Can natural section/retraction be checked pointwise?












6














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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    6














    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



























      6












      6








      6


      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















      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 Nov 22 at 14:53

























      asked Nov 22 at 13:24









      Gnampfissimo

      336




      336






















          3 Answers
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          6














          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
            Nov 22 at 15:38










          • ... 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
            Nov 22 at 15:50






          • 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
            Nov 22 at 16:05



















          5














          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



















          • 1




            Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
            – Peter LeFanu Lumsdaine
            Nov 23 at 10:50





















          2














          [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 $f : emptyset to 1$.






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          • 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
            Nov 22 at 14:47










          • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
            – Gnampfissimo
            Nov 22 at 14:58










          • I see. Perhaps you can make the question a bit more explicit then.
            – Andrej Bauer
            Nov 22 at 15:11











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          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
            Nov 22 at 15:38










          • ... 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
            Nov 22 at 15:50






          • 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
            Nov 22 at 16:05
















          6














          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
            Nov 22 at 15:38










          • ... 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
            Nov 22 at 15:50






          • 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
            Nov 22 at 16:05














          6












          6








          6






          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 Nov 22 at 15:03

























          answered Nov 22 at 14:53









          Reid Barton

          18.2k150104




          18.2k150104












          • 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
            Nov 22 at 15:38










          • ... 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
            Nov 22 at 15:50






          • 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
            Nov 22 at 16:05


















          • 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
            Nov 22 at 15:38










          • ... 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
            Nov 22 at 15:50






          • 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
            Nov 22 at 16:05
















          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
          Nov 22 at 15:38




          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
          Nov 22 at 15:38












          ... 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
          Nov 22 at 15:50




          ... 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
          Nov 22 at 15:50




          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
          Nov 22 at 16:05




          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
          Nov 22 at 16:05











          5














          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



















          • 1




            Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
            – Peter LeFanu Lumsdaine
            Nov 23 at 10:50


















          5














          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



















          • 1




            Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
            – Peter LeFanu Lumsdaine
            Nov 23 at 10:50
















          5












          5








          5






          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 Nov 22 at 21:46

























          answered Nov 22 at 21:38









          Qiaochu Yuan

          76.8k26317599




          76.8k26317599








          • 1




            Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
            – Peter LeFanu Lumsdaine
            Nov 23 at 10:50
















          • 1




            Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
            – Peter LeFanu Lumsdaine
            Nov 23 at 10:50










          1




          1




          Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
          – Peter LeFanu Lumsdaine
          Nov 23 at 10:50






          Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X to 1$ certainly has a “pointwise” section — any element $x in X$ gives a function $s_x : 1 to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X to 1$ has a pointwise section but no (natural) section.
          – Peter LeFanu Lumsdaine
          Nov 23 at 10:50













          2














          [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
            Nov 22 at 14:47










          • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
            – Gnampfissimo
            Nov 22 at 14:58










          • I see. Perhaps you can make the question a bit more explicit then.
            – Andrej Bauer
            Nov 22 at 15:11
















          2














          [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
            Nov 22 at 14:47










          • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
            – Gnampfissimo
            Nov 22 at 14:58










          • I see. Perhaps you can make the question a bit more explicit then.
            – Andrej Bauer
            Nov 22 at 15:11














          2












          2








          2






          [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 Nov 22 at 15:10

























          answered Nov 22 at 14:34









          Andrej Bauer

          30k477165




          30k477165












          • 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
            Nov 22 at 14:47










          • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
            – Gnampfissimo
            Nov 22 at 14:58










          • I see. Perhaps you can make the question a bit more explicit then.
            – Andrej Bauer
            Nov 22 at 15:11


















          • 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
            Nov 22 at 14:47










          • P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
            – Gnampfissimo
            Nov 22 at 14:58










          • I see. Perhaps you can make the question a bit more explicit then.
            – Andrej Bauer
            Nov 22 at 15:11
















          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
          Nov 22 at 14:47




          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
          Nov 22 at 14:47












          P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
          – Gnampfissimo
          Nov 22 at 14:58




          P.S.: I think in the end of the first paragraph you mean $epsilon_X$, right?
          – Gnampfissimo
          Nov 22 at 14:58












          I see. Perhaps you can make the question a bit more explicit then.
          – Andrej Bauer
          Nov 22 at 15:11




          I see. Perhaps you can make the question a bit more explicit then.
          – Andrej Bauer
          Nov 22 at 15:11


















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