About characterization of atomistic posets

Question

Fix some poset.

Let $\mathscr{A}$ be the map from elements of our poset into the set of atoms under this element.

Is it true, that injectivity of $\mathscr{A}$ implies that our poset is atomistic? (A poset is called atomistic iff every its element is a join of atoms under this element.)

Answer 1

Theorem A complete lattice is atomistic iff the function "$\mathscr{A}$" is injective.

Proof

$\Rightarrow$

Let our poset is atomistic. Then obviously $\mathscr{A} a \neq \mathscr{A} b$ for elements $a \neq b$.

$\Leftarrow$

Let "$\mathscr{A}$" be injective. Consider an element $a$ of our poset. Let $b = \bigsqcup \mathscr{A} a$ (where $\bigsqcup$ denotes the supremum). Obviously $b \sqsubseteq a$ and thus $\mathscr{A} b \subseteq \mathscr{A} a$. But if $x \in \mathscr{A} a$ then $x \sqsubseteq b$ and thus $x \in \mathscr{A} b$. So $\mathscr{A} a = \mathscr{A} b$. By injectivity $a = b$ that is $a = \bigsqcup \mathscr{A} a$.