Philosophy or meaning of adjoint functors












13














I have a question about the definition of the (left) adjoint of a functor.I am trying to understand the philosophy and reason of the definition of adjoint functors.



If I understand correctly the situation is as follows: Suppose one has the category $mathcal{C}$ with an object $A$ and a category $mathcal{D}$ with an object $B$ and functors $F: mathcal{C} to mathcal{D}$ and a functor $G: mathcal{D} to mathcal{C}$.



Now one can wonder how to make kind of cross morphism from $A$ to $B$. The two natural ways would be to push $A$ to $mathcal{D}$ using $F$ and take a morphism $phi$ there. But there is another equally natural way to this. Push $B$ to $mathcal{C}$ by $G$ and take a morphism $psi$ between $A$ and $GB$.



The natural equivalence then translates approach one in to approach two and back by making a correspondence between $phi$ and $psi$.



To make it a bit sensible one demands that the functors $F$ and $G$ are not arbitrary, but somehow linked together in the sense that linking $A_1 to A$ with $B to B_1$ is done on essentially( up to correspendance by a unit) the same way by pushing $A$ to $mathcal{D}$ or pushing $B$ to $mathcal{C}$.



Is this indeed the philosophy behind adjoint functors?










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  • Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
    – MrOperator
    Jan 3 '12 at 14:22






  • 2




    You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
    – Zhen Lin
    Jan 3 '12 at 14:24










  • Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
    – Qiaochu Yuan
    Apr 27 '12 at 20:40
















13














I have a question about the definition of the (left) adjoint of a functor.I am trying to understand the philosophy and reason of the definition of adjoint functors.



If I understand correctly the situation is as follows: Suppose one has the category $mathcal{C}$ with an object $A$ and a category $mathcal{D}$ with an object $B$ and functors $F: mathcal{C} to mathcal{D}$ and a functor $G: mathcal{D} to mathcal{C}$.



Now one can wonder how to make kind of cross morphism from $A$ to $B$. The two natural ways would be to push $A$ to $mathcal{D}$ using $F$ and take a morphism $phi$ there. But there is another equally natural way to this. Push $B$ to $mathcal{C}$ by $G$ and take a morphism $psi$ between $A$ and $GB$.



The natural equivalence then translates approach one in to approach two and back by making a correspondence between $phi$ and $psi$.



To make it a bit sensible one demands that the functors $F$ and $G$ are not arbitrary, but somehow linked together in the sense that linking $A_1 to A$ with $B to B_1$ is done on essentially( up to correspendance by a unit) the same way by pushing $A$ to $mathcal{D}$ or pushing $B$ to $mathcal{C}$.



Is this indeed the philosophy behind adjoint functors?










share|cite|improve this question
























  • Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
    – MrOperator
    Jan 3 '12 at 14:22






  • 2




    You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
    – Zhen Lin
    Jan 3 '12 at 14:24










  • Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
    – Qiaochu Yuan
    Apr 27 '12 at 20:40














13












13








13


10





I have a question about the definition of the (left) adjoint of a functor.I am trying to understand the philosophy and reason of the definition of adjoint functors.



If I understand correctly the situation is as follows: Suppose one has the category $mathcal{C}$ with an object $A$ and a category $mathcal{D}$ with an object $B$ and functors $F: mathcal{C} to mathcal{D}$ and a functor $G: mathcal{D} to mathcal{C}$.



Now one can wonder how to make kind of cross morphism from $A$ to $B$. The two natural ways would be to push $A$ to $mathcal{D}$ using $F$ and take a morphism $phi$ there. But there is another equally natural way to this. Push $B$ to $mathcal{C}$ by $G$ and take a morphism $psi$ between $A$ and $GB$.



The natural equivalence then translates approach one in to approach two and back by making a correspondence between $phi$ and $psi$.



To make it a bit sensible one demands that the functors $F$ and $G$ are not arbitrary, but somehow linked together in the sense that linking $A_1 to A$ with $B to B_1$ is done on essentially( up to correspendance by a unit) the same way by pushing $A$ to $mathcal{D}$ or pushing $B$ to $mathcal{C}$.



Is this indeed the philosophy behind adjoint functors?










share|cite|improve this question















I have a question about the definition of the (left) adjoint of a functor.I am trying to understand the philosophy and reason of the definition of adjoint functors.



