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

Many thanks for your detailed reply!

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.

Many thanks for your detailed reply!

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.