Equivalences of skeletal categories are isomorphisms












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As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.



Does this hold?










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    0












    $begingroup$


    As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.



    Does this hold?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.



      Does this hold?










      share|cite|improve this question











      $endgroup$




      As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C to D$ and $G:D to C$ with $FG simeq 1_D$ and $GF simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible.



      Does this hold?







      category-theory






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      edited Dec 5 '18 at 16:21







      Guido A.

















      asked Dec 5 '18 at 15:38









      Guido A.Guido A.

      7,3451730




      7,3451730






















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          $begingroup$

          It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.



          A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.



          So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks a lot!
            $endgroup$
            – Guido A.
            Dec 5 '18 at 18:54











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          $begingroup$

          It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.



          A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.



          So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks a lot!
            $endgroup$
            – Guido A.
            Dec 5 '18 at 18:54
















          1












          $begingroup$

          It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.



          A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.



          So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Got it, thanks a lot!
            $endgroup$
            – Guido A.
            Dec 5 '18 at 18:54














          1












          1








          1





          $begingroup$

          It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.



          A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.



          So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.






          share|cite|improve this answer









          $endgroup$



          It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.



          A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:Gto H$ seen as categories is indeed just an isomorphism, but a map $g:Hto G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.



          So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 18:18









          Kevin CarlsonKevin Carlson

          32.8k23372




          32.8k23372












          • $begingroup$
            Got it, thanks a lot!
            $endgroup$
            – Guido A.
            Dec 5 '18 at 18:54


















          • $begingroup$
            Got it, thanks a lot!
            $endgroup$
            – Guido A.
            Dec 5 '18 at 18:54
















          $begingroup$
          Got it, thanks a lot!
          $endgroup$
          – Guido A.
          Dec 5 '18 at 18:54




          $begingroup$
          Got it, thanks a lot!
          $endgroup$
          – Guido A.
          Dec 5 '18 at 18:54


















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