Equivalent statement for Borsuk-Ulam theorem I was going through this paper by Tanaka. In the introduction he says the following
"The classical Borsuk–Ulam theorem can be
restated as the point space is I-trivial."
I am not sure how to go about the proof of this equivalence, what I thought is that any bundle over a point space is just a vector space and it's sphere bundle is just the maximal sphere and we have an obvios $Z_2$ map between the spheres. I don't know where to go from here.
Can anyone suggest something?
thanks and regards
 A: Following the terminology of the paper, suppose that the point is not $I$-trivial. Then there exists some vector bundle over a point, say $\alpha$ such that $Ind(\alpha)> dim(\alpha)$, that is there exists $i \geq dim(\alpha)-1$ and a map $S^{i} \rightarrow S(\alpha)$ respecting the antipodal actions on both sides.
So in otherwords, we want to show that:
There is no $j>i$ and map $F: S^{j} \rightarrow S^{i}$ equivariant with respect to antipodal on both sides is equivelant to Borsuk-Ulam.
The $\mathbb{Z}_{2}$-equivariance can be restated as saying the $F$ is odd.
Quoting from the wikipedia page on the Borsuk Ulam theorem. "Define a retraction as a function $h:S^{n}\to S^{n-1}$. The Borsuk–Ulam theorem is equivalent to the following claim: there is no continuous odd retraction." A proof of this equivelance is also given there.
Note that the statement we want is in terms of functions $F:S^j \rightarrow S^i$ with $j>i$, but we can always restrict to a standard coordinate subsphere $S^{i+1} \subset S^{j}$ and the restriction will also be an odd function.
