The relation between the left- and right- adjoints participating in Galois connections w.r.t. a common functor

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Consider a monotone Galois connection $(F,G_r)$. Suppose that $(G_l,F)$ is another monotone Galois connection, where $F$ is the same functor in both connections. Has the relation between $G_l$ and $G_r$ been studied? Where can I read more about it?

abstract-algebra reference-request category-theory order-theory galois-connections

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edited Jul 28 at 7:52

asked Jul 28 at 7:44

Evan Aad

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Consider a monotone Galois connection $(F,G_r)$. Suppose that $(G_l,F)$ is another monotone Galois connection, where $F$ is the same functor in both connections. Has the relation between $G_l$ and $G_r$ been studied? Where can I read more about it?

abstract-algebra reference-request category-theory order-theory galois-connections

share|cite|improve this question

edited Jul 28 at 7:52

asked Jul 28 at 7:44

Evan Aad

5,36011748

add a comment | 

up vote
3
down vote

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up vote
3
down vote

favorite

Consider a monotone Galois connection $(F,G_r)$. Suppose that $(G_l,F)$ is another monotone Galois connection, where $F$ is the same functor in both connections. Has the relation between $G_l$ and $G_r$ been studied? Where can I read more about it?

abstract-algebra reference-request category-theory order-theory galois-connections

share|cite|improve this question

edited Jul 28 at 7:52

asked Jul 28 at 7:44

Evan Aad

5,36011748

Consider a monotone Galois connection $(F,G_r)$. Suppose that $(G_l,F)$ is another monotone Galois connection, where $F$ is the same functor in both connections. Has the relation between $G_l$ and $G_r$ been studied? Where can I read more about it?

abstract-algebra reference-request category-theory order-theory galois-connections

share|cite|improve this question

edited Jul 28 at 7:52

asked Jul 28 at 7:44

Evan Aad

5,36011748

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 7:52

edited Jul 28 at 7:52

edited Jul 28 at 7:52

asked Jul 28 at 7:44

Evan Aad

5,36011748

asked Jul 28 at 7:44

Evan Aad

5,36011748

asked Jul 28 at 7:44

Evan Aad

5,36011748

5,36011748

add a comment | 

add a comment | 

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As $F(x)leq F(x)leq F(x)$, we have $G_l(F(x)) leq x leq G_r(F(x))$ for any $x$. However I doubt we can say much more, because of the two following examples:

• If $L$ has a minimal and maximal element, then $Lto ast$ ($ast$ is the poset with only one element) has both an upper and lower adjoint. The upper one maps the unique element of $ast$ to the maximal element and the lower one to the minimal element. In that example, $G_l(F(x))$ is always the farthest of $G_r(F(x))$ possible. Actually we have $G_lleq G_r$.




• If $L$ as binary meet and joins, $Lto Ltimes L,,xmapsto (x,x)$ has both a lower and upper adjoint. The lower maps $(x,y)$ to the join $xvee y$ and the upper maps $(x,y)$ to the meet $xwedge y$. In this example, $G_l(F(x))=G_r(F(x))$ for any $x$. But for any $x,y$, one has $xwedge y leq x leq xvee y$ so in that example $G_rleq G_l$.

(If you try harder I'm pretty sure you can find examples when $G_l$ and $G_r$ are not comparable, globally but also even pointwise.)

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answered Jul 29 at 8:21

Pece

7,92211040

• 
  
  
  

  
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  (A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
  – amrsa
  Jul 29 at 9:20
  

  
  
  
  

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1 Answer
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1 Answer
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up vote
3
down vote

As $F(x)leq F(x)leq F(x)$, we have $G_l(F(x)) leq x leq G_r(F(x))$ for any $x$. However I doubt we can say much more, because of the two following examples:

• If $L$ has a minimal and maximal element, then $Lto ast$ ($ast$ is the poset with only one element) has both an upper and lower adjoint. The upper one maps the unique element of $ast$ to the maximal element and the lower one to the minimal element. In that example, $G_l(F(x))$ is always the farthest of $G_r(F(x))$ possible. Actually we have $G_lleq G_r$.




• If $L$ as binary meet and joins, $Lto Ltimes L,,xmapsto (x,x)$ has both a lower and upper adjoint. The lower maps $(x,y)$ to the join $xvee y$ and the upper maps $(x,y)$ to the meet $xwedge y$. In this example, $G_l(F(x))=G_r(F(x))$ for any $x$. But for any $x,y$, one has $xwedge y leq x leq xvee y$ so in that example $G_rleq G_l$.

