Upgrade adjunction to equivalence












8














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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  • 1




    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
    Dec 1 at 17:49








  • 3




    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
    Dec 2 at 2:02






  • 2




    The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
    – Dylan Wilson
    Dec 2 at 21:54






  • 1




    I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
    – Neil Strickland
    Dec 7 at 19:26






  • 2




    Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
    – Axel Boldt
    Dec 7 at 23:31
















8














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


















  • 1




    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
    Dec 1 at 17:49








  • 3




    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
    Dec 2 at 2:02






  • 2




    The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
    – Dylan Wilson
    Dec 2 at 21:54






  • 1




    I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
    – Neil Strickland
    Dec 7 at 19:26






  • 2




    Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
    – Axel Boldt
    Dec 7 at 23:31














8












8








8


3





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













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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asked Dec 1 at 16:10









BlackBrain

1463




1463








  • 1




    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
    Dec 1 at 17:49








  • 3




    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
    Dec 2 at 2:02






  • 2




    The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
    – Dylan Wilson
    Dec 2 at 21:54






  • 1




    I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
    – Neil Strickland
    Dec 7 at 19:26






  • 2




    Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
    – Axel Boldt
    Dec 7 at 23:31














  • 1




    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
    Dec 1 at 17:49








  • 3




    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
    Dec 2 at 2:02






  • 2




    The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
    – Dylan Wilson
    Dec 2 at 21:54






  • 1




    I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
    – Neil Strickland
    Dec 7 at 19:26






  • 2




    Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
    – Axel Boldt
    Dec 7 at 23:31








1




1




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
Dec 1 at 17:49






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
Dec 1 at 17:49






3




3




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
Dec 2 at 2:02




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
Dec 2 at 2:02




2




2




The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
– Dylan Wilson
Dec 2 at 21:54




The quote is pretty vague about what is meant by "suitable modification" so maybe it's good to advertise two other relevant results. (1) An adjunction induces a homotopy equivalence of classifying spaces, or on the categories obtained by inverting all morphisms on both sides, and (2) the Barr-Beck theorem (en.wikipedia.org/wiki/Beck%27s_monadicity_theorem) gives conditions on when a right adjoint induces an equivalence to the category of algebras over the associated monad on the target
– Dylan Wilson
Dec 2 at 21:54




1




1




I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
– Neil Strickland
Dec 7 at 19:26




I have made a comment on the Wikipedia talk page of the user who put in this comment (AxelBoldt).
– Neil Strickland
Dec 7 at 19:26




2




2




Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
– Axel Boldt
Dec 7 at 23:31




Just to set the record straight, when I wrote the above sentence, I had the following examples in mind that were already mentioned below: the Galois correspondence in field theory arising from an adjunction, and numerous equivalences of homotopy categories arising from Quillen adjunctions.
– Axel Boldt
Dec 7 at 23:31










4 Answers
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12














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.






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    10














    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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      6














      I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^to_leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.



      Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k subseteq L subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.



      This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k subseteq L subseteq K$.



      Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.






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        5














        I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:




        • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.


        • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.







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          4 Answers
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          12














          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


























            12














            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
























              12












              12








              12






              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 Dec 1 at 18:54









              Leonid Positselski

              10.7k13974




              10.7k13974























                  10














                  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


























                    10














                    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
























                      10












                      10








                      10






                      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 Dec 1 at 22:51









                      Leonid Positselski

                      10.7k13974




                      10.7k13974























                          6














                          I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^to_leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.



                          Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k subseteq L subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.



                          This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k subseteq L subseteq K$.



                          Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.






                          share|cite|improve this answer


























                            6














                            I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^to_leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.



                            Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k subseteq L subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.



                            This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k subseteq L subseteq K$.



                            Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.






                            share|cite|improve this answer
























                              6












                              6








                              6






                              I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^to_leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.



                              Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k subseteq L subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.



                              This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k subseteq L subseteq K$.



                              Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.






                              share|cite|improve this answer












                              I think the author of the wikipedia article probably had in mind Leonid Positselski's first answer, where one restricts to the full subcategory of fixed points of the adjunction. Beware there is no guarantee that the fixed points are nonempty! For example, if $F: Set^to_leftarrow Ab: U$ is the free/forgetful adjunction bewteen sets and abelian groups, the fixed points are empty.



                              Here's an illustrative example to have in mind which is not so degenerate. Let $K/k$ be a Galois extension. Then there is an adjuntion between the poset of intermediate subfields $k subseteq L subseteq K$ and the opposite of the poset of subgroups of of $Gal(K/k)$; in one direction we send a group to its field of fixed points and in the other direction we send a field to the group of automorphisms that fix it.



                              This adjunction is typically not an equivalence, but we can pass to the fixed points of the adjunction to get an equivalence between the poset of normal subgroups of $Gal(K/k)$ and the opposite of the poset of intermediate Galois extensions $k subseteq L subseteq K$.



                              Thus the fundamental theorem of Galois theory may be viewed as calculating the fixed point set of an adjunction, and thus as identifying where an adjunction restricts to an equivalence.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 7 at 18:08









                              Tim Campion

                              13.1k352120




                              13.1k352120























                                  5














                                  I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:




                                  • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.


                                  • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.







                                  share|cite|improve this answer




























                                    5














                                    I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:




                                    • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.


                                    • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.







                                    share|cite|improve this answer


























                                      5












                                      5








                                      5






                                      I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:




                                      • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.


                                      • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.







                                      share|cite|improve this answer














                                      I agree that Leonid Positselski’s first answer seems probably what the writer had in mind: given an adjunction, restricting to the categories of “fixed points” on each side yields an equivalence. Here are two important examples in nature, both involving the category of topological spaces:




                                      • There’s an adjunction between the categories of preordered sets and topological spaces, sending a preordered set $(X,leq)$ to $X$ with the topology of down-closed sets, and sending a topological space $Y$ to $Y$ with its specialisation order. All preorders are fixpoints; on the other side, the fixpoints are exactly the Alexandrov spaces, i.e. spaces where arbitrary intersections of opens are open. Restricting to this subcategory shows that the category of preorders is equivalent (in fact, isomorphic!) to the category of Alexandrov spaces.


                                      • The adjunction between the categories of topological spaces and locales, sending a topological space to its frame/locale of opens and sending a locale to its space of points, restricts to the equivalence between spatial locales and sober spaces.








                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 10 at 13:27









                                      John D. Cook

                                      3,3053556




                                      3,3053556










                                      answered Dec 7 at 18:23









                                      Peter LeFanu Lumsdaine

                                      7,94913766




                                      7,94913766






























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