The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes
Exercise: A $\textbf{semilattice}$ is a poset in which every finite subset has a least upper bound. Show that a semilattice is the same thing as an abelian monoid satisfying $x\cdot x=x$ for all $x$. [Hints: $\textit{LUB}\{x_1,\ldots,x_n\}=x_1+\ldots+x_n (=e\text{ if }n=0)$ where as $x\leq y$ iff $xy=x.$] Verify that a monoid homomorphism between semilattices preserves all finite $\textit{LUB}$'s and is automatically order-preserving.
Questions: I just few quick questions, the first is, I don't understand the how the hint resolves the question in particular where it states:
$\textit{LUB}\{x_1,\ldots,x_n\}=x_1+\ldots+x_n (=e\text{ if }n=0)$ where as $x\leq y$ iff $xy=x.$
I am guessing that for $\textit{LUB}\{x_1,\ldots,x_n\}=x_1+\ldots+x_n$, that would mean, I take the ${max}(x_1,\ldots,x_n)=x_i,$ and so $\textit{LUB}\{x_1,\ldots,x_n\}=x_1+\ldots+x_n=n\cdot x_i$? Also I don't understand the meaning of "$xy\leq x$ iff $xy=x$"?
Instead, can't I do the following: Suppose for any $a,b\in X$ and define the operation on the X with the binary operation: $a\cdot b={LUB}\{a,b\},$ and also show that it is a semilattice
Also, what does it mean to say "...monoid homomorphism between semilattices preserves all finite $\textit{LUB}$'s". I don't know what is meant by "preserves"?
Thank you in advance.
From the context (the hint $LUB\{x_1,\ldots,x_n\}=e$ if $n=0$), I assume that this author considers that, in a semilattice, every finite set (including the empty set) has an upper bound, which I think is not usual, but still a reasonable definition. Hence the correspondence with idempotent commutative monoids, and not just with idempotent commutative semigroups, as usual.
The goal is to prove that, given a commutative idempotent monoid $(M,+,e)$, and defining $$ x_1 \vee x_2 \vee \cdots \vee x_n = \begin{cases} x_1 + \cdots + x_n, &\text{if } n>0,\\ e, &\text{if } n=0, \end{cases} $$ then $(M,\vee,e)$ is a join-semilattice with a top element $e$.
Of course you can define the binary operation join as $x \vee y = x + y$, and then generalize to the definition above, knowing that $x+e=x$ corresponds to $x \vee e = x$, meaning that $e$ is the top element.
But to reach that conclusion you must start by showing that there is order relation defined by $$x \leq y \iff x \vee y = y \iff x + y = y,$$ where the operation $\vee$ is the one of the semilattice, and the $+$, the one of the monoid (indeed, they are the same; the distinction is only so that you don't get lost in the correspondence).
In this answer I showed that the above formula does, indeed, define an order relation, and further that it defines a semilattice.
There is a difference though, which is, as I put in the beginning, the definition of semilattice which is used. However, that is not a problem, since in your case, you already have, by definition that the join of an empty set of elements is a fixed element $e$.
The meaning of "monoid homomorphism between semilattices preserves all finite LUBs" is that, if $\psi:(M,+,e)\to(M',*,e')$ is a monoid homomorphism, that is, $\psi(a+b)=\psi(a)*\psi(b)$ and $\psi(e)=e'$, then, considering the corresponding semilattices $(M,\vee,e)$ and $(M',\oplus,e')$, with top elements $e$ and $e'$, respectively, we have that $\psi(a\vee b) = \psi(a) \oplus \psi(b)$ (which certainly follows by definition, i.e., these operations coincide with the ones of the monoids), but also that $$\psi(a_1 \vee \cdots \vee a_n) = \psi(a_1) \oplus \cdots \oplus \psi(a_n),$$ which is not difficult to show.
Moreover, show that $\psi$ is order-preserving, that is, $$a \leq b \implies \psi(a) \leq \psi(b),$$ which follows from \begin{align} a \leq b &\implies a + b = b\\ &\implies \psi(a) * \psi(b) = \psi(b)\\ &\implies \psi(a) \leq \psi(b). \end{align}