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Galois connection
The notion of a Galois connection has its root in Galois theory. By the fundamental theorem of Galois theory, there is a onetoone correspondence between the intermediate fields between a field $L$ and its subfield $F$ (with appropriate conditions imposed on the extension $L/F$), and the subgroups of the Galois group $\operatorname{Gal}(L/F)$ such that the bijection is inclusionreversing:
$\operatorname{Gal}(L/F)\supseteq H\supseteq\langle e\rangle\quad\mbox{ iff }% \quad F\subseteq L^{H}\subseteq L,\mbox{ and}$ 
$F\subseteq K\subseteq L\quad\mbox{ iff }\quad\operatorname{Gal}(L/F)\supseteq% \operatorname{Gal}(L/K)\supseteq\langle e\rangle.$ 
If the language of Galois theory is distilled from the above paragraph, what remains reduces to a more basic and general concept in the theory of orderedsets:
Definition. Let $(P,\leq_{P})$ and $(Q,\leq_{Q})$ be two posets. A Galois connection between $(P,\leq_{P})$ and $(Q,\leq_{Q})$ is a pair of functions $f:=(f^{*},f_{*})$ with $f^{*}\colon P\to Q$ and $f_{*}\colon Q\to P$, such that, for all $p\in P$ and $q\in Q$, we have
$f^{*}(p)\leq_{Q}q\quad\mbox{ iff }\quad p\leq_{P}f_{*}(q).$ 
We denote a Galois connection between $P$ and $Q$ by $P\stackrel{f}{\multimap}Q$, or simply $P\multimap Q$.
If we define $\leq_{P}^{{\prime}}$ on $P$ by $a\leq_{P}^{{\prime}}b$ iff $b\leq_{P}a$, and define $\leq_{Q}^{{\prime}}$ on $Q$ by $c\leq_{Q}^{{\prime}}d$ iff $d\leq_{Q}c$, then $(P,\leq_{P}^{{\prime}})$ and $(Q,\leq_{Q}^{{\prime}})$ are posets, (the duals of $(P,\leq_{P})$ and $(Q,\leq_{Q})$). The existence of a Galois connection between $(P,\leq_{P})$ and $(Q,\leq_{Q})$ is the same as the existence of a Galois connection between $(Q,\leq_{Q}^{{\prime}})$ and $(P,\leq_{P}^{{\prime}})$. In short, we say that there is a Galois connection between $P$ and $Q$ if there is a Galois connection between two posets $S$ and $T$ where $P$ and $Q$ are the underlying sets (of $S$ and $T$ respectively). With this, we may say without confusion that “a Galois connection exists between $P$ and $Q$ iff a Galois connection exists between $Q$ and $P$”.
Remarks.
1. Since $f^{*}(p)\leq_{Q}f^{*}(p)$ for all $p\in P$, then by definition, $p\leq_{P}f_{*}f^{*}(p)$. Alternatively, we can write
$\displaystyle 1_{P}\leq_{P}f_{*}f^{*},$ (1) where $1_{P}$ stands for the identity map on $P$. Similarly, if $1_{Q}$ is the identity map on $Q$, then
$\displaystyle f^{*}f_{*}\leq_{Q}1_{Q}.$ (2) 2. Suppose $a\leq_{P}b$. Since $b\leq_{P}f_{*}f^{*}(b)$ by the remark above, $a\leq_{P}f_{*}f^{*}(b)$ and so by definition, $f^{*}(a)\leq_{Q}f^{*}(b)$. This shows that $f^{*}$ is monotone. Likewise, $f_{*}$ is also monotone.
3. Now back to Inequality (1), $1_{P}\leq_{P}f_{*}f^{*}$ in the first remark. Applying the second remark, we obtain
$\displaystyle f^{*}\leq_{Q}f^{*}f_{*}f^{*}.$ (3) Next, according to Inequality (2), $f^{*}f_{*}(q)\leq_{Q}q$ for any $q\in Q$, it is true, in particular, when $q=f^{*}(p)$. Therefore, we also have
$\displaystyle f^{*}f_{*}f^{*}\leq_{Q}f^{*}.$ (4) Putting Inequalities (3) and (4) together we have
$\displaystyle f^{*}f_{*}f^{*}=f^{*}.$ (5) Similarly,
$\displaystyle f_{*}f^{*}f_{*}=f_{*}.$ (6) 4. If $(f,g)$ and $(f,h)$ are Galois connections between $(P,\leq_{P})$ and $(Q,\leq_{Q})$, then $g=h$. To see this, observe that $p\leq_{P}g(q)$ iff $f(p)\leq_{Q}q$ iff $p\leq_{P}h(q)$, for any $p\in P$ and $q\in Q$. In particular, setting $p=g(q)$, we get $g(q)\leq_{P}h(q)$ since $g(q)\leq_{P}g(q)$. Similarly, $h(q)\leq_{P}g(q)$, and therefore $g=h$. By a similarly argument, if $(g,f)$ and $(h,f)$ are Galois connections between $(P,\leq_{P})$ and $(Q,\leq_{Q})$, then $g=h$. Because of this uniqueness property, in a Galois connection $f=(f^{*},f_{*})$, $f^{*}$ is called the upper adjoint of $f_{*}$ and $f_{*}$ the lower adjoint of $f^{*}$.
