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I have been trying to solve this exercise from the book Fundamental Group and Covering Spaces written by Elon Lages Lima, chapter 3. It says that given $f,g : B^{2} \to S^{2}$ continuous such that for $(x,y) \in S^{1}$, $f(x,y) = (x,y,0)$ and $g(x,y) = (-y,x,0)$, there exists $(x,y) \in B^{2}$ with $f(x,y) = g(x,y)$ or $f(x,y) = -g(x,y)$.

I'm not really sure how to proceed. The definition of $f,g$ makes me think that I should try inner product or projective space. Or maybe i should simply proceed by contradiction and try to construct a function in contradiction with some application like Borsuk-Ulam theorem or that doesn't exists a not null tangent vectorial field on S^{2}. Neither of those have worked to me.

Any ideas, suggestions or answers? :)

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If neither $f(x,y)=g(x,y)$ nor $f(x,y)=-g(x,y)$ then there is a well defined nonzero tangent vector to the sphere pointing along the great circle containing $f(x,y)$ and $g(x,y)$ in the direction from $f(x,y)$ toward $g(x,y)$. This uses that $f(x,y) \neq g(x,y)$ to guarantee nonzero length, and $f(x,y) \neq -g(x,y)$ to guarantee there is only one such direction.

So if you can fill in from your assumptions about $f,g$ on the equator of the sphere to this vector field, you can finish with Borsuk-Ulam as you suggest. Thus the idea you had might be made to work.

EDIT: I think the way is to define $F(x,y,z)=f(x,y)$ and $G(x,y,z)=g(x,y),$ so that now $F,G$ are maps from the sphere $S^2$ to itself, where the sphere is considered as points in three space where $x^2+y^2+z^2=1$. In other words to get the image of a point on the sphere under $F$ or $G$ one projects onto the $xy$ plane and reads off the value of $f$ or $g$ respectively. This done we now have the functions $F,G$ defined on $S_2$ and I think the above idea produces a nonvanishing vector field.

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It's seems like a great idea!! But I don't see where are we using that the functions are already defined on the equator. I think is for the continuity of the vector field but I'm not really sure. – Frank Feb 2 '13 at 18:33
Now I have a doubt, when you find the direction from $f(x,y)$ towards $g(x,y)$ that tangent vector makes sense when you think of it like a vector starting at $f(x,y)$ but, how do we move it to the position $(x,y,z)$ ? – Frank Feb 2 '13 at 18:38
For your first comment: The use on the equator is that, given the definitions there, the points $f(x,y)=(x,y,0)$ and $g(x,y)=(-y,x,0)$ are neither equal nor antipodal. (One is rotated 90 degrees from the other.) I'll think more about second comment... – coffeemath Feb 3 '13 at 4:57
Yes but we are not really using that definition, we are only using that they are neither equal or antipodal, so the condition can be weakened, I think that condition must play an important role in the proof :S – Frank Feb 3 '13 at 14:28
@Frank : The more I think about it, my idea might not be right. The two given maps $f,g$ are defined on the 2-disk, and when "lifted" to the sphere as I suggested they might not work as intended. Because if I'm at say $(x,y,z)$ with $z>0$ all I have to work with is two points $f(x,y),g(x,y)$ somewhere on the sphere, not really related to the specific point $(x,y,z)$ where I'm located, but could be anywhere else. Using these I now don't see how to define a direction at the given point $(x,y,z)$. – coffeemath Feb 4 '13 at 0:30

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