# Fixed point relation $\ Fix\$ for pairs of manifolds

First the classical definition: a topological space $X$ has the fixed point property (fpp) $\ \Leftarrow:\Rightarrow\$ for every continuous $\ f : X\rightarrow X\$ there exists $\ p\in X\$ such that $\ f(p)=p$.

I've introduced (in the early 1960s) a related relation $\ X\, Fix\, Y\$ for pairs of topological spaces:

$$X\, Fix\, Y\quad\Leftarrow:\Rightarrow\quad \forall_{f:X\rightarrow Y,\ g:Y\rightarrow X}\,\exists_{p\in X}\ (g\circ f)(p) = p$$

where $\ f\ g\$ are continuous. Relation $\ Fix\$ is symmetric:

$$X\, Fix\, Y\quad\Leftrightarrow\quad Y\, Fix\, X$$

Furthermore, the following 3 properties of a space $X$ are equivalent:

• $\ X\$ has fpp
• $\ X\, Fix\, X$
• $\ \forall_Y\ (X\, Fix\, Y)$

where $Y$ is an arbitrary topological space.

QUESTION (old)   Do there exist connected Hausdorff compact manifolds $\ M'\ M''\$ (without boundary) of the same dimension,, which are not homotopically equivalent, and which do not satisfy $\ M' \, Fix\, M''\$ ? (See the newer question below)

Illustrations for $\ Fix$:

• If $\ X\$ is a retract of $\ X',\$ and $\ Y\$ of $\ Y'\,$ and if $\ X'\,Fix\,Y'\$ then $\ X\,Fix\,Y$.
• If covering $\ \dim(X) \lt S^n\$ then $\ X\, Fix\, S^n.\$ Thus every two spheres of different dimension $\ Fix\$ each other.
• If the homotopy group $\ \pi_n(X)\$ is trivial then $\ X\,Fix\,S^n$.
• $\ S^n\, Fix\ \mathbb{RP}^m\quad\quad S^{2\cdot n}\, Fix\ \mathbb{CP}^m\quad\quad S^{4\cdot n}\, Fix\ \mathbb{HP}^m\quad\quad$for every $\ n>1$.
• If $\ X\, Fix\, Y\$ then at least one of the spaces $\ X\ Y\$ is connected.
• If $\ X:=\bigcup_t X_t\$ represents a topological union, and if $\ Y\$ is connected, then $\ X\,Fix\,Y\ \Leftrightarrow\ \exists_t\ (X_t\,Fix\,Y)$

EDIT:   @Gabriel has easily provided (induced) a whole class of examples. It seems still of interest to answer the questions in my comment to Gabriel's answer below, even before one formulates an ultimate question (if it exists, or even if an ultimate question is not needed). After @Gabriel's answer I have added an illustration above (the first one); it supplies even more counter-examples. Thus, below, let me add a new version of my question.

QUESTION (2014-08-21)   First a standard definition: a topological space is called r-indecomposable $\ \Leftarrow:\Rightarrow\$ there exists a topological space $\ Y,\$ which is a non-trivial homotopic r-image of $\ X\$ (i.e. if there exist maps $\ f:X\rightarrow Y\ \ g:Y\rightarrow X\$ such that $\ f\circ g: Y\rightarrow Y\$ is homotopic to identity, while $\ Y\$ is not contractible nor homotopically equivalent to $\ X.\$.

Now the actual question: Do there exist Hausdorff compact manifolds (preferably without boundary), an r-indecomposable $\ X\$ and arbitrary $\ Y,\$ and of dimension $\dim(X)\ge\dim(Y),\$ which are not eqivalent homotopically, and such that $\ X\,Fix\,Y\$ fails?

If $X$ does not have the fixed point property then for any spaces $Y$ and $Z$, it is not the case that $X\times Y\ Fix\ X\times Z$.
So take for instance $S^2\times S^1$ and $S^1\times S^1\times S^1$.
• Gabriel, I don't have any (I did some about $S^1$, the fundamental group and covering spaces in 1960s but I don't remember anything, it's gone). – Włodzimierz Holsztyński Aug 27 '14 at 19:48