Does the set Hom(X,Y), in any category, always form an Abelian group?
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Given any category C and any two C-objects X and Y, is it always the case that the set Hom(X, Y) of morphisms from X to Y form an Abelian group? If so, and to be clear, is this always the case regardless of whether or not C itself is Abelian, Ab?
category-theory
asked Nov 15 at 16:25
Craig Dean
361
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Given any category C and any two C-objects X and Y, is it always the case that the set Hom(X, Y) of morphisms from X to Y form an Abelian group? If so, and to be clear, is this always the case regardless of whether or not C itself is Abelian, Ab?
category-theory
asked Nov 15 at 16:25
Craig Dean
361
1
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
2
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given any category C and any two C-objects X and Y, is it always the case that the set Hom(X, Y) of morphisms from X to Y form an Abelian group? If so, and to be clear, is this always the case regardless of whether or not C itself is Abelian, Ab?
category-theory
asked Nov 15 at 16:25
Craig Dean
361
Given any category C and any two C-objects X and Y, is it always the case that the set Hom(X, Y) of morphisms from X to Y form an Abelian group? If so, and to be clear, is this always the case regardless of whether or not C itself is Abelian, Ab?
category-theory
category-theory
asked Nov 15 at 16:25
Craig Dean
361
asked Nov 15 at 16:25
Craig Dean
361
asked Nov 15 at 16:25
Craig Dean
361
asked Nov 15 at 16:25
Craig Dean
361
asked Nov 15 at 16:25
Craig Dean
361
361
1
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
2
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38
add a comment |
1
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
2
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38
1
1
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
2
2
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
Other answers have argued, correctly, that there is in general no way to put a natural group structure on the set of morphisms between two objects.
To see that there is literally no way to do so, note that the set of morphisms from $X$ to $Y$ may be empty, and the empty set cannot be given an abelian group structure.
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
add a comment |
up vote
2
down vote
You can put an abelian group structure on pretty much any set you like, but it won't necessarily give you anything meaningful.
A category $mathcal{C}$ whose hom sets $mathcal{C}(X,Y)$ are endowed with an abelian group structure $(+,0)$. that interacts nicely with the categorical structure of $mathcal{C}$, is called an $mathbf{Ab}$-enriched category (or one of a number of other names, such as a ringoid).
By 'interacts nicely', I mean that composition $circ : mathcal{C}(X,Y) times mathcal{C}(Y,Z) to mathcal{C}(X,Z)$ is bilinear with respect to the three group operations involved.
This is a fairly strong condition that is not enjoyed by most categories. For example, in an $mathbf{Ab}$-enriched category, any initial object is also terminal and, more generally, finite products coincide with finite coproducts.
$mathbf{Ab}$-enriched categories are well studied. For example:
- Preadditive categories are $mathbf{Ab}$-enriched categories with a zero object (= initial and terminal)
- Additive categories are preadditive categories with finite biproducts;
- Abelian categories are additive categories with all kernels and cokernels, such that every monomorphism is a kernel and every epimorphism is a cokernel.
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
add a comment |
up vote
0
down vote
No. In general, there's no operation on $mathbf{Hom}(X,Y)$. To be specific: if we're working in the category of sets, what operation are you putting on $mathbf{Hom}({emptyset},{emptyset,{emptyset}})$? Obviously, you can put an abelian group structure on any set you like (or rather, I'll say on any reasonable set, to avoid messing around with large cardinal stuff), and Hom-sets are always sets, so you could always put an abelian group structure on it, it's just not useful and doesn't tell you anything about the categories or the objects that you didn't already know from properties of the underlying set.
It is true for Abelian categories, however (you can embed your category in $mathbf{Ab}$, and use the operations there).
answered Nov 15 at 16:30
user3482749
1,709411
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Other answers have argued, correctly, that there is in general no way to put a natural group structure on the set of morphisms between two objects.
To see that there is literally no way to do so, note that the set of morphisms from $X$ to $Y$ may be empty, and the empty set cannot be given an abelian group structure.
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
add a comment |
up vote
3
down vote
Other answers have argued, correctly, that there is in general no way to put a natural group structure on the set of morphisms between two objects.
To see that there is literally no way to do so, note that the set of morphisms from $X$ to $Y$ may be empty, and the empty set cannot be given an abelian group structure.
