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.










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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















up vote
3
down vote

favorite
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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.










share|cite|improve this question







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BlackBrain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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













up vote
3
down vote

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up vote
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down vote

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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.










share|cite|improve this question







New contributor




BlackBrain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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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  • 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


















  • 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
















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










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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.






share|cite|improve this answer




























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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.






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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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer























          up vote
          4
          down vote










          up vote
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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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          Leonid Positselski

          10.5k13872




          10.5k13872






















              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.






              share|cite|improve this answer

























                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.






                share|cite|improve this answer























                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Leonid Positselski

                  10.5k13872




                  10.5k13872






















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