Define ${\cal L}$ as in [this question](https://mathoverflow.net/questions/215582/how-big-is-the-lattice-of-all-functions): the set of functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(n)\leq f(n+1)\leq f(n)+1$, where two functions are considered equal if they differ at at most finitely many points, and $f\prec g$ is defined to mean $f(n)\leq g(n)$ for all but at most finitely many exceptions $n$. 


Let ${\cal P}(\omega)/fin$ be defined as in [this post](https://mathoverflow.net/questions/200769/antichain-on-mathcalp-omega-fin-of-cardinality-2-aleph-0). In [another post](https://mathoverflow.net/a/215820/70876), it was established that there is some kind of "natural" (not in the categorical sense) lattice embedding from ${\cal P}(\omega)/fin$ into ${\cal L}$.


My question is: Can the relationship between ${\cal P}(\omega)/fin$ and ${\cal L}$ be made more precise? Isn't ${\cal L}$ essentially ${\cal P}(\omega)/fin$ "stacked on top of itself $\omega$ times"? Or is ${\cal L}$ (a quotient of) a product of ${\cal P}(\omega)/fin$?



Relationship of ${\cal P}(\omega)/fin$ and ${\cal L}$
Look at your and my second answer to [the original question](https://mathoverflow.net/questions/215582/how-big-is-the-lattice-of-all-functions).
Take your map from $\mathcal{P}(\omega)$ into $\mathcal{L}$ (or rather the set of functions before identifying almost equal elements).
That is for $A\subseteq\omega$ define $f\_A$ as in your answer: $f\_A(n)=|n\cap A|$. What I showed is this: if $A\subset^\*B$ then $f\_B(n)-f\_A(n)$ diverges to infinity.
What you can also show is: if $A=^\*B$ then $f\_A-f\_B$ is constant on a tail. Indeed, if $A\setminus m=B\setminus m$ then for $n\ge m$ we have $f\_A(n)-f\_B(n)=f\_A(m)-f\_B(m)$. 
On the other hand, if $A\neq^\*B$ then $f\_A-f\_B$ is *not* constant on a tail: if $n\in A\setminus B$ then $f\_A(n)<f\_A(n+1)$ but $f\_B(n)=f\_B(n+1)$ and
vice versa.


This shows that $A\mapsto f\_A$ induces an injective map from $\mathcal{P}(\omega)/\mathit{fin}$ into $\mathcal{L}/{\equiv}$ where $f\equiv g$ means that $f-g$ is constant on a tail.
The map is also onto: given $f\in\mathcal{L}$ look at $g$ defined by $g(n)=f(n)-f(0)$; then $g=f\_A$, where $A=\{n:f(n+1)=f(n)+1\}$.


As shown in the other answer the map is order-preserving in the sense that $A\subset^\*B$ implies that $f\_B(n)-f\_A(n)$ diverges to infinity.


On the other hand disjoint sets can map to comparable functions: let $A$ consist of the even numbers and let $B=\{4n+1:n\in\omega\}$. Then $f\_A$ grows like $n/2$ and $f\_B$ grows like $n/4$ so that $f\_A(n)-f\_B(n)$ will diverge to infinity. 