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
add a comment |
up vote
3
down vote
up vote
3
down vote
Other answers have argued, correctly, that there is in general no way to put a natural group structure on the set of morphisms between two objects.
To see that there is literally no way to do so, note that the set of morphisms from $X$ to $Y$ may be empty, and the empty set cannot be given an abelian group structure.
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
Other answers have argued, correctly, that there is in general no way to put a natural group structure on the set of morphisms between two objects.
To see that there is literally no way to do so, note that the set of morphisms from $X$ to $Y$ may be empty, and the empty set cannot be given an abelian group structure.
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
answered Nov 15 at 20:40
Jeremy Rickard
15.9k11642
15.9k11642
add a comment |
add a comment |
up vote
2
down vote
You can put an abelian group structure on pretty much any set you like, but it won't necessarily give you anything meaningful.
A category $mathcal{C}$ whose hom sets $mathcal{C}(X,Y)$ are endowed with an abelian group structure $(+,0)$. that interacts nicely with the categorical structure of $mathcal{C}$, is called an $mathbf{Ab}$-enriched category (or one of a number of other names, such as a ringoid).
By 'interacts nicely', I mean that composition $circ : mathcal{C}(X,Y) times mathcal{C}(Y,Z) to mathcal{C}(X,Z)$ is bilinear with respect to the three group operations involved.
This is a fairly strong condition that is not enjoyed by most categories. For example, in an $mathbf{Ab}$-enriched category, any initial object is also terminal and, more generally, finite products coincide with finite coproducts.
$mathbf{Ab}$-enriched categories are well studied. For example:
- Preadditive categories are $mathbf{Ab}$-enriched categories with a zero object (= initial and terminal)
- Additive categories are preadditive categories with finite biproducts;
- Abelian categories are additive categories with all kernels and cokernels, such that every monomorphism is a kernel and every epimorphism is a cokernel.
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
add a comment |
up vote
2
down vote
You can put an abelian group structure on pretty much any set you like, but it won't necessarily give you anything meaningful.
A category $mathcal{C}$ whose hom sets $mathcal{C}(X,Y)$ are endowed with an abelian group structure $(+,0)$. that interacts nicely with the categorical structure of $mathcal{C}$, is called an $mathbf{Ab}$-enriched category (or one of a number of other names, such as a ringoid).
By 'interacts nicely', I mean that composition $circ : mathcal{C}(X,Y) times mathcal{C}(Y,Z) to mathcal{C}(X,Z)$ is bilinear with respect to the three group operations involved.
This is a fairly strong condition that is not enjoyed by most categories. For example, in an $mathbf{Ab}$-enriched category, any initial object is also terminal and, more generally, finite products coincide with finite coproducts.
$mathbf{Ab}$-enriched categories are well studied. For example:
- Preadditive categories are $mathbf{Ab}$-enriched categories with a zero object (= initial and terminal)
- Additive categories are preadditive categories with finite biproducts;
- Abelian categories are additive categories with all kernels and cokernels, such that every monomorphism is a kernel and every epimorphism is a cokernel.
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
add a comment |
up vote
2
down vote
up vote
2
down vote
You can put an abelian group structure on pretty much any set you like, but it won't necessarily give you anything meaningful.
A category $mathcal{C}$ whose hom sets $mathcal{C}(X,Y)$ are endowed with an abelian group structure $(+,0)$. that interacts nicely with the categorical structure of $mathcal{C}$, is called an $mathbf{Ab}$-enriched category (or one of a number of other names, such as a ringoid).
By 'interacts nicely', I mean that composition $circ : mathcal{C}(X,Y) times mathcal{C}(Y,Z) to mathcal{C}(X,Z)$ is bilinear with respect to the three group operations involved.
This is a fairly strong condition that is not enjoyed by most categories. For example, in an $mathbf{Ab}$-enriched category, any initial object is also terminal and, more generally, finite products coincide with finite coproducts.
$mathbf{Ab}$-enriched categories are well studied. For example:
- Preadditive categories are $mathbf{Ab}$-enriched categories with a zero object (= initial and terminal)
- Additive categories are preadditive categories with finite biproducts;
- Abelian categories are additive categories with all kernels and cokernels, such that every monomorphism is a kernel and every epimorphism is a cokernel.
