understanding natural transformations that are not natural isomorphisms











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What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?



A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.



This means that the ordinary natural transformation law



$$ eta_Y circ F(f) = G(f) circ eta_x $$



can be written as



$$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$



or, using $triangleright$ as reverse composition ...



$$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$



In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.



However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.



If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?










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    What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?



    A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.



    This means that the ordinary natural transformation law



    $$ eta_Y circ F(f) = G(f) circ eta_x $$



    can be written as



    $$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$



    or, using $triangleright$ as reverse composition ...



    $$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$



    In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.



    However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.



    If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?










    share|cite|improve this question


























      up vote
      4
      down vote

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      up vote
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      1





      What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?



      A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.



      This means that the ordinary natural transformation law



      $$ eta_Y circ F(f) = G(f) circ eta_x $$



      can be written as



      $$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$



      or, using $triangleright$ as reverse composition ...



      $$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$



      In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.



      However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.



      If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?










      share|cite|improve this question















      What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?



      A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.



      This means that the ordinary natural transformation law



      $$ eta_Y circ F(f) = G(f) circ eta_x $$



      can be written as



      $$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$



      or, using $triangleright$ as reverse composition ...



      $$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$



      In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.



      However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.



      If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?







      category-theory natural-transformations






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      edited Nov 19 at 0:31









      Erick Wong

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      20.1k22666










      asked Nov 17 at 3:54









      Gregory Nisbet

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          What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?




          No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.



          A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.






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            What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?




            No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.



            A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.






            share|cite|improve this answer

























              up vote
              4
              down vote



              accepted











              What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?




              No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.



              A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.






              share|cite|improve this answer























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted







                What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?




                No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.



                A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.






                share|cite|improve this answer













                What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?




                No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.



                A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Nov 17 at 4:14









                Qiaochu Yuan

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                275k32578914






























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