# Borsuk-Ulam Theorem for torus. [closed]

Is the Borsuk-Ulam theorem valid for a torus? In other words, for any map $f: S^1 \times S^1 \rightarrow \mathbb{R^2}$ there is a point $(x,y) \in S^1 \times S^1$ which $f(x,y)=f(-x,-y)$

I'm very stuck on this task. Can someone give a hint? Or there can be a detailed solution, if suddenly this task is easy enough.

Thank you in advance for help!

## closed as off-topic by user296602, Namaste, Saad, Xander Henderson, user223391 Jun 9 '18 at 14:49

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No. With the usual torus embedded in $\mathbb{R}^3$, lying on the $OXY$ plane, one has a natural projection onto that plane, $p:S^1×S^1\to \mathbb{R}^2$, which is continuous.
Two points on the torus have the same image if they are one above the other, in the same vertical line. In particular, they are in the same meridian of the torus, i.e. they have the same first coordinate. So, if $p(a,b) = p(c,d)$, $a = c$. This implies that the Borsuk-Ulam theorem fails on the torus because if $x=-x$, and then $x=0\notin S^1$.
• Could you, please, be more explicit - what "natural projection" are you talking about? UPD: Ah, I think I got it: you mean $S^1\times S^1$ embedded as a usual 3D torus into $\mathbb R^3$, and the projection goes onto the plane parallel to the big circle of the torus. It would help to mention this explicitly, I think. – lisyarus Jun 8 '18 at 11:21
• Yes, with the usual torus embedded in $\mathbb{R}^3$, lying on the $OXY$ plane, $p$ is just the projection onto that plane. I'll add it to the answer. – Javi Jun 8 '18 at 11:34
• My mistake; I wasn't familiar with the Borsuk-Ulam Theorem, which does specify that $-x$ is the antipodal point, not an opposite in $\mathbb R^1/(x+1\sim x)$.... I think the answer was better before the edit. – mr_e_man Jun 8 '18 at 19:29