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).
ct.category-theory
add a comment |
up vote
6
down vote
favorite
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
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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
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
ct.category-theory
edited Nov 22 at 14:53
asked Nov 22 at 13:24
Gnampfissimo
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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.
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
add a comment |
up vote
5
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.
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
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[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 $f : emptyset to 1$.
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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3 Answers
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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.
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
add a comment |
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.
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
add a comment |
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.
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.
edited Nov 22 at 15:03
answered Nov 22 at 14:53
Reid Barton
18.1k150103
18.1k150103
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
add a comment |
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
add a comment |
up vote
5
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.
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
add a comment |
up vote
5
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.
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
add a comment |
up vote
5
down vote
up vote
5
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.
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.
edited Nov 22 at 21:46
answered Nov 22 at 21:38
Qiaochu Yuan
76.7k25316597
76.7k25316597
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
add a comment |
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
add a comment |
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 $f : emptyset to 1$.
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
add a comment |
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 $f : emptyset to 1$.
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
add a comment |
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 $f : emptyset to 1$.
[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 $f : emptyset to 1$.
edited Nov 22 at 15:10
answered Nov 22 at 14:34
Andrej Bauer
29.8k477164
29.8k477164
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
add a comment |
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
add a comment |
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