Mixing
Composition of monadic functors isn't monadic
$begingroup$
Disclaimer: this question already has a solution here:
Composition of monadic functors may not be monadic
. However, I would like to understand how to solve this using another characterisation of monadic functors - Beck's monadicity theorem.
Let TFA denote the full subcategory of torsion-free abelian groups, A of abelian groups, and Set the usual category of sets.
Take it as given that the forgetful functor $F:$ A $to$ Set and the inclusion functor $i:$ TFA $to$ A are monadic.
Show that, however, the composition $F circ i:$ TFA $to$
Set, is not monadic, using Beck's monadicity theorem.
Recall that Beck's monadicity theorem states that a functor $G: A to B$ is monadic if and only if $G$ has a left adjoint, it reflects isomorphisms, and every $G$ split pair admits a coequalizer in $A$, which $G$ preserves.
Where I'm at:
It is clear that $F circ i$ has a left adjoint, and that it reflects isomorphisms, and so it is the last and final condition that must be false.
For it to be false, we must find a pair of group homomorphisms, between torsion-free abelian groups, $G_1overset{f}{underset{g}rightrightarrows}G_2$, such that as functions on the respective sets, they admit a split fork:
there is $C in $ Set, $e: G_2 rightleftarrows C:s$ and $t:G_2 to G_1$ s.t:
$e circ s = id_{C}$
$g circ t = s circ e$
$f circ t = id_{G_2}$
And that this set $C$ will not have the canonic structure of a torsion-free abelian group. (Or simply that $G_1, G_2$ have no coequalizer in TFA)
Note that, since A is cocomplete, this pair $f,g$ will admit a coequalizer (in A), so this set $C$ may have the structure of an abelian group.
So my question is, can you help find such an example? Or is my reasoning off?
I've tried several with $mathbb{Z}$, and variants of it, but in order to get that $C$ is not torsion free, I need that $f-g$ will have a big enough image so that $mathbb{Z}/Im(f-g)$ will be finite. I think this specific example is hopeless though.
category-theory adjoint-functors monads forgetful-functors
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
$endgroup$
add a comment |
$begingroup$
Disclaimer: this question already has a solution here:
Composition of monadic functors may not be monadic
. However, I would like to understand how to solve this using another characterisation of monadic functors - Beck's monadicity theorem.
Let TFA denote the full subcategory of torsion-free abelian groups, A of abelian groups, and Set the usual category of sets.
Take it as given that the forgetful functor $F:$ A $to$ Set and the inclusion functor $i:$ TFA $to$ A are monadic.
Show that, however, the composition $F circ i:$ TFA $to$
Set, is not monadic, using Beck's monadicity theorem.
Recall that Beck's monadicity theorem states that a functor $G: A to B$ is monadic if and only if $G$ has a left adjoint, it reflects isomorphisms, and every $G$ split pair admits a coequalizer in $A$, which $G$ preserves.
Where I'm at:
It is clear that $F circ i$ has a left adjoint, and that it reflects isomorphisms, and so it is the last and final condition that must be false.
For it to be false, we must find a pair of group homomorphisms, between torsion-free abelian groups, $G_1overset{f}{underset{g}rightrightarrows}G_2$, such that as functions on the respective sets, they admit a split fork:
there is $C in $ Set, $e: G_2 rightleftarrows C:s$ and $t:G_2 to G_1$ s.t:
$e circ s = id_{C}$
$g circ t = s circ e$
$f circ t = id_{G_2}$
And that this set $C$ will not have the canonic structure of a torsion-free abelian group. (Or simply that $G_1, G_2$ have no coequalizer in TFA)
Note that, since A is cocomplete, this pair $f,g$ will admit a coequalizer (in A), so this set $C$ may have the structure of an abelian group.
So my question is, can you help find such an example? Or is my reasoning off?
I've tried several with $mathbb{Z}$, and variants of it, but in order to get that $C$ is not torsion free, I need that $f-g$ will have a big enough image so that $mathbb{Z}/Im(f-g)$ will be finite. I think this specific example is hopeless though.
category-theory adjoint-functors monads forgetful-functors
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
$endgroup$
add a comment |
$begingroup$
Disclaimer: this question already has a solution here:
Composition of monadic functors may not be monadic
. However, I would like to understand how to solve this using another characterisation of monadic functors - Beck's monadicity theorem.
Let TFA denote the full subcategory of torsion-free abelian groups, A of abelian groups, and Set the usual category of sets.
Take it as given that the forgetful functor $F:$ A $to$ Set and the inclusion functor $i:$ TFA $to$ A are monadic.
Show that, however, the composition $F circ i:$ TFA $to$
Set, is not monadic, using Beck's monadicity theorem.
