Mapping the open ball to itself?

Question

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?

Answer 1

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 $x$.

Answer 2

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.