If I understand correctly the situation is as follows: Suppose one has the category $mathcal{C}$ with an object $A$ and a category $mathcal{D}$ with an object $B$ and functors $F: mathcal{C} to mathcal{D}$ and a functor $G: mathcal{D} to mathcal{C}$.



Now one can wonder how to make kind of cross morphism from $A$ to $B$. The two natural ways would be to push $A$ to $mathcal{D}$ using $F$ and take a morphism $phi$ there. But there is another equally natural way to this. Push $B$ to $mathcal{C}$ by $G$ and take a morphism $psi$ between $A$ and $GB$.



The natural equivalence then translates approach one in to approach two and back by making a correspondence between $phi$ and $psi$.



To make it a bit sensible one demands that the functors $F$ and $G$ are not arbitrary, but somehow linked together in the sense that linking $A_1 to A$ with $B to B_1$ is done on essentially( up to correspendance by a unit) the same way by pushing $A$ to $mathcal{D}$ or pushing $B$ to $mathcal{C}$.



Is this indeed the philosophy behind adjoint functors?







category-theory






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edited Jan 3 '12 at 15:15









Martin Sleziak

44.6k7115270




44.6k7115270










asked Jan 3 '12 at 14:17









MrOperator

1258




1258












  • Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
    – MrOperator
    Jan 3 '12 at 14:22






  • 2




    You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
    – Zhen Lin
    Jan 3 '12 at 14:24










  • Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
    – Qiaochu Yuan
    Apr 27 '12 at 20:40


















  • Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
    – MrOperator
    Jan 3 '12 at 14:22






  • 2




    You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
    – Zhen Lin
    Jan 3 '12 at 14:24










  • Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
    – Qiaochu Yuan
    Apr 27 '12 at 20:40
















Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
– MrOperator
Jan 3 '12 at 14:22




Thanks @usero2139 for correcting my spelling mistakes. A bit of my native langue Dutch that was seeping trough.
– MrOperator
Jan 3 '12 at 14:22




2




2




You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
– Zhen Lin
Jan 3 '12 at 14:24




You could view it that way if you like. But the usual motivation behind adjunctions is that they capture the idea of "most universal solution to a problem". See here.
– Zhen Lin
Jan 3 '12 at 14:24












Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
– Qiaochu Yuan
Apr 27 '12 at 20:40




Related: math.stackexchange.com/questions/20364/… , math.stackexchange.com/questions/25455/… .
– Qiaochu Yuan
Apr 27 '12 at 20:40










2 Answers
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11














As the link of Zhen Lin says adjoint functors deal with universals.



Usually I think to adjointness situations in two different ways: as conditions of global existence of universals (related to the comment of Zhen Lin above) and as a conditions of global representability.
(By Yoneda lemma these things are essentially the same but they are different, from a psychological point of view).



Let's see the first: the global existence of universals.



The following fact holds: a functor $mathcal G colon mathbf A to mathbf X$ has a left adjoint if and only if for each $x in mathbf X$ exists a morphisms $eta_x colon x to mathcal G(mathcal F(x))$, where $mathcal F(x) in mathbf A$, which is universal from $x$ to $mathcal G$ [i.e. for every other object $a in mathbf A$ and morphism $tau colon x to mathcal G(a)$ exists a unique $h colon mathcal F(x) to a$ such that $mathcal tau=G(h)circ eta_x$].



In this case it can be show that the induced function $mathcal F$ from the set of objects of $mathbf X$ to the the set of objects of $mathbf A$ can be extended to a functor $mathcal F colon mathbf X to mathbf A$ (unique up to natural isomorphism). This condition of adjointness, the existence of one universal morphism from every object in $mathbf A$, justifies the phrase global existance of universals.



There's also a dual result that says that a functor $mathcal F colon mathbf X to mathbf A$ has a right adjoint $mathcal G colon mathbf A to mathbf X$ if and only if exists a family of morphisms $epsilon_a colon mathcal F(mathcal G(a)) to a$ each one being universal from $mathcal F$ to $a in mathbf A$, i.e. for each other morphisms $sigma colon mathcal F(x) to a$ exists a unique $k colon x to mathcal G(a)$ such that $sigma = epsilon_a circ mathcal F(k)$.



