0
$\begingroup$

Take two posets $A, B$ (partially ordered sets). Now consider these posets to be categories $Cat(A), Cat(B)$ respectively. Consider a map from $A$ to $B$, $f: A \rightarrow B$. This can be seen as a functor $F : Cat(A) \rightarrow Cat(B)$. Under what conditions on the posets $A, B$ and the map $f$, will the functor $F$ have an adjoint? I would like to say that if $f$ is Scott-continuous, then $F$ will have an adjoint. Is this true?

$\endgroup$

2 Answers 2

3
$\begingroup$

I would like to say that if f is Scott-continuous, then F will have an adjoint. Is this true?

No.

If $A$ and $B$ are complete (semi)lattices then $f:A\to B$ has a right adjoint $f\dashv g$ iff $f$ preserves all joins, and then $g:B\to A$ that preserves all meets, and vice versa.

We could call this the (Special) Adjoint Function/or Theorem. However, the result is actually so familiar that it is built into the language of mathematics, if not everyday life, for example in phrases like least common multiple.

If $A$ and $B$ are finite lattices, every order-preserving function between them is Scott continuous, but $f:A\to B$ has a right adjoint iff it preserves $\bot$ and $\lor$.

Of course, posets or dcpos that are not lattices can have or fail to have adjoints too, but it is really not worth the trouble of formulating (necessarily very complicated) theorems about this.

Contravariant adjunctions between lattices arise extremely commonly just from an arbitrary binary relation. I feel that this justifies retaining the name Galois connection for the contravariant situation.

Let $R\subset X\times Y$ be any relation between two sets. For any subsets $A\subset X$ and $B\subset Y$, write $$ A^+ = \{ b | \forall a\in A.R(a,b) \} \text{ and } B^+ = \{ a | \forall b\in B.R(a,b) \}. $$ Then this defines a contravariant adjunction between the powersets of $X$ and $Y$.

In the particular case of Galois theory, let $X$ be the field, $Y$ its automorphism group and $R$ the relation that a particular automorphism fixes a particular element of the field. Then as a corallary and not as a necessary hypothesis, $A^+$ is a subgroup for any subset $B$ and similarly $B^+$ is a subfield for any subset $A$. Under the usual conditions (and only now do we introduce non-trivial algebra) the $(-)^+$ constructions restrict to a bijection between subgroups and subfields.

$\endgroup$
3
$\begingroup$

An adjunction between posets is precisely a Galois connection. (Paul has characterized these in the case where the posets are both complete.) English Wikipedia; nLab.

$\endgroup$
6
  • $\begingroup$ I would say that a Galois connection is a contravariant adjunction $A\to B^{op}$. There are also co-Galois connections, but after a glass of wine or two this evening I am not going to commit myself to saying which is which (see my book). $\endgroup$ Aug 17, 2014 at 20:03
  • $\begingroup$ Wikipedia calls the covariant version a monotone Galois connection and the contravariant version an antitone Galois connection. The nLab mentions both but only discusses the contravariant case in detail. Of course, it's just a matter of an op, so once you understand one of these, you should understand the other. $\endgroup$ Aug 20, 2014 at 3:16
  • $\begingroup$ Wikipedia is an unquestioned authority? In Galois theory the relationship between subfields and subgroups is contravariant. $\endgroup$ Aug 20, 2014 at 6:44
  • $\begingroup$ Of course Wikipedia is not an unquestioned authority! It is merely one example of a usage of terminology. I have no wish to claim that this terminology is authoritative, only that it has been used. But it is the last sentence in my previous comment that really matters. $\endgroup$ Aug 22, 2014 at 4:19
  • $\begingroup$ If we just want to see that both versions of the terminology are used, then the references at the bottom of the Wikipedia article are neatly divided into those that use each convention. (No reference is given for the adjectives ‘monotone’ and ‘antitone’, although they are pretty straightforward to justify.) $\endgroup$ Aug 22, 2014 at 4:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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