Examples.

The most famous example is already mentioned in the first paragraph above: let $L$ is a finitedimensional Galois extension of a field $F$, and $G:=\operatorname{Gal}(L/F)$ is the Galois group of $L$ over $F$. If we define
 a.
$P=\{K\mid K\mbox{ is a field such that }F\subseteq K\subseteq L\},$ with $\leq_{P}=\subseteq$,
 b.
$Q=\{H\mid H\mbox{ is a subgroup of }G\},$ with $\leq_{Q}=\supseteq$,
 c.
$f^{*}:P\to Q$ by $f^{*}(K)=\operatorname{Gal}(L/K)$, and
 d.
$f_{*}:Q\to P$ by $f_{*}(H)=L^{H}$, the fixed field of $H$ in $L$.
Then, by the fundamental theorem of Galois theory, $f^{*}$ and $f_{*}$ are bijections, and $(f^{*},f_{*})$ is a Galois connection between $P$ and $Q$.
 a.

Let $X$ be a topological space. Define $P$ be the set of all open subsets of $X$ and $Q$ the set of all closed subsets of $X$. Turn $P$ and $Q$ into posets with the usual settheoretic inclusion. Next, define $f^{*}:P\to Q$ by $f^{*}(U)=\overline{U}$, the closure of $U$, and $f_{*}:Q\to P$ by $f_{*}(V)=\operatorname{int}(V)$, the interior of $V$. Then $(f^{*},f_{*})$ is a Galois connection between $P$ and $Q$. Incidentally, those elements fixed by $f_{*}f^{*}$ are precisely the regular open sets of $X$, and those fixed by $f^{*}f_{*}$ are the regular closed sets.
Remark. The pair of functions in a Galois connection are order preserving as shown above. One may also define a Galois connection as a pair of maps $f^{*}:P\to Q$ and $f_{*}:Q\to P$ such that $f^{*}(p)\leq_{Q}q$ iff $f_{*}(q)\leq_{P}p$, so that the pair $f^{*},f_{*}$ are order reversing. In any case, the two definitions are equivalent in that one may go from one definition to another, (simply exchange $Q$ with $Q^{{\partial}}$, the dual of $Q$).
References
 1 T.S. Blyth, Lattices and Ordered Algebraic Structures, Springer, New York (2005).
 2 B. A. Davey, H. A. Priestley, Introduction to Lattices and Order, 2nd Edition, Cambridge (2003)
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Comments
What does this mean?
I'm having a problem with the following sentence in this article:
"The existence of a Galois connection between P and Q is the same as the existence of a Galois connection between Q and P so the definition is unambiguous."
Is this right? Does it use the same maps? f^* and f_*, just interchanged?
I can't seem to prove this.
Perhaps I am just misundertanding.
Mike Zabrocki
Re: What does this mean?
Yes, use the duals of the orderings (you can define $a {\le}^{\prime} b$ if and only if $b {\le} a$) on each of the posets, and switch the maps $f^*$ and $f_*$.
Does this answer your question?
Chi
Re: What does this mean?
I guess. But what this says to me is
"The existence of a Galois connection between P and Q is the same as the existence of a Galois connection between Q^D and P^D."
where P^D means the dual posets. The way I am reading the sentence in
the article says that there is a Galois connection between P and Q iff there
is a Galois connection between Q and P. Is this right? Thanks.
Mike Zabrocki
Re: What does this mean?
I clarified the wording in the entry a little. Does this help?
Chi
Re: What does this mean?
I think that it is clear now.
BTW, you are missing an open parenthesis at "Q, \leq_Q')".
The confusion for me came up because I identify P with
the poset (P, \le_P) which you were not doing explictly. So I interpret
"a Galois connection between P and Q" to mean
"a Galois connection between (P, \le_P) and (Q, \le_Q)"
When you say there is a
"We denote a Galois connection between P and Q" you mean,
"there exists orders \le_P and \le_Q such that there
is a Galois connection between (P, \le_P) and (Q, \le_Q)."
right?
Mike
Re: What does this mean?
That's right. And thanks for the comment.
Chi