@[Pete Morcos](https://forum.azimuthproject.org/profile/2373/Pete%20Morcos)

I have a questions about :
\$$f(b)=\bigwedge_Q \emptyset=\top_Q=3\$$

How do we compare the elements of the empty set to any other elements?

Is the above statement is true because all \$$q \in Q \$$ satisfy \$$q \le q' \$$ for any \$$q' \in \emptyset \$$ ?

Using this logic I could also say that all \$$q \in Q \$$ satisfy \$$q' \le q \$$ for any \$$q' \in \emptyset \$$ ,
which means all \$$q \in Q \$$ are \$$q \equiv q' \$$ for any \$$q' \in \emptyset \$$, so using transitivity of equivalence,
we have that all \$$q \in Q \$$ are equivalent, isn't it?

Another question arises when we take another preorder, thanks to [Grant Roy](https://forum.azimuthproject.org/profile/1768/Grant%20Roy),
taking

![](https://i.imgur.com/ByTrB3Q.png)

as our \$$P \$$, Q is still \$$1 \to 2 \to 3 \$$ as above, and \$$g : Q \to P \$$:

\$g(3) = d \$
\$g(2) = b \$
\$g(1) = a \$

\$$g\$$ is monotone, meet preserving, and even the empty set should find a meet in our new \$$P\$$, which is \$$e\$$.

The construction for a left adjoint would assign
\$$f(e) = \bigwedge_Q \emptyset = 1 \$$

and the Galois connection is not satisfied again?!

I'd be happy for your insights.