Apologizing for looking into pre-history, I have a question about a theorem from Chapter 1 ;
this came up in a discussion with [Grant Roy](https://forum.azimuthproject.org/profile/1768/Grant%20Roy) at the [Caltech Study Group](https://forum.azimuthproject.org/discussion/2066/applied-category-theory-course-caltech-study-group)

#### Theorem (Adjoint functor theorem for preorders).
> Suppose Q is a preorder that has all meets and let P be any preorder. A monotone map g : Q → P preserves meets if and only if it is a right adjoint.
Similarly, if P has all joins and Q is any preorder, a monotone map f : P → Q preserves joins if and only if it is a left adjoint.

The question deals with a case where \$$P \$$ does not have all joins, and we have a right adjoint which is not a surjection; specifically, when \$$P \$$ has two elements which are _symmetric_ (in a sense which should be clearer in the example below), and the right adjoint maps to one, but not the other.

Based on the theorem part for right adjoints, let's take the ["bowtie" poset](https://forum.azimuthproject.org/discussion/comment/17902/#Comment_17902)

![](https://i.imgur.com/Km2PosD.png)

as our \$$P\$$, and \$$1 \to 2 \to 3 \$$ as our \$$Q\$$.

Choose \$$g(1) = g(2) = g(3) = a \$$ for \$$g : Q \to P \$$ .

To the best of my understanding, \$$Q \$$ has all meets and \$$g \$$ is a monotone map which preserves them, therefore \$$g \$$ is a right adjoint according to the above stated theorem.

Constructing its left adjoint, based on the candidate in the proof:
\$f(p) = \bigwedge \\{q \in Q : \; p \le_P g(q) \\} \$
we find \$$f(a) = f(c) = f(d) = 1 \$$ and \$$f(b) = \bigwedge \emptyset = ? \$$;

Applying brute force search for an assignment for \$$f(b)\$$, I could not find one satisfying a Galois Connection in this case;
however, this case seems to meet the settings of the theorem, isn't it?

Could anyone help me identify my mistake?