Upgrade adjunction to equivalence
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I'm studying category theory by myself and I just came across this sentence from Wikipedia:
An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors.
Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence.
ct.category-theory adjoint-functors
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I'm studying category theory by myself and I just came across this sentence from Wikipedia:
An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors.
Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence.
ct.category-theory adjoint-functors
New contributor
Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm studying category theory by myself and I just came across this sentence from Wikipedia:
An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors.
Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence.
ct.category-theory adjoint-functors
New contributor
I'm studying category theory by myself and I just came across this sentence from Wikipedia:
An adjunction between categories C and D is somewhat akin to a "weak form" of an equivalence between C and D, and indeed every equivalence is an adjunction. In many situations, an adjunction can be "upgraded" to an equivalence, by a suitable natural modification of the involved categories and functors.
Can someone provide an example of an "upgrade" of an adjunction to an equivalence? I'm interested in understanding why I could intuitively think of an adjunction as a weak form of an equivalence.
ct.category-theory adjoint-functors
ct.category-theory adjoint-functors
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BlackBrain
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Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago
add a comment |
Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago
Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago
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Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$ (as mentioned in a comment above).
Let $mathcal Asubsetmathcal C$ be the full subcategory consisting of all objects $Ainmathcal C$ for which the natural morphism $Alongrightarrow GF(A)$ is an isomorphism. Similarly, let $mathcal Bsubsetmathcal D$ be the full subcategory consisting of all objects $Binmathcal D$ for which the natural morphism $FG(B)longrightarrow B$ is an isomorphism.
Then one can check that $F(mathcal A)subsetmathcal B$ and $G(mathcal B)subsetmathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $mathcal Asubsetmathcal C$ and $mathcal Bsubsetmathcal D$ are again adjoint functors: the functor $F|_{mathcal A}colon mathcal Alongrightarrowmathcal B$ is left adjoint to the functor $G|_{mathcal B}colon mathcal Blongrightarrowmathcal A$. The adjunction between the functors $F|_{mathcal A}$ and $G|_{mathcal B}$ is an equivalence between the categories $mathcal A$ and $mathcal B$,
$$
F|_{mathcal A}colonmathcal A,simeq,mathcal B:!G|_{mathcal B}.
$$
This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.
I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{mathcal A},,G|_{mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".
Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.
add a comment |
up vote
2
down vote
Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$, as above.
Denote by $mathcal S$ the multiplicative class of morphisms in $mathcal C$ generated by all the morphisms $Clongrightarrow GF(C)$, where $C$ ranges over the objects of $mathcal C$. Similarly, denote by $mathcal T$ the multiplicative class of morphisms in $mathcal D$ generated by all the morphisms $FG(D)longrightarrow D$, where $D$ ranges over the objects of $mathcal D$.
Then one can check that the composition $mathcal Clongrightarrow mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ of the functor $F$ with the localization functor $mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ takes all the morphisms from $mathcal S$ to isomorphisms in $mathcal D[mathcal T^{-1}]$. So the functor $Fcolonmathcal Clongrightarrowmathcal D$ descends to a functor $overline Fcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]$; and similarly the functor $Gcolonmathcal Dlongrightarrowmathcal C$ descends to a functor $overline Gcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal S^{-1}]$.
The functors $overline F$ and $overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories:
$$
overline Fcolonmathcal C[mathcal S^{-1}],simeq,mathcal D[mathcal T^{-1}]:!overline G.
$$
Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $mathcal C[mathcal S^{-1}]$ and $mathcal D[mathcal T^{-1}]$ of the given two categories $mathcal C$ and $mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $mathcal Ssubsetmathcal C$ and $mathcal Tsubsetmathcal D$.
But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor
$$
mathbb LFcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]
$$
and a right derived functor
$$
mathbb RGcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal T^{-1}].
$$
Then the functor $mathbb LF$ is usually left adjoint to the functor $mathbb RG$, and under certain assumptions they are even adjoint equivalences.
I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.
add a comment |
2 Answers
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2 Answers
2
active
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active
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Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$ (as mentioned in a comment above).
Let $mathcal Asubsetmathcal C$ be the full subcategory consisting of all objects $Ainmathcal C$ for which the natural morphism $Alongrightarrow GF(A)$ is an isomorphism. Similarly, let $mathcal Bsubsetmathcal D$ be the full subcategory consisting of all objects $Binmathcal D$ for which the natural morphism $FG(B)longrightarrow B$ is an isomorphism.
