# Can a composition with itself of a universal self-map be non-universal?

I have formulated (and published) the notion of a universal map (and of a universal morphism), and the problems below, in the early 1960-ies.

DEFINITION   A continuous map   $u: X\rightarrow Y$   is called universal   $\Leftarrow:\Rightarrow$   for every continuous map   $f:X\rightarrow Y$   there exists a point   $p\in X$   such that   $f(p)=u(p)$.

Questions:

• Does there exist a topological space   $X$,   and a universal map   $u:X\rightarrow X$   such that   $u\circ u : X\rightarrow X$   is not universal?
• Can   $X$   in the question above be a 2-dimensional finite polyhedron?

I am convinced that the answer to the first question is yes. Most likely $X$ can be a finite polyhedron. I have formulated also a categorical version of this question, and a more special for monoids too (a monoid is a category with just one object after all). The one for monoids was solved by Ralph N McKenzie in May of 2006 (private communication). He considered the free type of a construction. There should be a natural way to topologize McKenzie's example or any monoid example, i.e. to get a topological space with the desired properties out of the respective monoid. But to obtain a polyhedron is another matter.

Within topology the theory of universal maps includes the topological dimension theory and the theory of the fixed point property; it is connected to the theory of manifolds, and to the stable cohomotopy groups. Couniversal morphisms would be perhaps of interest to the theory of Banach Algebras. While applying a version of the fixed point property in functional analysis and to differential equations may require an odd introductory step, it may be more natural to apply universal maps (in at least one case it is indeed).

EXAMPLE (late 1960ies)   The composition of two universal mappings between 2-dimensional polyhedra does not have to be universal. Indeed, let   $M$ be the Möbius strip (compact, with its boundary). Let   $f : M \rightarrow D$   be the map corresponding to the identification of the equator points, so that it sends the equator to the center of the unit complex disk  $D$.   Let   $g : D\rightarrow D$   be given by   $\forall_{z\in D}\ g(z) := z^2$.   Then   $f\ \ g$   are universal, while   $g\circ f$   is not.

Justification:   all three maps   $f\ \ g\ \ \ g\circ f$   are into disk   $D$.   The universality of the first two, and the non-universality of the composition, follows from the following elementary characterization of the universal mappings into a Euclidean ball (or cube):

Let   $B$   be a closed ball in an $n$-dimensional Euclidean space. Let   $S$   be the boundary of   $B$.   Consider an arbitrary continuous function   $f:X\rightarrow B$,   where   $X$   is an arbitrary topological space, and let   $F := f^{-1}(S)$.   Then the following two properties of   $f$   are equivalent:

• $f$   is universal;
• there does not exist a continuous function   $g:X\rightarrow B$   such that   $g(X)\subseteq S$   and   $g|F=f|F.$

For instance, let's go back to the map   $f:M\rightarrow D$   of the above example. We can start with a definition of the Mobius strip   $M$   as a topological quotient of the following subspace   $M'$   of the complex plane   $\mathbb C$:

$$M'\ :=\ \{z\in\mathbb C: \frac 12\le |z|\le 1\}$$

Space   $M$   (the Mobius strip) is obtained from   $M'$   by identifying all pairs of points   $z\ \ w\ \in\ M'$   such that   $w=-z$   and   $|z|=|w|=\frac 12$.   Now   $f :M \rightarrow D$   is induced by   $\phi:M'\rightarrow D$   given by:

$$\forall_{z\in M'}\quad \phi(z) := (2\cdot |z|-1)\cdot z$$

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Can you say more on why f and g in your example are universal? Also, can you say something more about McKenzie's example? Gerhard "Interested Minds Wish To Know" Paseman, 2013.05.12 – Gerhard Paseman May 13 '13 at 0:59
@Gerhard "universal" Paseman: I'll expand the text of my "question" to provide a justification of "EXAMPLE" (I'll also try to include the reference to my corresponding "Fund. Math" paper.). About the McKenzie's example, I have an email from him with a pdf attachment. I wish he showed up on MO and told his story himself. I have no qualms to inform the public about his result, but I feel uneasy about writing more about it. – Włodzimierz Holsztyński May 13 '13 at 3:58
OK. When I was Ralph's student, he spelled his last name as McKenzie, as opposed to Mckenzie. I would be surprised if that has changed. Gerhard "More General Than Universal Now" Paseman, 2013.05.12 – Gerhard Paseman May 13 '13 at 4:06
I apologize for my initial misspellings of Ralph's last name. In my email there were several instances of his name appearing, and each time, when there was a mixture of the lower and upper characters, the letter k/K appeared in the lower case (which I carefully checked before posting my question). Then seeing your @Gerhard "correct spelling" Paseman I trusted you, of course, but checked the state of affairs with Google anyway, and indeed you and Google agree. – Włodzimierz Holsztyński May 13 '13 at 4:20
Could you provide a link to McKenzie's example? Is a note made of it somewhere? – Todd Trimble Dec 11 '14 at 2:08