Can natural section/retraction be checked pointwise?











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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).










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    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).










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    Gnampfissimo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      5
      down vote

      favorite
      1









      up vote
      5
      down vote

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      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






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      edited 8 hours ago





















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          3 Answers
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          up vote
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          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
            7 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
            7 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
            6 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
            8 hours ago










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










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


















          up vote
          2
          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.






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            3 Answers
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            up vote
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            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
              7 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
              7 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
              6 hours ago















            up vote
            5
            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
              7 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
              7 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
              6 hours ago













            up vote
            5
            down vote



            accepted







            up vote
            5
            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 8 hours ago

























            answered 8 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
              7 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
              7 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
              6 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
              7 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
              7 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
              6 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
            7 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
            7 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
            7 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
            7 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
            6 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
            6 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
              8 hours ago










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










            • I see. Perhaps you can make the question a bit more explicit then.
              – Andrej Bauer
              7 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
              8 hours ago










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










            • I see. Perhaps you can make the question a bit more explicit then.
              – Andrej Bauer
              7 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 7 hours ago

























            answered 8 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
              8 hours ago










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










            • I see. Perhaps you can make the question a bit more explicit then.
              – Andrej Bauer
              7 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
              8 hours ago










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










            • I see. Perhaps you can make the question a bit more explicit then.
              – Andrej Bauer
              7 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
            8 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
            8 hours ago












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




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












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




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










            up vote
            2
            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













              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










                up vote
                2
                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 1 hour ago

























                answered 1 hour ago









                Qiaochu Yuan

                76.2k25314595




                76.2k25314595






















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