Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

I think this is a counterexample:

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 $g(x) \cap^{\mathfrak C} g(a) \neq g(y) \cap^{\mathfrak C} g(a)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.