Philosophy or meaning of adjoint functors
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
add a comment |
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
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
add a comment |
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
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
category-theory
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
add a comment |
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
add a comment |
2 Answers
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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.
add a comment |
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.
add a comment |
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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.
add a comment |
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.
add a comment |
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.
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.
edited Nov 28 at 21:32
answered Jan 3 '12 at 21:26
Giorgio Mossa
13.8k11749
13.8k11749
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Apr 27 '12 at 20:32
Ronnie Brown
11.9k12938
11.9k12938
add a comment |
add a comment |
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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