# Homeo-Fixed point property

Edit: According to comment of Michał Kukieła I revised the question

A topological space $X$ satisfies "Homeo-fixed point" property if every homeomorphism $f$ on $X$ possess a fixed point.

Is there an example of a connected manifold with this property but does not satisfies fixed point property?

Edit Now i found a related paper, so I add the link to this question:

• Notation: FPP = fixed point property. Disjoint union of a manifold with FPP and a manifold without FPP does not have FPP, but it has FPP with respect to homeomorphisms. So you probably want to assume your manifold is connected. – Michał Kukieła Oct 7 '14 at 10:24
• What if one connect the two spaces X and Y of Michał Kukieła's example by an arc (say with some more convenient assumption)? – Pietro Majer Oct 7 '14 at 19:08
• E.g. a topological space $P$ homeomorphic to the letter "P". Every homeo $P\to P$ has at least 2 fixed points, while a continuous map $P\to P$ may have none. ($P$ is not a topological manifold of course) – Pietro Majer Oct 7 '14 at 19:26
• @RyanBudney: "Is there an example of something with property X but not with property FPP?" In other words, is there a connected topological space (or even manifold) such that each homeomorphism has fixed point but not every continuous mapping has fixed point? – Vít Tuček Oct 12 '14 at 21:45
• @RyanBudney I just offered my interpretation of the question, whose wording can (and should) certainly be improved. I'm a topological outsider (ehm) and my understanding is that FPP is a standard term which means that all continuous mappings of the space to itself have a fixed point. Since homeomorphism is a special case of a continuous mapping it makes sense to ask for which spaces all homeorphisms have a fixed point whereas some continuous mappings don't. – Vít Tuček Oct 13 '14 at 1:54

consider the connected sum of the klein bottle with the projective plane

there is a map with no fixed point collapse to the klein bottle and rotate

any homeo induces a map on mod two homology which is an isomorphism respecting the intersection form

i didn't calculate but if the trace mod two of any such 3x3 matrix is non zero then any homeomorphism has a fixed point

in any case this is an idea to construct an example

Dennis Sullivan

Let me construct a non-paracompact counterexample. Of course, a paracompact counterexample is better in many respects since differential topology is usually restricted to paracompact manifolds and because the manifolds that naturally occur and the ones sold here are all paracompact.

Let $L=([0,1)\times\omega_{1})\setminus\{(0,0)\}$ be the long line. Let $M$ be any paracompact manifold with the fixed point property for homeomorphisms. For instance, $M$ could be a projective space over the reals, the complex numbers, or over the quaternions. Let $N=L\times M$.

I claim that $N$ has the fixed point property for homeomorphisms but not for continuous maps.

Clearly, there is a retraction $f:N\rightarrow S$ where $S\simeq M\times(0,1]$. The set $M\times(0,1]$ does not have the fixed point property since the map $M\times(0,1]\rightarrow M\times(0,1],(x,y)\mapsto(x,y/2)$ has no fixed point. Therefore there is some map $g:S\rightarrow S$ without a fixed point. Thus $g\circ f:N\rightarrow S$ has no fixed point.

Now assume that $h:N\rightarrow N$ is a homeomorphism.

If $x\in M$, then let $\Gamma_{x}:L\rightarrow L,\Delta_{x}:L\rightarrow L$ be the maps where $\Gamma_{x}(y)=\pi_{1}(h(y,x)),\Delta_{x}(y)=\pi_{1}(h(y,x))$. Let $C_{x}=\{\alpha\in\omega_{1}|\Gamma_{x}(\alpha,0)=(\alpha,0)\},D_{x}= \{\alpha\in\omega_{1}|\Delta_{x}(\alpha,0)=(\alpha,0)\}.$

I claim that the sets $C_{x},D_{x}$ are club sets (club stands for closed unbounded). It is clear that the sets $C_{x},D_{x}$ are closed, so we now need to show $C_{x},D_{x}$ are unbounded.

It is easy to see that for all $\alpha\in\omega_{1}$, we have $\Gamma_{x}(z)>(0,\alpha)$ for sufficiently large $z$. Therefore, for all $\alpha\in\omega_{1}$ there is a sequence $(y_{n})_{n}$ with $(0,\alpha)<y_{0}$ and where $y_{n}<y_{n+1},y_{n}<\Gamma_{x}(y_{n+1}),\Gamma_{x}(y_{n})<y_{n+1},\Gamma_{x}(y_{n})<y_{n+1}$. Furthermore, one can choose the sequence $(y_{n})_{n}$ so that $\lim_{n\rightarrow\infty}y_{n}=(0,\gamma)$ for some $\gamma$. Therefore, we have $\Gamma_{x}(0,\gamma)=\lim_{n\rightarrow\infty}\Gamma_{x}(y_{n})=(0,\gamma)$. We conclude that $\gamma\in C_{x}$, so $C_{x}$ is unbounded. Thus $C_{x}$ is a club set. The set $D_{x}$ is a club set as well by an identical argument.

Let $A\subseteq M$ be a countable dense subset. Then let $E=\bigcap_{x\in X}C_{x}\cap D_{x}$. Then $E$ is a club set being the countable intersection of club sets.

Now assume that $\alpha\in E$. Then since $(\alpha,0)=\Delta_{x}(\alpha,0)=\pi_{1}(h((\alpha,0),x)$ for $x\in X$, we have $h[\{(\alpha,0)\}\times X]\subseteq\{(\alpha,0)\}\times M$. Therefore, we have $h[\{(\alpha,0)\}\times M]=h[\overline{\{(\alpha,0)\}\times X}]\subseteq\{(\alpha,0)\}\times M$. Similarly, we have $h^{-1}[\{(\alpha,0)\}\times M]\subseteq\{(\alpha,0)\}\times M$. Therefore, $\{(\alpha,0)\}\times M\subseteq h[\{(\alpha,0)\}\times M]$. We conclude that $h[\{(\alpha,0)\}\times M]=\{(\alpha,0)\}\times M$.

However, since $M$ has the fixed point property for homeomorphisms, and $h[\{(\alpha,0)\}\times M]=\{(\alpha,0)\}\times M$, the mapping $h$ has some fixed point on $\{(\alpha,0)\}\times M$.