Let $A$ and $B$ be two ordered vector space equipped with the order topology. Let $f:A\to B$ be an increasing function (if $x\leq_A y$ then $f(x)\leq_B f(y)$).
Is there two increasing functions $g:A\to B$ and $h:B\to B$ such that $g$ is continuous and surjective with $f=h\circ g$.
Is there some related continuous/discontinuous decomposition of monotone functions ?
This question arise from my research in information theory, the fact that $f$ is increasing is related to the data processing inequality for some information measure, I also typically use $B=\mathbb R$ with it's natural ordering so I am interested in that too.
I think the statement is equivalent to there exist an increasing bijective function $h:B\to B$ such that $h\circ f$ is an increasing and continuous function. If we take in the previous statement $h^{-1}$ then this works and then $g=h\circ f$.
There may be a problem with that if the domain of $f$ is for instance $[a,b[\cup ]b,a]$ in the reals, however I'm not sure this can happen when $f$ is increasing.
No, this is not true. Since $B$ is equipped with the order topology, an increasing bijection $h:B\to B$ is a homeomorphism, so if $f=hg$ then $g$ is continuous iff $f$ is continuous. So, you are just asking whether an increasing function is automatically continuous. This is not true; for instance, you could have $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x$ for $x\leq 0$ and $f(x)=x+1$ for $x>0$.