Can natural section/retraction be checked pointwise?











up vote
5
down vote

favorite
1












Analogously to
this old question,
I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation.



For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been:



Given a natural transformation $varepsilon: F Rightarrow G$ between functors $F, G: C to D$, if its components are right invertible with sections $eta_C: GC to FC$, naturality of $eta$ comes down to $Gc circ eta_C = p_{C'} circ Gc circ eta_C$ for any morphism $c: C to C'$, where I defined the idempotents $p_C := eta_C circ varepsilon_C$. At this point I don't see under which conditions there is a choice of $eta_C$ and $eta_{C'}$ such that the above equation is fulfilled (apart from $p_C = text{id}_{FC}$, i.e. natural isomorphisms, of course).



I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms).



Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way...



If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it).










share|cite|improve this question









New contributor




Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    5
    down vote

    favorite
    1












    Analogously to
    this old question,
    I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation.



    For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been:



    Given a natural transformation $varepsilon: F Rightarrow G$ between functors $F, G: C to D$, if its components are right invertible with sections $eta_C: GC to FC$, naturality of $eta$ comes down to $Gc circ eta_C = p_{C'} circ Gc circ eta_C$ for any morphism $c: C to C'$, where I defined the idempotents $p_C := eta_C circ varepsilon_C$. At this point I don't see under which conditions there is a choice of $eta_C$ and $eta_{C'}$ such that the above equation is fulfilled (apart from $p_C = text{id}_{FC}$, i.e. natural isomorphisms, of course).



    I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms).



    Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way...



    If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it).










    share|cite|improve this question









    New contributor




    Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      5
      down vote

      favorite
      1









      up vote
      5
      down vote

      favorite
      1






      1





      Analogously to
      this old question,
      I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation.



      For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been:



      Given a natural transformation $varepsilon: F Rightarrow G$ between functors $F, G: C to D$, if its components are right invertible with sections $eta_C: GC to FC$, naturality of $eta$ comes down to $Gc circ eta_C = p_{C'} circ Gc circ eta_C$ for any morphism $c: C to C'$, where I defined the idempotents $p_C := eta_C circ varepsilon_C$. At this point I don't see under which conditions there is a choice of $eta_C$ and $eta_{C'}$ such that the above equation is fulfilled (apart from $p_C = text{id}_{FC}$, i.e. natural isomorphisms, of course).



      I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms).



      Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way...



      If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it).










      share|cite|improve this question









      New contributor




      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Analogously to
      this old question,
      I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation.



      For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been:



      Given a natural transformation $varepsilon: F Rightarrow G$ between functors $F, G: C to D$, if its components are right invertible with sections $eta_C: GC to FC$, naturality of $eta$ comes down to $Gc circ eta_C = p_{C'} circ Gc circ eta_C$ for any morphism $c: C to C'$, where I defined the idempotents $p_C := eta_C circ varepsilon_C$. At this point I don't see under which conditions there is a choice of $eta_C$ and $eta_{C'}$ such that the above equation is fulfilled (apart from $p_C = text{id}_{FC}$, i.e. natural isomorphisms, of course).



      I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms).



      Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way...



      If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it).







      ct.category-theory






      share|cite|improve this question









      New contributor




      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago





















      New contributor




      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 14 hours ago









      Gnampfissimo

      286




      286




      New contributor




      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          6
          down vote



          accepted










          No, you cannot check the property of being a section/retraction pointwise. Take $C = {cdot to cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A to B)$ to the object $(1 : B to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.



          This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G to *$, which has no section; or in the category of simplicial sets, the map $partial Delta^1 to Delta^1$, which is not the inclusion of a retraction.






          share|cite|improve this answer























          • This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
            – Gnampfissimo
            12 hours ago










          • ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
            – Gnampfissimo
            11 hours ago






          • 1




            Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
            – Reid Barton
            11 hours ago


















          up vote
          3
          down vote













          Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.



          Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.






          share|cite|improve this answer






























            up vote
            2
            down vote













            [Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]



            To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.



            Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
            $$eta_X(0) =
            begin{cases}
            0 & text{if $X = emptyset$} \
            1 & text{otherwise}.
            end{cases}
            $$

            Then naturality of $eta$ fails for the map $f : emptyset to 1$.






            share|cite|improve this answer























            • Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
              – Gnampfissimo
              12 hours ago










            • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
              – Gnampfissimo
              12 hours ago










            • I see. Perhaps you can make the question a bit more explicit then.
              – Andrej Bauer
              12 hours ago











            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: "504"
            };
            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
            });


            }
            });






            Gnampfissimo is a new contributor. Be nice, and check out our Code of Conduct.










             

            draft saved


            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315955%2fcan-natural-section-retraction-be-checked-pointwise%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            6
            down vote



            accepted










            No, you cannot check the property of being a section/retraction pointwise. Take $C = {cdot to cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A to B)$ to the object $(1 : B to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.



            This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G to *$, which has no section; or in the category of simplicial sets, the map $partial Delta^1 to Delta^1$, which is not the inclusion of a retraction.






            share|cite|improve this answer























            • This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
              – Gnampfissimo
              12 hours ago










            • ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
              – Gnampfissimo
              11 hours ago






            • 1




              Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
              – Reid Barton
              11 hours ago















            up vote
            6
            down vote



            accepted










            No, you cannot check the property of being a section/retraction pointwise. Take $C = {cdot to cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A to B)$ to the object $(1 : B to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.