Then one can check that $F(mathcal A)subsetmathcal B$ and $G(mathcal B)subsetmathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $mathcal Asubsetmathcal C$ and $mathcal Bsubsetmathcal D$ are again adjoint functors: the functor $F|_{mathcal A}colon mathcal Alongrightarrowmathcal B$ is left adjoint to the functor $G|_{mathcal B}colon mathcal Blongrightarrowmathcal A$. The adjunction between the functors $F|_{mathcal A}$ and $G|_{mathcal B}$ is an equivalence between the categories $mathcal A$ and $mathcal B$,
$$
F|_{mathcal A}colonmathcal A,simeq,mathcal B:!G|_{mathcal B}.
$$
This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.
I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{mathcal A},,G|_{mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".
Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.
add a comment |
up vote
4
down vote
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$ (as mentioned in a comment above).
Let $mathcal Asubsetmathcal C$ be the full subcategory consisting of all objects $Ainmathcal C$ for which the natural morphism $Alongrightarrow GF(A)$ is an isomorphism. Similarly, let $mathcal Bsubsetmathcal D$ be the full subcategory consisting of all objects $Binmathcal D$ for which the natural morphism $FG(B)longrightarrow B$ is an isomorphism.
Then one can check that $F(mathcal A)subsetmathcal B$ and $G(mathcal B)subsetmathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $mathcal Asubsetmathcal C$ and $mathcal Bsubsetmathcal D$ are again adjoint functors: the functor $F|_{mathcal A}colon mathcal Alongrightarrowmathcal B$ is left adjoint to the functor $G|_{mathcal B}colon mathcal Blongrightarrowmathcal A$. The adjunction between the functors $F|_{mathcal A}$ and $G|_{mathcal B}$ is an equivalence between the categories $mathcal A$ and $mathcal B$,
$$
F|_{mathcal A}colonmathcal A,simeq,mathcal B:!G|_{mathcal B}.
$$
This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.
I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{mathcal A},,G|_{mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".
Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.
add a comment |
up vote
4
down vote
up vote
4
down vote
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$ (as mentioned in a comment above).
Let $mathcal Asubsetmathcal C$ be the full subcategory consisting of all objects $Ainmathcal C$ for which the natural morphism $Alongrightarrow GF(A)$ is an isomorphism. Similarly, let $mathcal Bsubsetmathcal D$ be the full subcategory consisting of all objects $Binmathcal D$ for which the natural morphism $FG(B)longrightarrow B$ is an isomorphism.
Then one can check that $F(mathcal A)subsetmathcal B$ and $G(mathcal B)subsetmathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $mathcal Asubsetmathcal C$ and $mathcal Bsubsetmathcal D$ are again adjoint functors: the functor $F|_{mathcal A}colon mathcal Alongrightarrowmathcal B$ is left adjoint to the functor $G|_{mathcal B}colon mathcal Blongrightarrowmathcal A$. The adjunction between the functors $F|_{mathcal A}$ and $G|_{mathcal B}$ is an equivalence between the categories $mathcal A$ and $mathcal B$,
$$
F|_{mathcal A}colonmathcal A,simeq,mathcal B:!G|_{mathcal B}.
$$
This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.
I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{mathcal A},,G|_{mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".
Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$ (as mentioned in a comment above).
Let $mathcal Asubsetmathcal C$ be the full subcategory consisting of all objects $Ainmathcal C$ for which the natural morphism $Alongrightarrow GF(A)$ is an isomorphism. Similarly, let $mathcal Bsubsetmathcal D$ be the full subcategory consisting of all objects $Binmathcal D$ for which the natural morphism $FG(B)longrightarrow B$ is an isomorphism.
Then one can check that $F(mathcal A)subsetmathcal B$ and $G(mathcal B)subsetmathcal A$. The restrictions of the adjoint functors $F$ and $G$ to the full subcategories $mathcal Asubsetmathcal C$ and $mathcal Bsubsetmathcal D$ are again adjoint functors: the functor $F|_{mathcal A}colon mathcal Alongrightarrowmathcal B$ is left adjoint to the functor $G|_{mathcal B}colon mathcal Blongrightarrowmathcal A$. The adjunction between the functors $F|_{mathcal A}$ and $G|_{mathcal B}$ is an equivalence between the categories $mathcal A$ and $mathcal B$,
$$
F|_{mathcal A}colonmathcal A,simeq,mathcal B:!G|_{mathcal B}.
$$
This result can be found in the paper A. Frankild, P. Jorgensen, "Foxby equivalence, complete modules, and torsion modules", J. Pure Appl. Algebra 174 #2, p.135-147, 2002, https://doi.org/10.1016/S0022-4049(02)00043-9 , Theorem 1.1.
I am not sure whether this should be properly called "upgrading an adjuction to an equivalence", though. The passage from the adjoint pair $(F,G)$ to the equivalence $(F|_{mathcal A},,G|_{mathcal B})$ entails losing rather than gaining information. Perhaps it would be better to call it "restricting an adjunction to an equivalence".
