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Let $X,Y$ be topological spaces. We call a continuous map $u:X\to Y$ universal if for every continous map $f:X\to Y$ there is $x\in X$ such that $f(x) = u(x)$.

If $u:X\to Y$ and $v:Y\to Z$ are universal, is $v\circ u: X\to Z$ universal?

(I am thinking about this question because of an inspiration from Fixed points and universal maps for posets ).

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  • $\begingroup$ I must be stupid, but isn't "universal" equivalent to surjective? $\endgroup$ – Oblomov Nov 19 '14 at 15:08
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    $\begingroup$ @Oblomov Surjectivity is clearly necessary but universality is much stronger (the identity on $\mathbb R$ is surjective but not universal with $f(x)=x+1$). $\endgroup$ – Jochen Wengenroth Nov 19 '14 at 15:15
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    $\begingroup$ That's right, @JochenWengenroth. The identity map $\textrm{id}: X\to X$ on a space $X$ is universal if and only if $X$ has the fixed point property (every continous self-map has a fixed point), which is quite a strong requirement for topological spaces. If $u:X\to Y$ is universal then $Y$ has the fixed point property. $\endgroup$ – Dominic van der Zypen Nov 19 '14 at 15:22
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    $\begingroup$ Remark: if $g \circ f$ is universal, then $g$ is universal. For example, whenever $u: X \to Y$ is universal, it follows that $1_Y$ is universal ($Y$ has the fixed-point property) since $u = 1_Y \circ u$. $\endgroup$ – Todd Trimble Nov 19 '14 at 16:08
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    $\begingroup$ @Oblomov What is mildly nice is that if $X$ is connected, then $u: X \to [0, 1]$ is universal precisely when $u$ is surjective. (It then follows from my previous comment that for any $X$, as long as $u(C) = [0, 1]$ for some connected component $C$ of $X$, then $u$ is universal.) $\endgroup$ – Todd Trimble Nov 19 '14 at 16:10
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This seems to be answered negatively in "On the composition and products of universal mappings" by W. Holsztyński (Fundamenta Mathematicae 64(2) (1969), 181-188).

I've not looked at the proof or really tried to figure out why it works, but if I understand correctly what is claimed, then one counterexample from the paper is as follows:

$M$ is a Möbius band, regarded as the annulus $\left\{z\in\mathbb{C}:\frac{1}{2}\leq\vert z\vert\leq 1\right\}$ with $z$ and $-z$ identified for $\vert z\vert=\frac{1}{2}$.

$Q$ is the closed unit disc $\left\{z\in\mathbb{C}:\vert z\vert\leq 1\right\}$.

Apparently the map $u:M\to Q$ induced by $z\mapsto\left(2-\frac{1}{\vert z\vert}\right)z$, which identifies the inner boundary of the annulus to a point, and the map $v:Q\to Q$ given by $z\mapsto z^2$ are both universal, but $v\circ u$ is not.

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