            This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G to *$, which has no section; or in the category of simplicial sets, the map $partial Delta^1 to Delta^1$, which is not the inclusion of a retraction.






            share|cite|improve this answer























            • This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
              – Gnampfissimo
              12 hours ago










            • ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
              – Gnampfissimo
              11 hours ago






            • 1




              Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
              – Reid Barton
              11 hours ago













            up vote
            6
            down vote



            accepted







            up vote
            6
            down vote



            accepted






            No, you cannot check the property of being a section/retraction pointwise. Take $C = {cdot to cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A to B)$ to the object $(1 : B to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.



            This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G to *$, which has no section; or in the category of simplicial sets, the map $partial Delta^1 to Delta^1$, which is not the inclusion of a retraction.






            share|cite|improve this answer














            No, you cannot check the property of being a section/retraction pointwise. Take $C = {cdot to cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A to B)$ to the object $(1 : B to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.



            This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G to *$, which has no section; or in the category of simplicial sets, the map $partial Delta^1 to Delta^1$, which is not the inclusion of a retraction.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 12 hours ago

























            answered 12 hours ago









            Reid Barton

            18k149103




            18k149103












            • This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
              – Gnampfissimo
              12 hours ago










            • ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
              – Gnampfissimo
              11 hours ago






            • 1




              Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
              – Reid Barton
              11 hours ago


















            • This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
              – Gnampfissimo
              12 hours ago










            • ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
              – Gnampfissimo
              11 hours ago






            • 1




              Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
              – Reid Barton
              11 hours ago
















            This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
            – Gnampfissimo
            12 hours ago




            This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me.
            – Gnampfissimo
            12 hours ago












            ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
            – Gnampfissimo
            11 hours ago




            ... and is wrong, since for any morphism $f: A to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: text{id}_A to text{id}_B$ in the arrow category. What did you mean with "required" then?
            – Gnampfissimo
            11 hours ago




            1




            1




            Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
            – Reid Barton
            11 hours ago




            Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete).
            – Reid Barton
            11 hours ago










            up vote
            3
            down vote













            Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.



            Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.






            share|cite|improve this answer



























              up vote
              3
              down vote













              Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.



              Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.






              share|cite|improve this answer

























                up vote
                3
                down vote










                up vote
                3
                down vote









                Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.



                Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.






                share|cite|improve this answer














                Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.



                Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 5 hours ago

























                answered 6 hours ago









                Qiaochu Yuan

                76.2k25314595




                76.2k25314595






















                    up vote
                    2
                    down vote













                    [Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]



                    To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.



                    Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
                    $$eta_X(0) =
                    begin{cases}
                    0 & text{if $X = emptyset$} \
                    1 & text{otherwise}.
                    end{cases}
                    $$

                    Then naturality of $eta$ fails for the map $f : emptyset to 1$.






                    share|cite|improve this answer























                    • Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                      – Gnampfissimo
                      12 hours ago










                    • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                      – Gnampfissimo
                      12 hours ago










                    • I see. Perhaps you can make the question a bit more explicit then.
                      – Andrej Bauer
                      12 hours ago















                    up vote
                    2
                    down vote













                    [Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]



                    To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.



                    Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
                    $$eta_X(0) =
                    begin{cases}
                    0 & text{if $X = emptyset$} \
                    1 & text{otherwise}.
                    end{cases}
                    $$

                    Then naturality of $eta$ fails for the map $f : emptyset to 1$.






                    share|cite|improve this answer























                    • Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                      – Gnampfissimo
                      12 hours ago










                    • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                      – Gnampfissimo
                      12 hours ago










                    • I see. Perhaps you can make the question a bit more explicit then.
                      – Andrej Bauer
                      12 hours ago













                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    [Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]



                    To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.



                    Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
                    $$eta_X(0) =
                    begin{cases}
                    0 & text{if $X = emptyset$} \
                    1 & text{otherwise}.
                    end{cases}
                    $$

                    Then naturality of $eta$ fails for the map $f : emptyset to 1$.






                    share|cite|improve this answer














                    [Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]



                    To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.



                    Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
                    $$eta_X(0) =
                    begin{cases}
                    0 & text{if $X = emptyset$} \
                    1 & text{otherwise}.
                    end{cases}
                    $$

                    Then naturality of $eta$ fails for the map $f : emptyset to 1$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 12 hours ago

























                    answered 13 hours ago









                    Andrej Bauer

                    29.5k477162




                    29.5k477162












                    • Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                      – Gnampfissimo
                      12 hours ago










                    • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                      – Gnampfissimo
                      12 hours ago










                    • I see. Perhaps you can make the question a bit more explicit then.
                      – Andrej Bauer
                      12 hours ago


















                    • Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                      – Gnampfissimo
                      12 hours ago










                    • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                      – Gnampfissimo
                      12 hours ago










                    • I see. Perhaps you can make the question a bit more explicit then.
                      – Andrej Bauer
                      12 hours ago
















                    Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                    – Gnampfissimo
                    12 hours ago




                    Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation").
                    – Gnampfissimo
                    12 hours ago












                    P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                    – Gnampfissimo
                    12 hours ago




                    P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
                    – Gnampfissimo
                    12 hours ago












                    I see. Perhaps you can make the question a bit more explicit then.
                    – Andrej Bauer
                    12 hours ago




                    I see. Perhaps you can make the question a bit more explicit then.
                    – Andrej Bauer
                    12 hours ago










                    Gnampfissimo is a new contributor. Be nice, and check out our Code of Conduct.










                     

                    draft saved


                    draft discarded


















                    Gnampfissimo is a new contributor. Be nice, and check out our Code of Conduct.













                    Gnampfissimo is a new contributor. Be nice, and check out our Code of Conduct.












                    Gnampfissimo is a new contributor. Be nice, and check out our Code of Conduct.















                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315955%2fcan-natural-section-retraction-be-checked-pointwise%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

                    Costa Masnaga

                    Fotorealismo

                    Sidney Franklin