These facts can be rephrased in term of representable functors.
So the existance of the family of universal morphisms $langle eta_x colon x to mathcal G(mathcal F(x))rangle_{x in mathbf X}$ is equivalent via Yoneda's lemma to the representability, meaning that the functor $mathcal G$ has a left adjoint if each functor $mathbf X(x,mathcal G(-)) colon mathbf A to mathbf {Set}$ is representable [that's because each universal morphism $eta_x colon x to mathcal G(mathcal F(x))$ gives a natural isomorphism $mathbf X(x,mathcal G(-)) cong mathbf A(mathcal F(x),-)$]. Similarly the existance of the family of universal morphisms $langle epsilon_a colon mathcal F(mathcal G(a)) to arangle_{a in mathbf A}$ implies that each of the functors $mathbf A(mathcal F(-),a)$ is representable [because the universal morphism $epsilon_a colon mathcal F(mathcal G(a)) to a$ give a natural isomorphism $mathbf A(mathcal F(-),a) cong mathbf X(-,mathcal G(a))$].



Representability says that each fact about morphisms of type $x to mathcal G(a)$ is equivalent to a fact about morphisms of type $mathcal F(x) to a$. Simplifying this means that we can traduce facts/problems in the category $mathbf X$ in other facts/problems in the category $mathbf A$, and the opposite. This enables us to transport problems in the most convenient setting.



I hope this answer was useful.






share|cite|improve this answer































    9














    One should also mention the utility of adjoint functors for explicit computation, which happens because left adjoints preserve colimits and right adjoints preserve limits.



    Consider for example the forgetful functor $Ob$ from groupoids to sets. This has a right adjoint, which to a set $X$ assigns what is called sometimes the indiscrete or coarse groupoid, say $Ind(X)$, on $X$, which has exactly one morphism between any two objects.



    Now suppose we want to compute $L=$colim$G_i$ of some diagram of groupoids. Because of the above, we know that $Ob(L)=$colim$Ob(G_i)$. That is a start.






    share|cite|improve this answer





















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      11














      As the link of Zhen Lin says adjoint functors deal with universals.



      Usually I think to adjointness situations in two different ways: as conditions of global existence of universals (related to the comment of Zhen Lin above) and as a conditions of global representability.
      (By Yoneda lemma these things are essentially the same but they are different, from a psychological point of view).



      Let's see the first: the global existence of universals.



      The following fact holds: a functor $mathcal G colon mathbf A to mathbf X$ has a left adjoint if and only if for each $x in mathbf X$ exists a morphisms $eta_x colon x to mathcal G(mathcal F(x))$, where $mathcal F(x) in mathbf A$, which is universal from $x$ to $mathcal G$ [i.e. for every other object $a in mathbf A$ and morphism $tau colon x to mathcal G(a)$ exists a unique $h colon mathcal F(x) to a$ such that $mathcal tau=G(h)circ eta_x$].



      In this case it can be show that the induced function $mathcal F$ from the set of objects of $mathbf X$ to the the set of objects of $mathbf A$ can be extended to a functor $mathcal F colon mathbf X to mathbf A$ (unique up to natural isomorphism). This condition of adjointness, the existence of one universal morphism from every object in $mathbf A$, justifies the phrase global existance of universals.



      There's also a dual result that says that a functor $mathcal F colon mathbf X to mathbf A$ has a right adjoint $mathcal G colon mathbf A to mathbf X$ if and only if exists a family of morphisms $epsilon_a colon mathcal F(mathcal G(a)) to a$ each one being universal from $mathcal F$ to $a in mathbf A$, i.e. for each other morphisms $sigma colon mathcal F(x) to a$ exists a unique $k colon x to mathcal G(a)$ such that $sigma = epsilon_a circ mathcal F(k)$.



      These facts can be rephrased in term of representable functors.
      So the existance of the family of universal morphisms $langle eta_x colon x to mathcal G(mathcal F(x))rangle_{x in mathbf X}$ is equivalent via Yoneda's lemma to the representability, meaning that the functor $mathcal G$ has a left adjoint if each functor $mathbf X(x,mathcal G(-)) colon mathbf A to mathbf {Set}$ is representable [that's because each universal morphism $eta_x colon x to mathcal G(mathcal F(x))$ gives a natural isomorphism $mathbf X(x,mathcal G(-)) cong mathbf A(mathcal F(x),-)$]. Similarly the existance of the family of universal morphisms $langle epsilon_a colon mathcal F(mathcal G(a)) to arangle_{a in mathbf A}$ implies that each of the functors $mathbf A(mathcal F(-),a)$ is representable [because the universal morphism $epsilon_a colon mathcal F(mathcal G(a)) to a$ give a natural isomorphism $mathbf A(mathcal F(-),a) cong mathbf X(-,mathcal G(a))$].



