understanding natural transformations that are not natural isomorphisms











up vote
4
down vote

favorite
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




























    up vote
    4
    down vote

    favorite
    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


























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 0:31









      Erick Wong

      20.1k22666




      20.1k22666










      asked Nov 17 at 3:54









      Gregory Nisbet

      381211




      381211






















          1 Answer
          1






          active

          oldest

          votes

















          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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001947%2funderstanding-natural-transformations-that-are-not-natural-isomorphisms%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            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











              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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 4:14









                Qiaochu Yuan

                275k32578914




                275k32578914






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001947%2funderstanding-natural-transformations-that-are-not-natural-isomorphisms%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Plaza Victoria

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...