How to prove that there exists a continuous function $f:B^2 \to B^2$ without constant points? Here, $B^2$ is the unit open ball. I guess $f$ can be for example like this $f: re^{iax} \to re^{ibx} $ but in what sense it does not include constant points?
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When you say "constant point" do you mean a solution to $f(x)=x$? – Milo Brandt Nov 23 '14 at 18:10
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This is something not clear to me either. – user39726 Nov 23 '14 at 18:13
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2Your example function has $0$ as a fixed point. – TonyK Nov 23 '14 at 18:13
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So, my example works indeed, right? Also, for any x I get some constant point, no? – user39726 Nov 23 '14 at 18:14
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My comment shows precisely that your example doesn't work. – TonyK Nov 23 '14 at 20:52
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@TonyK: I'm pretty sure when the OP says "my example works", E means the same thing that you do when you say "your example doesn't work". – Eric Stucky Nov 24 '14 at 02:04
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Let $b$ be the bottom point of the sphere, i.e. $(0,0...,0,1)$. For any $x$ in the ball let $f(x)$ be the midpoint of the line segment with endpoints $b$ and $x$.
Mirko
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I would presume that "constant point" refers to a solution to $f(x)=x$ - that is, what would typically be called a fixed point. The function $re^{iax}\mapsto re^{ibx}$ certainly does have a constant point at $0$, so it wouldn't work.
A really simple example would be to consider that the operation of contraction towards a point (i.e. scaling) is continuous, and the only constant point therein would be the center of the contraction. So, for instance $z\mapsto \frac{z}2$ has a fixed point at $z=0$. However, if we put the constant point on the boundary of the ball, which is not included, we could have a function like $$z\mapsto \frac{1+z}2$$ would be a continuous function mapping the ball to itself without a constant point (since its "constant point" would be on the boundary, and hence not included).
You could do a little better and at least establish that a bijective, continuous $f$ exists, noting that an open ball is homeomorphic to an open square, and a map like $$f(x,y)=(x^2,y)$$ on the square $(0,1)\times (0,1)$ has no constant points, but is continuous and bijective; thus, if we map the ball onto the square, apply this, then map back to the ball, we would get a homeomorphism with no constant points from the ball to itself.
Milo Brandt
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Could you make it a bit more "formal/rigorous"? It is a nice answer. In what sense there is no constant point in your $f(x,y)$? – user39726 Nov 23 '14 at 18:19
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1A constant point would be a solution $$(x,y)=(x^2,y)$$ or, equivalently, to the two simultaneous equations $$x=x^2$$ $$y=y^2.$$ Since we are working with $x$ in $(0,1)$ and the first equation has solutions only for $x=0$ or $x=1$, there can be no constant point. – Milo Brandt Nov 23 '14 at 18:20
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Why is the there $y=y^2$ in the second line of your last set of equations? – user39726 Nov 23 '14 at 18:29
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Ok, thanks. So, to say it in my words, because the open unit ball is homomorphic to the open unit square we consider the mapping $f(x,y)=(x^2,y)$ which has a constant point only for $x=0,1$ points that are not included in the unit square. Could I not use $f(x,y)=(x^2,y^2)$? – user39726 Nov 23 '14 at 18:34
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Yes, that's correct. You could use that $f$ as well; there are a lot of possible solutions. – Milo Brandt Nov 23 '14 at 18:47
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4Of course one can also choose a homeomorphism between the open disk and the plane. Then the plane allows translations (additions of a constant vector) which clearly are continuous and without fixed points. – Jeppe Stig Nielsen Nov 23 '14 at 22:07