Consider the closed unit ball $\overline{\mathbb B ^n}$, and let $\sim$ be some equivalence relation on $\overline{\mathbb B ^n}$ that only identifies points on the boundary $\mathbb S ^{n-1}$ of $\overline{\mathbb B ^n}$, that is, if $x,y\in \overline{\mathbb B ^n}$ are such that $x\sim y$ and $x\neq y$, then $x,y\in \mathbb S ^{n-1}$.
Let $F: \overline{\mathbb B ^n}/\sim \hspace{0.2em}\to \overline{\mathbb B ^n}/\sim$ be a continuous function that sends $\lbrace [x]:x\in \mathbb S ^{n-1}\rbrace$ into itself. Does $F$ have fixed points?
Notice that if $n=1$, then the answer is yes. This is because $\overline{\mathbb B ^1}/\sim$ is homeomorphic to $[-1,1]$ or to a circle; in the first case the result follows by using Brouwer's fixed point theorem and in the second case the result is true simply because $F$ fixes the point in $\overline{\mathbb B ^1}/\sim$ corresponding to the identification of the endpoints of $\overline{\mathbb B ^1}$.
I'm not exactly sure if the result is true in general. I've been trying to adapt Milnor's elementary proof of Brouwer's fixed theorem to this specific case, which is discussed here https://people.math.sc.edu/howard/Notes/brouwer.pdf, but I'm not seeing if this will be fruitful of not.
Does the problem become easier if we assume that $F$ is a homeomorphism that sends $\lbrace [x]:x\in \mathbb S ^{n-1}\rbrace$ onto itself?
Any help or suggestions are appreciated!
Best Answer
This is false in the general setup, even in very low dimension $2$. I don't know about $F$ being homeomorphism case.
Consider the two dimensional unit disc $B^2=\{(x,y)\in\mathbb{R}^2\ |\ \lVert (x,y)\rVert\leq 1\}$ and take $(x,y)\sim (x,-y)$ for points on the boundary, plus $(-1,0)\sim (0,1)$. This glueing results in a two-dimensional sphere with two points glued (i.e. a torus but with glued hole), and the boundary becomes a small circle. See this picture:
One of those small circles is the image of $S^1$ under the quotient map. Lets say it is the bottom one.
Now the thing is that each of those circles is a retract of whole space. Indeed, we first retract the upper hemisphere onto the lower (by simply taking $-z$ on $z$ coordinate, after embedding it correctly into $\mathbb{R}^3$), and then squash with rotation until we hit the bottom circle. Like this:
So this retraction onto the bottom circle together with rotation of the circle will not have a fixed point. Even though it does map the circle onto itself.
Unfortunately this is not a good example if we want $F$ to be a homeomorphism (although it is a homeomorphism on the image of the boundary into itself). Because every self homeomorphism of that space has to map the special glueing point into itself (because it is the only point without euclidean neighborhood), and thus has to be a fixed point.