd continuous function$S^n \to S^{n-1}\$, hence can not exist by Borsuk-Ulam.

• @ThomasRot Yes this was the aim of the question, provided the answer would be affirmative. But, as the answer shows, it is not the case. – Ali Taghavi Aug 14 '17 at 8:11
• Sorry i need to stop commenting on my phone – Thomas Rot Aug 14 '17 at 10:14