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 $S^1\times S^1\times S^1$.

  • $\begingroup$ Thank you @Gabriel. I have overstated--so-to-speak--my conjecture. I'll need to modify it. I like this topic but worked very little on it, too bad. Right away one may ask about different 2-dim compact manifolds. Then, following your observation, one may restrict the above general question to manifolds such that at least one of them is not a non-trivial Cartesian product (especially when one of them is), etc. It'd be neat to get a still more elegant hupothesis. $\endgroup$ – Włodzimierz Holsztyński Aug 21 '14 at 20:07
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    $\begingroup$ But all the non-sphere surfaces are r-decomposable, right? They all retract to the circle. So none of them should Fix each other. $\endgroup$ – Gabriel C. Drummond-Cole Aug 26 '14 at 5:12
  • $\begingroup$ right, thank you (a bit embarrassed). $\endgroup$ – Włodzimierz Holsztyński Aug 27 '14 at 6:43
  • $\begingroup$ And now that we're past surfaces I think it'll be much harder for me to think of examples. It seems like r-indecomposability is a pretty strong requirement, do you happen to know anything about that? $\endgroup$ – Gabriel C. Drummond-Cole Aug 27 '14 at 16:45
  • $\begingroup$ 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). $\endgroup$ – Włodzimierz Holsztyński Aug 27 '14 at 19:48

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