      Representability says that each fact about morphisms of type $x to mathcal G(a)$ is equivalent to a fact about morphisms of type $mathcal F(x) to a$. Simplifying this means that we can traduce facts/problems in the category $mathbf X$ in other facts/problems in the category $mathbf A$, and the opposite. This enables us to transport problems in the most convenient setting.



      I hope this answer was useful.






      share|cite|improve this answer




























        11














        As the link of Zhen Lin says adjoint functors deal with universals.



        Usually I think to adjointness situations in two different ways: as conditions of global existence of universals (related to the comment of Zhen Lin above) and as a conditions of global representability.
        (By Yoneda lemma these things are essentially the same but they are different, from a psychological point of view).



        Let's see the first: the global existence of universals.



        The following fact holds: a functor $mathcal G colon mathbf A to mathbf X$ has a left adjoint if and only if for each $x in mathbf X$ exists a morphisms $eta_x colon x to mathcal G(mathcal F(x))$, where $mathcal F(x) in mathbf A$, which is universal from $x$ to $mathcal G$ [i.e. for every other object $a in mathbf A$ and morphism $tau colon x to mathcal G(a)$ exists a unique $h colon mathcal F(x) to a$ such that $mathcal tau=G(h)circ eta_x$].



        In this case it can be show that the induced function $mathcal F$ from the set of objects of $mathbf X$ to the the set of objects of $mathbf A$ can be extended to a functor $mathcal F colon mathbf X to mathbf A$ (unique up to natural isomorphism). This condition of adjointness, the existence of one universal morphism from every object in $mathbf A$, justifies the phrase global existance of universals.



        There's also a dual result that says that a functor $mathcal F colon mathbf X to mathbf A$ has a right adjoint $mathcal G colon mathbf A to mathbf X$ if and only if exists a family of morphisms $epsilon_a colon mathcal F(mathcal G(a)) to a$ each one being universal from $mathcal F$ to $a in mathbf A$, i.e. for each other morphisms $sigma colon mathcal F(x) to a$ exists a unique $k colon x to mathcal G(a)$ such that $sigma = epsilon_a circ mathcal F(k)$.



        These facts can be rephrased in term of representable functors.
        So the existance of the family of universal morphisms $langle eta_x colon x to mathcal G(mathcal F(x))rangle_{x in mathbf X}$ is equivalent via Yoneda's lemma to the representability, meaning that the functor $mathcal G$ has a left adjoint if each functor $mathbf X(x,mathcal G(-)) colon mathbf A to mathbf {Set}$ is representable [that's because each universal morphism $eta_x colon x to mathcal G(mathcal F(x))$ gives a natural isomorphism $mathbf X(x,mathcal G(-)) cong mathbf A(mathcal F(x),-)$]. Similarly the existance of the family of universal morphisms $langle epsilon_a colon mathcal F(mathcal G(a)) to arangle_{a in mathbf A}$ implies that each of the functors $mathbf A(mathcal F(-),a)$ is representable [because the universal morphism $epsilon_a colon mathcal F(mathcal G(a)) to a$ give a natural isomorphism $mathbf A(mathcal F(-),a) cong mathbf X(-,mathcal G(a))$].



        Representability says that each fact about morphisms of type $x to mathcal G(a)$ is equivalent to a fact about morphisms of type $mathcal F(x) to a$. Simplifying this means that we can traduce facts/problems in the category $mathbf X$ in other facts/problems in the category $mathbf A$, and the opposite. This enables us to transport problems in the most convenient setting.



        I hope this answer was useful.






        share|cite|improve this answer


























          11












          11








          11






          As the link of Zhen Lin says adjoint functors deal with universals.



          Usually I think to adjointness situations in two different ways: as conditions of global existence of universals (related to the comment of Zhen Lin above) and as a conditions of global representability.
          (By Yoneda lemma these things are essentially the same but they are different, from a psychological point of view).