(If you try harder I'm pretty sure you can find examples when $G_l$ and $G_r$ are not comparable, globally but also even pointwise.)

share|cite|improve this answer

answered Jul 29 at 8:21

Pece

7,92211040

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
  – amrsa
  Jul 29 at 9:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

As $F(x)leq F(x)leq F(x)$, we have $G_l(F(x)) leq x leq G_r(F(x))$ for any $x$. However I doubt we can say much more, because of the two following examples:

• If $L$ has a minimal and maximal element, then $Lto ast$ ($ast$ is the poset with only one element) has both an upper and lower adjoint. The upper one maps the unique element of $ast$ to the maximal element and the lower one to the minimal element. In that example, $G_l(F(x))$ is always the farthest of $G_r(F(x))$ possible. Actually we have $G_lleq G_r$.




• If $L$ as binary meet and joins, $Lto Ltimes L,,xmapsto (x,x)$ has both a lower and upper adjoint. The lower maps $(x,y)$ to the join $xvee y$ and the upper maps $(x,y)$ to the meet $xwedge y$. In this example, $G_l(F(x))=G_r(F(x))$ for any $x$. But for any $x,y$, one has $xwedge y leq x leq xvee y$ so in that example $G_rleq G_l$.

(If you try harder I'm pretty sure you can find examples when $G_l$ and $G_r$ are not comparable, globally but also even pointwise.)

share|cite|improve this answer

answered Jul 29 at 8:21

Pece

7,92211040

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
  – amrsa
  Jul 29 at 9:20
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

As $F(x)leq F(x)leq F(x)$, we have $G_l(F(x)) leq x leq G_r(F(x))$ for any $x$. However I doubt we can say much more, because of the two following examples:

• If $L$ has a minimal and maximal element, then $Lto ast$ ($ast$ is the poset with only one element) has both an upper and lower adjoint. The upper one maps the unique element of $ast$ to the maximal element and the lower one to the minimal element. In that example, $G_l(F(x))$ is always the farthest of $G_r(F(x))$ possible. Actually we have $G_lleq G_r$.




• If $L$ as binary meet and joins, $Lto Ltimes L,,xmapsto (x,x)$ has both a lower and upper adjoint. The lower maps $(x,y)$ to the join $xvee y$ and the upper maps $(x,y)$ to the meet $xwedge y$. In this example, $G_l(F(x))=G_r(F(x))$ for any $x$. But for any $x,y$, one has $xwedge y leq x leq xvee y$ so in that example $G_rleq G_l$.

(If you try harder I'm pretty sure you can find examples when $G_l$ and $G_r$ are not comparable, globally but also even pointwise.)

share|cite|improve this answer

answered Jul 29 at 8:21

Pece

7,92211040

As $F(x)leq F(x)leq F(x)$, we have $G_l(F(x)) leq x leq G_r(F(x))$ for any $x$. However I doubt we can say much more, because of the two following examples:

• If $L$ has a minimal and maximal element, then $Lto ast$ ($ast$ is the poset with only one element) has both an upper and lower adjoint. The upper one maps the unique element of $ast$ to the maximal element and the lower one to the minimal element. In that example, $G_l(F(x))$ is always the farthest of $G_r(F(x))$ possible. Actually we have $G_lleq G_r$.




• If $L$ as binary meet and joins, $Lto Ltimes L,,xmapsto (x,x)$ has both a lower and upper adjoint. The lower maps $(x,y)$ to the join $xvee y$ and the upper maps $(x,y)$ to the meet $xwedge y$. In this example, $G_l(F(x))=G_r(F(x))$ for any $x$. But for any $x,y$, one has $xwedge y leq x leq xvee y$ so in that example $G_rleq G_l$.

(If you try harder I'm pretty sure you can find examples when $G_l$ and $G_r$ are not comparable, globally but also even pointwise.)

share|cite|improve this answer

answered Jul 29 at 8:21

Pece

7,92211040

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 8:21

Pece

7,92211040

answered Jul 29 at 8:21

Pece

7,92211040

answered Jul 29 at 8:21

Pece

7,92211040

7,92211040

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
  – amrsa
  Jul 29 at 9:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
  – amrsa
  Jul 29 at 9:20
  

  
  
  
  

1

1

(A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
– amrsa
Jul 29 at 9:20

(A little bit more generally) it also follows from your formula in the first paragraph that if $F$ is onto, then $G_lleq G_r$. Another valid formula is that $F(G_r(y))leq yleq F(G_l(y))$, and if $F$ is one to one, then $G_rleq G_l$. So if $F$ is a bijection, $G_l=G_r$; otherwise these ideas might help to find counter-examples.
– amrsa
Jul 29 at 9:20

add a comment | 

 

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