Let $\omega_1$ be the first uncountable ordinal, 
same as the set of all countable ordinals. 
Let $F$ be the set of all regressive functions 
$f$ from $\omega_1$ minus singleton $0$ into $\omega_1$, 
that is $f(\alpha) < \alpha$ for $0 < \alpha < \omega_1$. 
Define a partial order on $F$ by $f \leq g$ if 
$f(\alpha) \leq g(\alpha)$ for all $0<\alpha < \omega_1$. 
Let $K$ be the subset of $F$, of functions with 
a finite range. 

Question: 
Is there an order-preserving homomorphism 
$h : F \to K$, i.e if $f \leq g$ then $h(f) \leq h(g)$, 
and with the additional property that $f \leq h(f)$ ? 

I had dropped one requirement (and comments below 
show that it is essential), but one must also assume 
that the functions $f\in F$ preserve the order, that is 
$f(\alpha)\leq f(\beta)$ if $\alpha \leq \beta$. (Perhaps 
the latter would be correctly, but confusingly to me, 
termed non-decreasing: I prefer to say each $f$ is order-preserving. 
I did mean the functions $f$ must be regressive, when 
I had imprecisely said decreasing, in the original statement.)