Recall that Beck's monadicity theorem states that a functor $G: A to B$ is monadic if and only if $G$ has a left adjoint, it reflects isomorphisms, and every $G$ split pair admits a coequalizer in $A$, which $G$ preserves.
Where I'm at:
It is clear that $F circ i$ has a left adjoint, and that it reflects isomorphisms, and so it is the last and final condition that must be false.
For it to be false, we must find a pair of group homomorphisms, between torsion-free abelian groups, $G_1overset{f}{underset{g}rightrightarrows}G_2$, such that as functions on the respective sets, they admit a split fork:
there is $C in $ Set, $e: G_2 rightleftarrows C:s$ and $t:G_2 to G_1$ s.t:
$e circ s = id_{C}$
$g circ t = s circ e$
$f circ t = id_{G_2}$
And that this set $C$ will not have the canonic structure of a torsion-free abelian group. (Or simply that $G_1, G_2$ have no coequalizer in TFA)
Note that, since A is cocomplete, this pair $f,g$ will admit a coequalizer (in A), so this set $C$ may have the structure of an abelian group.
So my question is, can you help find such an example? Or is my reasoning off?
I've tried several with $mathbb{Z}$, and variants of it, but in order to get that $C$ is not torsion free, I need that $f-g$ will have a big enough image so that $mathbb{Z}/Im(f-g)$ will be finite. I think this specific example is hopeless though.
category-theory adjoint-functors monads forgetful-functors
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
$endgroup$
Disclaimer: this question already has a solution here:
Composition of monadic functors may not be monadic
. However, I would like to understand how to solve this using another characterisation of monadic functors - Beck's monadicity theorem.
Let TFA denote the full subcategory of torsion-free abelian groups, A of abelian groups, and Set the usual category of sets.
Take it as given that the forgetful functor $F:$ A $to$ Set and the inclusion functor $i:$ TFA $to$ A are monadic.
Show that, however, the composition $F circ i:$ TFA $to$
Set, is not monadic, using Beck's monadicity theorem.
Recall that Beck's monadicity theorem states that a functor $G: A to B$ is monadic if and only if $G$ has a left adjoint, it reflects isomorphisms, and every $G$ split pair admits a coequalizer in $A$, which $G$ preserves.
Where I'm at:
It is clear that $F circ i$ has a left adjoint, and that it reflects isomorphisms, and so it is the last and final condition that must be false.
For it to be false, we must find a pair of group homomorphisms, between torsion-free abelian groups, $G_1overset{f}{underset{g}rightrightarrows}G_2$, such that as functions on the respective sets, they admit a split fork:
there is $C in $ Set, $e: G_2 rightleftarrows C:s$ and $t:G_2 to G_1$ s.t:
$e circ s = id_{C}$
$g circ t = s circ e$
$f circ t = id_{G_2}$
And that this set $C$ will not have the canonic structure of a torsion-free abelian group. (Or simply that $G_1, G_2$ have no coequalizer in TFA)
Note that, since A is cocomplete, this pair $f,g$ will admit a coequalizer (in A), so this set $C$ may have the structure of an abelian group.
So my question is, can you help find such an example? Or is my reasoning off?
I've tried several with $mathbb{Z}$, and variants of it, but in order to get that $C$ is not torsion free, I need that $f-g$ will have a big enough image so that $mathbb{Z}/Im(f-g)$ will be finite. I think this specific example is hopeless though.
category-theory adjoint-functors monads forgetful-functors
category-theory adjoint-functors monads forgetful-functors
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
edited Jan 15 at 13:39
Arnaud D.
15.9k52443
15.9k52443
asked Jan 14 at 19:59
MariahMariah
1,5051618
asked Jan 14 at 19:59
MariahMariah
1,5051618
asked Jan 14 at 19:59
MariahMariah
1,5051618
1,5051618
add a comment |
add a comment |
1 Answer
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$begingroup$
Your reasoning is correct, and your idea to look for a case where the coequalizer in $mathbf{A}$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $mathbb{Z}/nmathbb{Z}$ is the quotient of $mathbb{Z}$ by the subgroup $nmathbb{Z}$. This means that its underlying set is the set of equivalence classes of $mathbb{Z}$ under the relation $R$ defined (as a subset of $mathbb{Z}times mathbb{Z}$) by
$$R={(x,y)in Bbb Ztimes Bbb Z mid x-yin nmathbb{Z}};$$
this set of equivalence classes is the coequalizer of the restrictions of the product projections to $R$, and it splits in $mathbf{Set}$.
In fact $R$ is a subgroup of $mathbb{Z}times mathbb{Z}$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $mathbf{Set}$ (or in $mathbf{A}$) is $mathbb{Z}/nmathbb{Z}$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $mathbf{TFA}to mathbf{Set}$ (but it does exist : in fact it is the trivial group).
