# Congruence for embeddings of a poset into semilattices

Let $\mathfrak{A}$ be a poset, $\mathfrak{B}$ and $\mathfrak{C}$ be meet-semilattices with least elements. Let $f:\mathfrak{A}\rightarrow\mathfrak{B}$ and $g:\mathfrak{A}\rightarrow\mathfrak{C}$ are order embeddings.

Can we warrant that $f(x)\cap^{\mathfrak{B}}f(a) = f(y)\cap^{\mathfrak{B}}f(a) \Leftrightarrow g(x)\cap^{\mathfrak{C}}g(a) = g(y)\cap^{\mathfrak{C}}g(a)$ for every $x,y,a\in\mathfrak{A}$? How to prove this?

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The posets are $\mathfrak A, \mathfrak B$ and $\mathfrak C$ respectively, from left to right, with the obvious order embeddings $f: \mathfrak A \hookrightarrow \mathfrak B$ and $g: \mathfrak A \hookrightarrow \mathfrak C$. Let $a \in \mathfrak A$ be the element in the middle and $x,y \in \mathfrak A$ the other two elements to the left and right of $a$. Then $f(x) \cap^{\mathfrak B} f(a) = f(y) \cap^{\mathfrak B} f(a)$, but $g(x) \cap^{\mathfrak C} g(a) \neq g(y) \cap^{\mathfrak C} g(a)$.