Do adjunctions map to every object in a category?
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I am trying to show that right adjoints preserve limits. So suppose we have $F: C rightarrow D$ and $G: D rightarrow C$ be 2 adjoint functors with $G$ the right adjoint. So suppose we have a limit $(N, phi)$ of the diagram $F: C rightarrow D$ where $phi$ is indexed by objects $X$ in $C$ with $phi_X: N rightarrow F(X).$ Now I want to show that $(G(N), G(phi))$ is a limit of $G: D rightarrow C.$ However, something that confuses me is that $G(phi_X): G(N) rightarrow G(F(X))$ are the morphisms but can this be indexed by every object in $D?$ The family of morphisms for a cone for the object $G(N)$ should be $psi: G(N) rightarrow G(Y)$ for every object $Y$ in $D.$ So does $F(X)$ hit every object in $D?$ That is, for every object $Y$ in $D$ does there exist $X$ in $C$ such that $F(X) = Y?$
Do we need every object to be hit?
category-theory
asked Nov 16 at 4:22
伽罗瓦
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I am trying to show that right adjoints preserve limits. So suppose we have $F: C rightarrow D$ and $G: D rightarrow C$ be 2 adjoint functors with $G$ the right adjoint. So suppose we have a limit $(N, phi)$ of the diagram $F: C rightarrow D$ where $phi$ is indexed by objects $X$ in $C$ with $phi_X: N rightarrow F(X).$ Now I want to show that $(G(N), G(phi))$ is a limit of $G: D rightarrow C.$ However, something that confuses me is that $G(phi_X): G(N) rightarrow G(F(X))$ are the morphisms but can this be indexed by every object in $D?$ The family of morphisms for a cone for the object $G(N)$ should be $psi: G(N) rightarrow G(Y)$ for every object $Y$ in $D.$ So does $F(X)$ hit every object in $D?$ That is, for every object $Y$ in $D$ does there exist $X$ in $C$ such that $F(X) = Y?$
Do we need every object to be hit?
category-theory
asked Nov 16 at 4:22
伽罗瓦
1,043615
1
An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
1
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
1
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to show that right adjoints preserve limits. So suppose we have $F: C rightarrow D$ and $G: D rightarrow C$ be 2 adjoint functors with $G$ the right adjoint. So suppose we have a limit $(N, phi)$ of the diagram $F: C rightarrow D$ where $phi$ is indexed by objects $X$ in $C$ with $phi_X: N rightarrow F(X).$ Now I want to show that $(G(N), G(phi))$ is a limit of $G: D rightarrow C.$ However, something that confuses me is that $G(phi_X): G(N) rightarrow G(F(X))$ are the morphisms but can this be indexed by every object in $D?$ The family of morphisms for a cone for the object $G(N)$ should be $psi: G(N) rightarrow G(Y)$ for every object $Y$ in $D.$ So does $F(X)$ hit every object in $D?$ That is, for every object $Y$ in $D$ does there exist $X$ in $C$ such that $F(X) = Y?$
Do we need every object to be hit?
category-theory
asked Nov 16 at 4:22
伽罗瓦
1,043615
I am trying to show that right adjoints preserve limits. So suppose we have $F: C rightarrow D$ and $G: D rightarrow C$ be 2 adjoint functors with $G$ the right adjoint. So suppose we have a limit $(N, phi)$ of the diagram $F: C rightarrow D$ where $phi$ is indexed by objects $X$ in $C$ with $phi_X: N rightarrow F(X).$ Now I want to show that $(G(N), G(phi))$ is a limit of $G: D rightarrow C.$ However, something that confuses me is that $G(phi_X): G(N) rightarrow G(F(X))$ are the morphisms but can this be indexed by every object in $D?$ The family of morphisms for a cone for the object $G(N)$ should be $psi: G(N) rightarrow G(Y)$ for every object $Y$ in $D.$ So does $F(X)$ hit every object in $D?$ That is, for every object $Y$ in $D$ does there exist $X$ in $C$ such that $F(X) = Y?$
Do we need every object to be hit?
category-theory
category-theory
asked Nov 16 at 4:22
伽罗瓦
1,043615
asked Nov 16 at 4:22
伽罗瓦
1,043615
asked Nov 16 at 4:22
伽罗瓦
1,043615
asked Nov 16 at 4:22
伽罗瓦
1,043615
asked Nov 16 at 4:22
伽罗瓦
1,043615
1,043615
1
An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
1
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
1
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40
|
show 1 more comment
1
An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
1
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
1
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40
1
1
An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
1
1
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
1
1
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40
|
show 1 more comment
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An easy counterexample is to take practically any reflective subcategory. The abelianization functor on the category of groups is definitely not surjective (or essentially surjective).
– Malice Vidrine
Nov 16 at 4:41
1
In most basic examples of adjunction, you can see that the answer is no, e.g. in most (but not all) free/forgetful adjunctions for algebras, the answer is no
– Max
Nov 16 at 10:17
@MaliceVidrine The abelianization functor is essentially surjective, and in fact any reflector to a full subcategory is, since $epsilon_X: FGXto X$ Is an isomorphism.
– Arnaud D.
Nov 16 at 11:11
@ArnaudD. It seems rather clear that Malice was talking about the one of the two adjoints in that adjunction which is not surjective.
– Kevin Carlson
Nov 16 at 15:28
1
In @ArnaudD's defense, I did phrase that in a maximally confusing way.
– Malice Vidrine
Nov 16 at 16:40