Note: It humble request to everyone please don't downvote my post .Im not genius or intelligent.Im new to the Algebraic Topology.Try to understand my level of understanding.I feel discourage whenever i get downvote
I have some confusion in this answer
Question : 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 $f(x,y)=f(-x,-y)?$
Here is the outlined of the given answer
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 $x=0\notin S^1$.
My confusion : Im not getting the statement "they are in the same meridian of the torus, i.e. they have the same first coordinate".
My attempt :If they are in the same meridian of the torus then $p(\theta_1, \theta_2)=p(\theta_1 +\frac{1}{2} , \theta_2 +\frac{1}{2})$
Now take $a=\theta_1$ and $c=\theta_1 +\frac{1}{2} $
$\theta_1=\theta_1 +\frac{1}{2} \implies \theta_1-\theta_1 =\frac{1}{2} \neq 0 \implies a \ne c$
and $\theta_1=\theta_1 +\frac{1}{2} \implies \frac{1}{2} \in S^1$
My question : why they have the same first coordinate in torus ?
Imagine projecting this torus on the horizontal plane:
Points that are mapped on the same point on the plane lie one above another, and so they have to both be on the same little circle (like the red one). E.g. here the upper point of the red circle will have the same image as its lower point. Points of other small circles, however, will have different projections on the plane.
The first coordinate of a point on a torus is specifying exactly the meridian (small circle) on which the point lies.