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
add a comment |
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$begingroup$
Your reasoning is correct, and your idea to look for a case where the coequalizer in $mathbf{A}$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $mathbb{Z}/nmathbb{Z}$ is the quotient of $mathbb{Z}$ by the subgroup $nmathbb{Z}$. This means that its underlying set is the set of equivalence classes of $mathbb{Z}$ under the relation $R$ defined (as a subset of $mathbb{Z}times mathbb{Z}$) by
$$R={(x,y)in Bbb Ztimes Bbb Z mid x-yin nmathbb{Z}};$$
this set of equivalence classes is the coequalizer of the restrictions of the product projections to $R$, and it splits in $mathbf{Set}$.
In fact $R$ is a subgroup of $mathbb{Z}times mathbb{Z}$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $mathbf{Set}$ (or in $mathbf{A}$) is $mathbb{Z}/nmathbb{Z}$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $mathbf{TFA}to mathbf{Set}$ (but it does exist : in fact it is the trivial group).
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
add a comment |
$begingroup$
Your reasoning is correct, and your idea to look for a case where the coequalizer in $mathbf{A}$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $mathbb{Z}/nmathbb{Z}$ is the quotient of $mathbb{Z}$ by the subgroup $nmathbb{Z}$. This means that its underlying set is the set of equivalence classes of $mathbb{Z}$ under the relation $R$ defined (as a subset of $mathbb{Z}times mathbb{Z}$) by
$$R={(x,y)in Bbb Ztimes Bbb Z mid x-yin nmathbb{Z}};$$
this set of equivalence classes is the coequalizer of the restrictions of the product projections to $R$, and it splits in $mathbf{Set}$.
In fact $R$ is a subgroup of $mathbb{Z}times mathbb{Z}$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $mathbf{Set}$ (or in $mathbf{A}$) is $mathbb{Z}/nmathbb{Z}$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $mathbf{TFA}to mathbf{Set}$ (but it does exist : in fact it is the trivial group).
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
add a comment |
$begingroup$
Your reasoning is correct, and your idea to look for a case where the coequalizer in $mathbf{A}$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $mathbb{Z}/nmathbb{Z}$ is the quotient of $mathbb{Z}$ by the subgroup $nmathbb{Z}$. This means that its underlying set is the set of equivalence classes of $mathbb{Z}$ under the relation $R$ defined (as a subset of $mathbb{Z}times mathbb{Z}$) by
$$R={(x,y)in Bbb Ztimes Bbb Z mid x-yin nmathbb{Z}};$$
this set of equivalence classes is the coequalizer of the restrictions of the product projections to $R$, and it splits in $mathbf{Set}$.
In fact $R$ is a subgroup of $mathbb{Z}times mathbb{Z}$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $mathbf{Set}$ (or in $mathbf{A}$) is $mathbb{Z}/nmathbb{Z}$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $mathbf{TFA}to mathbf{Set}$ (but it does exist : in fact it is the trivial group).
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
Your reasoning is correct, and your idea to look for a case where the coequalizer in $mathbf{A}$ is not torsion-free is the right one. Here's a simple way to obtain this : you know that the group $mathbb{Z}/nmathbb{Z}$ is the quotient of $mathbb{Z}$ by the subgroup $nmathbb{Z}$. This means that its underlying set is the set of equivalence classes of $mathbb{Z}$ under the relation $R$ defined (as a subset of $mathbb{Z}times mathbb{Z}$) by
$$R={(x,y)in Bbb Ztimes Bbb Z mid x-yin nmathbb{Z}};$$
this set of equivalence classes is the coequalizer of the restrictions of the product projections to $R$, and it splits in $mathbf{Set}$.
In fact $R$ is a subgroup of $mathbb{Z}times mathbb{Z}$, and in particular it's a torsion-free group ! Moreover the restriction of the projections are group homomorphisms. So you have two group homomorphisms between torsion-free abelian groups, and their coequalizer in $mathbf{Set}$ (or in $mathbf{A}$) is $mathbb{Z}/nmathbb{Z}$, which is not torsion-free. As a consequence, the coequalizer of the two morphisms will not be preserved by the forgetful functor $mathbf{TFA}to mathbf{Set}$ (but it does exist : in fact it is the trivial group).
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
edited Jan 15 at 13:43
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
answered Jan 15 at 13:38
Arnaud D.Arnaud D.
15.9k52443
15.9k52443
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
add a comment |
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
$begingroup$
Thank you, I was hovering around this idea, but couldn't make it work. Taking $R$ itself as one of the groups is what makes this work easily, and what I didn't think of doing.
$endgroup$
– Mariah
Jan 15 at 14:25
add a comment |
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