Let $\omega_1$ be the first uncountable ordinal, 
same as the set of all countable ordinals. 
Let $F$ be the set of all functions 
$f$ from $\omega_1$ minus singleton $0$ into $\omega_1$ that 
are (a) regressive i.e. $f(\alpha) < \alpha$ for $0 < \alpha < \omega_1$, 
and (b) order-preserving (same as non-decreasing) 
i.e. $f(\alpha)\leq f(\beta)$ if $\alpha \leq \beta$. 
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 $f \le g$ then $h(f) \le h(g)$, 
and with the additional property that $f \le h(f)$ ? 

(I had dropped (b) in my first post, but comments below 
show that it is essential. Also, I did mean the functions $f$ must be regressive, when I had imprecisely said decreasing, in the original statement. I added the gn tag since the question stated is an order-theoretic translation of a question from general topology: Whether $\omega_1$ has a monotone interior-preserving open operator.) 

Edit April 28, 2014: The answer by Noah S below is correct but _incomplete_ (for each $f\in F$ it finds $h(f)\in K$ with $f \le h(f)$ but does not consider whether $h(f) \le h(g)$ when $f\le g \in F$). The question is open. Thank you 

Edit April 12, 2015. I [reposted at MSE](https://math.stackexchange.com/questions/1231431/order-preserving-map-of-regressive-functions-on-omega-1)