Then again, I do not know what the author of the passage in the Wikipedia article might have had in mind.
answered 7 hours ago
Leonid Positselski
10.5k13872
10.5k13872
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up vote
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Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$, as above.
Denote by $mathcal S$ the multiplicative class of morphisms in $mathcal C$ generated by all the morphisms $Clongrightarrow GF(C)$, where $C$ ranges over the objects of $mathcal C$. Similarly, denote by $mathcal T$ the multiplicative class of morphisms in $mathcal D$ generated by all the morphisms $FG(D)longrightarrow D$, where $D$ ranges over the objects of $mathcal D$.
Then one can check that the composition $mathcal Clongrightarrow mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ of the functor $F$ with the localization functor $mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ takes all the morphisms from $mathcal S$ to isomorphisms in $mathcal D[mathcal T^{-1}]$. So the functor $Fcolonmathcal Clongrightarrowmathcal D$ descends to a functor $overline Fcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]$; and similarly the functor $Gcolonmathcal Dlongrightarrowmathcal C$ descends to a functor $overline Gcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal S^{-1}]$.
The functors $overline F$ and $overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories:
$$
overline Fcolonmathcal C[mathcal S^{-1}],simeq,mathcal D[mathcal T^{-1}]:!overline G.
$$
Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $mathcal C[mathcal S^{-1}]$ and $mathcal D[mathcal T^{-1}]$ of the given two categories $mathcal C$ and $mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $mathcal Ssubsetmathcal C$ and $mathcal Tsubsetmathcal D$.
But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor
$$
mathbb LFcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]
$$
and a right derived functor
$$
mathbb RGcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal T^{-1}].
$$
Then the functor $mathbb LF$ is usually left adjoint to the functor $mathbb RG$, and under certain assumptions they are even adjoint equivalences.
I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.
add a comment |
up vote
2
down vote
Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$, as above.
Denote by $mathcal S$ the multiplicative class of morphisms in $mathcal C$ generated by all the morphisms $Clongrightarrow GF(C)$, where $C$ ranges over the objects of $mathcal C$. Similarly, denote by $mathcal T$ the multiplicative class of morphisms in $mathcal D$ generated by all the morphisms $FG(D)longrightarrow D$, where $D$ ranges over the objects of $mathcal D$.
Then one can check that the composition $mathcal Clongrightarrow mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ of the functor $F$ with the localization functor $mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ takes all the morphisms from $mathcal S$ to isomorphisms in $mathcal D[mathcal T^{-1}]$. So the functor $Fcolonmathcal Clongrightarrowmathcal D$ descends to a functor $overline Fcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]$; and similarly the functor $Gcolonmathcal Dlongrightarrowmathcal C$ descends to a functor $overline Gcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal S^{-1}]$.
The functors $overline F$ and $overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories:
$$
overline Fcolonmathcal C[mathcal S^{-1}],simeq,mathcal D[mathcal T^{-1}]:!overline G.
$$
Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $mathcal C[mathcal S^{-1}]$ and $mathcal D[mathcal T^{-1}]$ of the given two categories $mathcal C$ and $mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $mathcal Ssubsetmathcal C$ and $mathcal Tsubsetmathcal D$.
But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor
$$
mathbb LFcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]
$$
and a right derived functor
$$
mathbb RGcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal T^{-1}].
$$
Then the functor $mathbb LF$ is usually left adjoint to the functor $mathbb RG$, and under certain assumptions they are even adjoint equivalences.
I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.
add a comment |
up vote
2
down vote
up vote
2
down vote
Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$, as above.
Denote by $mathcal S$ the multiplicative class of morphisms in $mathcal C$ generated by all the morphisms $Clongrightarrow GF(C)$, where $C$ ranges over the objects of $mathcal C$. Similarly, denote by $mathcal T$ the multiplicative class of morphisms in $mathcal D$ generated by all the morphisms $FG(D)longrightarrow D$, where $D$ ranges over the objects of $mathcal D$.
Then one can check that the composition $mathcal Clongrightarrow mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ of the functor $F$ with the localization functor $mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ takes all the morphisms from $mathcal S$ to isomorphisms in $mathcal D[mathcal T^{-1}]$. So the functor $Fcolonmathcal Clongrightarrowmathcal D$ descends to a functor $overline Fcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]$; and similarly the functor $Gcolonmathcal Dlongrightarrowmathcal C$ descends to a functor $overline Gcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal S^{-1}]$.