          Let's see the first: the global existence of universals.



          The following fact holds: a functor $mathcal G colon mathbf A to mathbf X$ has a left adjoint if and only if for each $x in mathbf X$ exists a morphisms $eta_x colon x to mathcal G(mathcal F(x))$, where $mathcal F(x) in mathbf A$, which is universal from $x$ to $mathcal G$ [i.e. for every other object $a in mathbf A$ and morphism $tau colon x to mathcal G(a)$ exists a unique $h colon mathcal F(x) to a$ such that $mathcal tau=G(h)circ eta_x$].



          In this case it can be show that the induced function $mathcal F$ from the set of objects of $mathbf X$ to the the set of objects of $mathbf A$ can be extended to a functor $mathcal F colon mathbf X to mathbf A$ (unique up to natural isomorphism). This condition of adjointness, the existence of one universal morphism from every object in $mathbf A$, justifies the phrase global existance of universals.



          There's also a dual result that says that a functor $mathcal F colon mathbf X to mathbf A$ has a right adjoint $mathcal G colon mathbf A to mathbf X$ if and only if exists a family of morphisms $epsilon_a colon mathcal F(mathcal G(a)) to a$ each one being universal from $mathcal F$ to $a in mathbf A$, i.e. for each other morphisms $sigma colon mathcal F(x) to a$ exists a unique $k colon x to mathcal G(a)$ such that $sigma = epsilon_a circ mathcal F(k)$.



          These facts can be rephrased in term of representable functors.
          So the existance of the family of universal morphisms $langle eta_x colon x to mathcal G(mathcal F(x))rangle_{x in mathbf X}$ is equivalent via Yoneda's lemma to the representability, meaning that the functor $mathcal G$ has a left adjoint if each functor $mathbf X(x,mathcal G(-)) colon mathbf A to mathbf {Set}$ is representable [that's because each universal morphism $eta_x colon x to mathcal G(mathcal F(x))$ gives a natural isomorphism $mathbf X(x,mathcal G(-)) cong mathbf A(mathcal F(x),-)$]. Similarly the existance of the family of universal morphisms $langle epsilon_a colon mathcal F(mathcal G(a)) to arangle_{a in mathbf A}$ implies that each of the functors $mathbf A(mathcal F(-),a)$ is representable [because the universal morphism $epsilon_a colon mathcal F(mathcal G(a)) to a$ give a natural isomorphism $mathbf A(mathcal F(-),a) cong mathbf X(-,mathcal G(a))$].



          Representability says that each fact about morphisms of type $x to mathcal G(a)$ is equivalent to a fact about morphisms of type $mathcal F(x) to a$. Simplifying this means that we can traduce facts/problems in the category $mathbf X$ in other facts/problems in the category $mathbf A$, and the opposite. This enables us to transport problems in the most convenient setting.



          I hope this answer was useful.






          share|cite|improve this answer














          As the link of Zhen Lin says adjoint functors deal with universals.



          Usually I think to adjointness situations in two different ways: as conditions of global existence of universals (related to the comment of Zhen Lin above) and as a conditions of global representability.
          (By Yoneda lemma these things are essentially the same but they are different, from a psychological point of view).



          Let's see the first: the global existence of universals.



          The following fact holds: a functor $mathcal G colon mathbf A to mathbf X$ has a left adjoint if and only if for each $x in mathbf X$ exists a morphisms $eta_x colon x to mathcal G(mathcal F(x))$, where $mathcal F(x) in mathbf A$, which is universal from $x$ to $mathcal G$ [i.e. for every other object $a in mathbf A$ and morphism $tau colon x to mathcal G(a)$ exists a unique $h colon mathcal F(x) to a$ such that $mathcal tau=G(h)circ eta_x$].



          In this case it can be show that the induced function $mathcal F$ from the set of objects of $mathbf X$ to the the set of objects of $mathbf A$ can be extended to a functor $mathcal F colon mathbf X to mathbf A$ (unique up to natural isomorphism). This condition of adjointness, the existence of one universal morphism from every object in $mathbf A$, justifies the phrase global existance of universals.



