Trying to get a different angle for the question [Fixed points and universal maps for posets](https://mathoverflow.net/questions/180252/fixed-points-and-universal-maps-for-posets), I want to compare universal maps to a different kind of functions.


First recall that for posets $P,Q$ an order-preserving map $u:P\to Q$ is **universal** if for any order-preserving map $f:P\to Q$ there is $x\in P$ such that $f(x) = u(x)$.


Let $\mathcal{C}$ be any category and let $A, B$ be objects. A morphism $l: A\to B$ is said to be **left-factoring** if for any $Z\in \mathbf{Ob}(\mathcal{C})$ and any morphism $f: Z\to B$ there is $h: Z\to A$ such that $f = l\circ h$. (I didn't find the correct terminology for what I call left-factoring maps; any help is appreciated.)


In the category of posets, it is easy to see that universal maps, as well as left-factoring maps, are surjective. (The converse does not hold for either universal or left-factoring maps.)


**Question**: Are universal maps left-factoring?


(Note that left-factoring maps need not be universal: the identity on $\omega$ with the usual well-ordering is left-factoring, but not universal, because $\textsf{id}\_\omega$ and the successor map $n\mapsto n+1$ do not agree on any point.)



Universal and left-factoring order-preserving maps
I claim that not every universal map is left-factoring. Let me first give a proposition that allows us to construct many universal maps, and then I will give a counterexample from the proposition. A poset $X$ shall be called a DCPO (directed complete partial order) if every directed subset of $X$ has a least upper bound. It is well known and not to hard to show that a poset is a DCPO if and only if every well ordered chain in $X$ has a least upper bound. Recall that the Knaster-Tarski fixed point theorem states that if $L$ is a complete lattice and $f:L\rightarrow L$, then $f$ has a fixed point. The following proposition is a generalization of the Knaster-Tarski fixed point theorem that allows us to construct universal mappings.


$\mathbf{Proposition}$ Suppose that $f,g:X\rightarrow Y$ are order preserving maps where


1. $f(0)=0$
2. $X$ is a DCPO
3. whenever $L\subseteq X$ is a chain, then $f[L]$ has a least upper bound and $f(\bigvee L)=\bigvee f[L]$.
4. whenever $x\in L$ and $f(x)<r$, there is some $y>x$ with $f(y)\leq r$.


Then there is some $c$ with $f(c)=g(c)$.


$\mathbf{Proof}$ We shall construct a transfinite strictly increasing sequence $(x\_{\alpha})\_{\alpha}$ in $X$ such that $f(x\_{\alpha})\leq g(x\_{\alpha})$ until we obtain an $x\_{\delta}$ with $f(x\_{\delta})=g(x\_{\delta})$. Let $x\_{0}=0$. If $\gamma$ is a limit ordinal and $x\_{\alpha}$ has been obtained for $\alpha<\gamma$, then let $x\_{\gamma}=\bigvee\_{\alpha<\gamma}x\_{\alpha}$. Then since $f(x\_{\alpha})\leq g(x\_{\alpha})$ for all $\alpha<\gamma$, we have
$f(x\_{\gamma})=f(\bigvee\_{\alpha<\gamma}x\_{\alpha})=\bigvee\_{\alpha<\gamma}f(x\_{\alpha})\leq g(x\_{\gamma})$.


Now suppose that $x\_{\alpha}$ has been constructed and $f(x\_{\alpha})\leq g(x\_{\alpha})$. If $f(x\_{\alpha})=g(x\_{\alpha})$, then we discontinue the transfinite process. If $f(x\_{\alpha})<g(x\_{\alpha})$, then there is some $x\in X$ with $x>x\_{\alpha}$ and $f(x)\leq g(x\_{\alpha})$. Therefore select some $x\_{\alpha+1}\in X$ such that 
$x\_{\alpha}<x\_{\alpha+1}$ and $f(x\_{\alpha+1})\leq g(x\_{\alpha})$. Then we have $f(x\_{\alpha+1})\leq g(x\_{\alpha+1})$, and the transfinite process continues. This transfinite process however must terminate at some ordinal $x\_{\delta}$, so we have $f(x\_{\delta})=g(x\_{\delta})$. $\mathbf{QED}$.


Let's now use the above proposition to construct a universal map that is not left-factoring. Suppose that $T$ is a the tree which consists of all strictly increasing functions $f:\alpha\rightarrow[0,1]$ for some ordinal $\alpha$. Let $L:T\rightarrow[0,1]$ be the mapping where $L(f)=\sup\{f(\beta)|\beta\in\textrm{Dom}(f)\}$. Then $L$ is universal. However, the mapping $L$ is not left-factoring. In particular, there is no order preserving map $h:[0,1]\rightarrow T$ with $L\circ h=1\_{[0,1]}$ where $1\_{[0,1]}$ denotes the identity function on $[0,1]$. If $h$ were such a mapping with $L\circ h=1\_{[0,1]}$, then whenever $r<s$, then $r=L(h(r))<L(h(s))=s$, so $h(r)<h(s)$, hence $\textrm{Dom}(h(r))<\textrm{Dom}(h(s))$. However, if $(x\_{n})\_{n}$ is an infinite descending sequence of real numbers in $[0,1]$, then $(\textrm{Dom}(h(x\_{n})))\_{n}$ would be an infinite descending sequence of ordinals, a contradiction.