The functors $overline F$ and $overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories:
$$
overline Fcolonmathcal C[mathcal S^{-1}],simeq,mathcal D[mathcal T^{-1}]:!overline G.
$$
Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $mathcal C[mathcal S^{-1}]$ and $mathcal D[mathcal T^{-1}]$ of the given two categories $mathcal C$ and $mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $mathcal Ssubsetmathcal C$ and $mathcal Tsubsetmathcal D$.
But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor
$$
mathbb LFcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]
$$
and a right derived functor
$$
mathbb RGcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal T^{-1}].
$$
Then the functor $mathbb LF$ is usually left adjoint to the functor $mathbb RG$, and under certain assumptions they are even adjoint equivalences.
I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.
Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $mathcal C$ and $mathcal D$ be two categories, and let $Fcolonmathcal Clongrightarrow mathcal D$ and $Gcolonmathcal Dlongrightarrowmathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $Fcirc Glongrightarrow operatorname{Id}_{mathcal D}$ and $operatorname{Id}_{mathcal C}longrightarrow Gcirc F$, as above.
Denote by $mathcal S$ the multiplicative class of morphisms in $mathcal C$ generated by all the morphisms $Clongrightarrow GF(C)$, where $C$ ranges over the objects of $mathcal C$. Similarly, denote by $mathcal T$ the multiplicative class of morphisms in $mathcal D$ generated by all the morphisms $FG(D)longrightarrow D$, where $D$ ranges over the objects of $mathcal D$.
Then one can check that the composition $mathcal Clongrightarrow mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ of the functor $F$ with the localization functor $mathcal Dlongrightarrow mathcal D[mathcal T^{-1}]$ takes all the morphisms from $mathcal S$ to isomorphisms in $mathcal D[mathcal T^{-1}]$. So the functor $Fcolonmathcal Clongrightarrowmathcal D$ descends to a functor $overline Fcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]$; and similarly the functor $Gcolonmathcal Dlongrightarrowmathcal C$ descends to a functor $overline Gcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal S^{-1}]$.
The functors $overline F$ and $overline G$ are still adjoint to each other, and this adjunction is an equivalence between the two localized categories:
$$
overline Fcolonmathcal C[mathcal S^{-1}],simeq,mathcal D[mathcal T^{-1}]:!overline G.
$$
Yet another and perhaps even more natural interpretation of what may be meant by the sentence in Wikipedia also involves passing to localizations $mathcal C[mathcal S^{-1}]$ and $mathcal D[mathcal T^{-1}]$ of the given two categories $mathcal C$ and $mathcal D$ with respect to some natural multiplicative classes of morphisms (often called the classes of weak equivalences) $mathcal Ssubsetmathcal C$ and $mathcal Tsubsetmathcal D$.
But, rather than hoping that the functors $F$ and $G$ would simply descend to functors between the localized categories, one derives them in some way, producing a left derived functor
$$
mathbb LFcolonmathcal C[mathcal S^{-1}]longrightarrowmathcal D[mathcal T^{-1}]
$$
and a right derived functor
$$
mathbb RGcolonmathcal D[mathcal T^{-1}]longrightarrowmathcal C[mathcal T^{-1}].
$$
Then the functor $mathbb LF$ is usually left adjoint to the functor $mathbb RG$, and under certain assumptions they are even adjoint equivalences.
I would not go into further details on derived functors etc. in this answer, but rather suggest some keywords or a key sentence which you could look up: a Quillen equivalence between two model categories induces an equivalence between their homotopy categories.
answered 3 hours ago
Leonid Positselski
10.5k13872
10.5k13872
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Abstractly speaking an equivalence is just an adjunction with the unit (or counit) a natural isomorphism; an adjunction is a pair $F,G:mathcal{C}rightleftarrowsmathcal{D}$ together with a natural transformation $eta:1_mathcal{C}rightarrow Gcirc F$ satisfying a universal property, while an equivalence is all this plus the requirement that $eta$ be a natural isomorphism (we can prove the existence of the second natural isomorphism $epsilon:Fcirc Grightarrow 1_mathcal{D}$ from just the above). For a specific example I will think on it, but this is the essence of the difference.
– Alec Rhea
8 hours ago
See here (ncatlab.org/nlab/show/adjoint+equivalence) for a detailed discussion; I would also encourage you to view an equivalence as a 'strong adjunction' rather than the other way around -- pretty much every construction in category theory can be thought of as a 'strong adjunction' of some sort, from equivalences to limits/colimits to Kan extensions. I would post an answer, but I think this question might be a better fit over at math.stackexchange as a general question about mathematics.
– Alec Rhea
7 hours ago
Although the answers below have done an excellent job of trying to interpret the quote as written, I would argue that the quote is actually rather misleading, and if any of these examples are what its author had in mind then it could be better phrased to convey that.
– Mike Shulman
21 mins ago