          There's also a dual result that says that a functor $mathcal F colon mathbf X to mathbf A$ has a right adjoint $mathcal G colon mathbf A to mathbf X$ if and only if exists a family of morphisms $epsilon_a colon mathcal F(mathcal G(a)) to a$ each one being universal from $mathcal F$ to $a in mathbf A$, i.e. for each other morphisms $sigma colon mathcal F(x) to a$ exists a unique $k colon x to mathcal G(a)$ such that $sigma = epsilon_a circ mathcal F(k)$.



          These facts can be rephrased in term of representable functors.
          So the existance of the family of universal morphisms $langle eta_x colon x to mathcal G(mathcal F(x))rangle_{x in mathbf X}$ is equivalent via Yoneda's lemma to the representability, meaning that the functor $mathcal G$ has a left adjoint if each functor $mathbf X(x,mathcal G(-)) colon mathbf A to mathbf {Set}$ is representable [that's because each universal morphism $eta_x colon x to mathcal G(mathcal F(x))$ gives a natural isomorphism $mathbf X(x,mathcal G(-)) cong mathbf A(mathcal F(x),-)$]. Similarly the existance of the family of universal morphisms $langle epsilon_a colon mathcal F(mathcal G(a)) to arangle_{a in mathbf A}$ implies that each of the functors $mathbf A(mathcal F(-),a)$ is representable [because the universal morphism $epsilon_a colon mathcal F(mathcal G(a)) to a$ give a natural isomorphism $mathbf A(mathcal F(-),a) cong mathbf X(-,mathcal G(a))$].



          Representability says that each fact about morphisms of type $x to mathcal G(a)$ is equivalent to a fact about morphisms of type $mathcal F(x) to a$. Simplifying this means that we can traduce facts/problems in the category $mathbf X$ in other facts/problems in the category $mathbf A$, and the opposite. This enables us to transport problems in the most convenient setting.



          I hope this answer was useful.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 21:32

























          answered Jan 3 '12 at 21:26









          Giorgio Mossa

          13.8k11749




          13.8k11749























              9














              One should also mention the utility of adjoint functors for explicit computation, which happens because left adjoints preserve colimits and right adjoints preserve limits.



              Consider for example the forgetful functor $Ob$ from groupoids to sets. This has a right adjoint, which to a set $X$ assigns what is called sometimes the indiscrete or coarse groupoid, say $Ind(X)$, on $X$, which has exactly one morphism between any two objects.



              Now suppose we want to compute $L=$colim$G_i$ of some diagram of groupoids. Because of the above, we know that $Ob(L)=$colim$Ob(G_i)$. That is a start.






              share|cite|improve this answer


























                9














                One should also mention the utility of adjoint functors for explicit computation, which happens because left adjoints preserve colimits and right adjoints preserve limits.



                Consider for example the forgetful functor $Ob$ from groupoids to sets. This has a right adjoint, which to a set $X$ assigns what is called sometimes the indiscrete or coarse groupoid, say $Ind(X)$, on $X$, which has exactly one morphism between any two objects.



                Now suppose we want to compute $L=$colim$G_i$ of some diagram of groupoids. Because of the above, we know that $Ob(L)=$colim$Ob(G_i)$. That is a start.






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                  One should also mention the utility of adjoint functors for explicit computation, which happens because left adjoints preserve colimits and right adjoints preserve limits.



                  Consider for example the forgetful functor $Ob$ from groupoids to sets. This has a right adjoint, which to a set $X$ assigns what is called sometimes the indiscrete or coarse groupoid, say $Ind(X)$, on $X$, which has exactly one morphism between any two objects.



                  Now suppose we want to compute $L=$colim$G_i$ of some diagram of groupoids. Because of the above, we know that $Ob(L)=$colim$Ob(G_i)$. That is a start.






                  share|cite|improve this answer












                  One should also mention the utility of adjoint functors for explicit computation, which happens because left adjoints preserve colimits and right adjoints preserve limits.



                  Consider for example the forgetful functor $Ob$ from groupoids to sets. This has a right adjoint, which to a set $X$ assigns what is called sometimes the indiscrete or coarse groupoid, say $Ind(X)$, on $X$, which has exactly one morphism between any two objects.



                  Now suppose we want to compute $L=$colim$G_i$ of some diagram of groupoids. Because of the above, we know that $Ob(L)=$colim$Ob(G_i)$. That is a start.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 27 '12 at 20:32









                  Ronnie Brown

                  11.9k12938




                  11.9k12938






























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