# Special Galois connections

Let $(f;g)$ is a Galois connection between two posets.

Consider the special case $g\circ f = \operatorname{id}_{\operatorname{dom} f}$.

What can be said about this special case of Galois connections? Does there exist a special term for this kind of Galois connections?

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I think its called galois insertion – Belgi Dec 17 '12 at 22:24
Search by "A primer on Galois connections". I think you can download it for free. – Makoto Kato Dec 17 '12 at 22:48
@MakotoKato: Thanks but there are no word "insertion" in "A primer on Galois connections". – porton Dec 17 '12 at 22:53
@porton I don't think the term "Galois insertion" is a standard one. Anyway, whatever they call it, I think the point is what can be said about it. Look for (co)reflections and interior(or closure) connections in that book. – Makoto Kato Dec 17 '12 at 22:57
This set of slides seems to use "Galois injection" and "Galois insertion/surjection" as different. Not sure which is which, however. See slide 21 in cs.au.dk/~jmi/AbsInt/week3.pdf – Thomas Andrews Dec 18 '12 at 14:53

From the definitions on slide 21 in this PDF, the terminology seems to call this a "Galois injection."

The dual concept, where $f\circ g=1_B$, is called a "Galois surjection" or "Galois insertion." (The slides use $\alpha$ and $\gamma$, but I ran back a few slides to the definitions to make sure, and $(\alpha,\gamma)$ corresponds to $(f,g)$.

I have no idea if this is standard. Googling "Galois injection" seems to find this definition in a number of places. All of the papers I have found that use this term are in the area of theoretical computer science and logic, so you might have to be explicit outside that area.

There is a way to classify all such "Galois injections," up to isomporphism, in terms of a single operator $h$ on the right poset of the connection: $h:B\to B$.

Let $A,B$ be your two posets, with $(f,g)$ a Galois connection: $f:A\to B$ and $g:B\to A$ and $f(a)\leq b$ iff $a\leq g(b)$.

If $g(f(a))=a$, then we define $h:B\to B$ by $h(b)=f(g(b))$. Then you can show that

• $b\leq b'$ implies $h(b)\leq h(b')$ (Since: $f,g$ are monotonic)
• $h(b)\leq b$ (Since: $g(b)\leq g(b)$ implies $h(b)=f(g(b))\leq b$)
• $h(h(b))=h(b)$ (Since: $h\circ h = f\circ g\circ f\circ g= f\circ 1_A\circ g = h$)

So $h$ is a closure operator of some sort on $B^{op}$.

On the other hand, if you have an operator $h$ which has these properties on $B$, define $A$ to be the image of $h$, and define $f:A\to B$ to be the natural inclusion and $g:B\to A$ to be $g=h$. Then $(f,g)$ is a Galois connection with $g\circ f = 1_A$.

Basically, such an $h$ determines $A$ up to isomorphism.

So any such Galois connection, up to isomorphism, comes from a closure operator on $B^{op}$. According Wikipedia, this is variously called a "kernel operator," an "interior operator" or a "dual closure operator."

[Aside: Note that if you had the opposite starting condition, that $f\circ g = 1_B$, then the pair would, up to isomorphism, arise from a closure operator on $A$.]

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