Do these sets have the fixed-point property?

Question

I need some help determining if the following sets have the "fixed-point property" (A topological space $X$ has this property if for every continuous function $f: X \to Y,$ there exists an $x_0 \in X$ such that $f(x_0)=x_0).$

$X = (0,1) \times (0,1)$

The continuous function $f:(0,1) \times (0,1) \to Y, \hspace{0.4mm} (x,y) \mapsto (\frac{1}{2}x, \frac{1}{2}y)$ has no fixed point $(x_0,y_0) \in (0,1) \times (0,1)$ such that $f(x_0,y_0) = (x_0,y_0),$ so $X$ does not have the fixed-point property.

$D = \big \{(x,y):(x-1)^2+y^2 < 1 \big \} $

An open disk centered at $(1,0):$ The continuous function $f(x,y) = (\frac{1}{2}x,\frac{1}{2}y)$ has no fixed point $(x_0,y_0) \in D$ such that $f(x_0,y_0) = (x_0,y_0),$ so $D$ does not have the fixed-point property.

$A = \big \{(x,y):1 \leqslant x^2+y^2 \leqslant 2 \big \} $

Define $f:A \to Y$ by $(x,y) \mapsto (x+1,y+1).$ Then $f$ has no fixed points in $A$ and so $A$ does not have the fixed point property.

Questions: Do my functions work here? Do they show that these sets do not have the fixed-point property? (I'm pretty sure all of them don't have it)

Answer 1

The target space "$Y$" of these maps is supposed to be the same as the domain (otherwise, it would be quite a coincidence if $f(x_0)=x_0$, since $x_0$ is a point of the domain and $f(x_0)$ is a point of $Y$!). For your first two examples, it turns out that range of your map is contained in the domain so they actually do work, but your last one doesn't.

Answer 2

$f(x,y)=(-x,-y)$ does fix all points with $x=0$ or $y=0$. Consider using polar coordinates instead of Cartesian coordinates to find an example which doesn't fix any points.