For every continuous map $f$ : $S^2$$\rightarrow$$\Bbb{R}P^2$ (from sphere to real projective plane) does there exist a pair of antipodal points that landed together? A.k.a. there exist $$x,-x \in S^2$$ such that $$f(x) = f(-x)$$
Is the Borsuk-Ulam theorem still true if we replace $\Bbb{R^2}$ with $\Bbb{R}P^2$??
By the construction of Projective plane as a quotient space of Sphere this is trivially true because every antipodal pair is a uniqe point in $\Bbb{R}P^2$
I think it might be true for any continuous map.I don’t have any rigorous proof but i have an intuition!
Lets imagine $\Bbb{R}P^2$ as set of all lines in $\Bbb{R^3}$ that goes through origin and an unit Sphere $S^2$ embedded in $\Bbb{R^3}$. No matter where we put the Sphere in $\Bbb{R^3}$ there always some lines that goes through orgin and goes through the sphere (or touch tangentially) thus it defines a mapping from Sphere to projective plane. Now wherever we put the Sphere there is one line that goes through the centre of the Sphere and the origin And we can interprete this as antipodal points with same image, thus proving the theorem.
If anyone is not sure what i said feel free to ask again. I wish I could make an animation clip to show what I described . Of course, i can be wrong. Please correct me then. Thanks very much.
Yes, this is true. Note that your map $f : S^2 \mapsto P^2({\bf R})$ lifts to a map $\tilde{f} : S^2 \rightarrow S^2$ because $S^2$ is simply connected.
There is a general result that asserts that any continuous map $\tilde{f} : S^{2n} \rightarrow S^{2n}$ either has a fixed point or sends some point to its antipod. This is exercice 16.9 in the book of Greenberg, Harper, algebraic topology.
The proof is broken in several part. If $\tilde{f}$ does not have fixed point, it is homotopic to the antipodal map. Composing then by the antipodal map, we obtain a map that is homotopic to a constant. All these maps have a fixed point.
EDIT: this answer is incomplete as pointed out in the comments. I still leave it here in the hope it may inspire someone else.