I'm looking for any hint on proving that the Warsaw circle has the fixed point property. By this I mean that for every continuous function $f$ from the Warsaw circle onto itself there exists a point $x$ on it such that $f(x)=x$.
I've been trying to find a contradiction using the fact that the Warsaw circle is simply connected but i haven't been able to. Any hint would be appreciated, i don't want anyone to give me a full solution, just to push me in the correct direction or motivate the right ideas.
This question appears as an exercise on a first course on algebraic topology which has been centered on the fundamental group. We've touched on the Warsaw circle only superficially and have only proven that it's simply connected. The tools that we've primarily developed on the course are path homotopies and the path lifting technique.
Theorem: If $f$ is a continuous map from a compact set to itself and $f$ has no fixed point then there exists some $k>0$ such that $d(x, f(x))>k$ $\forall x $.
Now consider the closure of topologist's sine curve with an arc from $(0,1)$ to $(\frac1\pi, 0)$.
call the arc segment of the $y$ axis from $(0,0)$ to $(0,1)$ as $A$.
Note that every point of $A$ is a limit point of some sequence of points $x_1,x_2,\dots$ of the topologist's sine curve. Now if $f$ has no fixed point then using continuity and the above theorem, show that the limit of the sequence $f(x_1),f(x_2),\dots$ lies in $A$.
Can you do the rest from here?