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
You can put an abelian group structure on pretty much any set you like, but it won't necessarily give you anything meaningful.
A category $mathcal{C}$ whose hom sets $mathcal{C}(X,Y)$ are endowed with an abelian group structure $(+,0)$. that interacts nicely with the categorical structure of $mathcal{C}$, is called an $mathbf{Ab}$-enriched category (or one of a number of other names, such as a ringoid).
By 'interacts nicely', I mean that composition $circ : mathcal{C}(X,Y) times mathcal{C}(Y,Z) to mathcal{C}(X,Z)$ is bilinear with respect to the three group operations involved.
This is a fairly strong condition that is not enjoyed by most categories. For example, in an $mathbf{Ab}$-enriched category, any initial object is also terminal and, more generally, finite products coincide with finite coproducts.
$mathbf{Ab}$-enriched categories are well studied. For example:
- Preadditive categories are $mathbf{Ab}$-enriched categories with a zero object (= initial and terminal)
- Additive categories are preadditive categories with finite biproducts;
- Abelian categories are additive categories with all kernels and cokernels, such that every monomorphism is a kernel and every epimorphism is a cokernel.
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
edited Nov 15 at 23:45
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
answered Nov 15 at 17:03
Clive Newstead
49.3k472132
49.3k472132
add a comment |
add a comment |
up vote
0
down vote
No. In general, there's no operation on $mathbf{Hom}(X,Y)$. To be specific: if we're working in the category of sets, what operation are you putting on $mathbf{Hom}({emptyset},{emptyset,{emptyset}})$? Obviously, you can put an abelian group structure on any set you like (or rather, I'll say on any reasonable set, to avoid messing around with large cardinal stuff), and Hom-sets are always sets, so you could always put an abelian group structure on it, it's just not useful and doesn't tell you anything about the categories or the objects that you didn't already know from properties of the underlying set.
It is true for Abelian categories, however (you can embed your category in $mathbf{Ab}$, and use the operations there).
answered Nov 15 at 16:30
user3482749
1,709411
add a comment |
up vote
0
down vote
No. In general, there's no operation on $mathbf{Hom}(X,Y)$. To be specific: if we're working in the category of sets, what operation are you putting on $mathbf{Hom}({emptyset},{emptyset,{emptyset}})$? Obviously, you can put an abelian group structure on any set you like (or rather, I'll say on any reasonable set, to avoid messing around with large cardinal stuff), and Hom-sets are always sets, so you could always put an abelian group structure on it, it's just not useful and doesn't tell you anything about the categories or the objects that you didn't already know from properties of the underlying set.
It is true for Abelian categories, however (you can embed your category in $mathbf{Ab}$, and use the operations there).
answered Nov 15 at 16:30
user3482749
1,709411
add a comment |
up vote
0
down vote
up vote
0
down vote
No. In general, there's no operation on $mathbf{Hom}(X,Y)$. To be specific: if we're working in the category of sets, what operation are you putting on $mathbf{Hom}({emptyset},{emptyset,{emptyset}})$? Obviously, you can put an abelian group structure on any set you like (or rather, I'll say on any reasonable set, to avoid messing around with large cardinal stuff), and Hom-sets are always sets, so you could always put an abelian group structure on it, it's just not useful and doesn't tell you anything about the categories or the objects that you didn't already know from properties of the underlying set.
It is true for Abelian categories, however (you can embed your category in $mathbf{Ab}$, and use the operations there).
answered Nov 15 at 16:30
user3482749
1,709411
No. In general, there's no operation on $mathbf{Hom}(X,Y)$. To be specific: if we're working in the category of sets, what operation are you putting on $mathbf{Hom}({emptyset},{emptyset,{emptyset}})$? Obviously, you can put an abelian group structure on any set you like (or rather, I'll say on any reasonable set, to avoid messing around with large cardinal stuff), and Hom-sets are always sets, so you could always put an abelian group structure on it, it's just not useful and doesn't tell you anything about the categories or the objects that you didn't already know from properties of the underlying set.
It is true for Abelian categories, however (you can embed your category in $mathbf{Ab}$, and use the operations there).
answered Nov 15 at 16:30
user3482749
1,709411
answered Nov 15 at 16:30
user3482749
1,709411
answered Nov 15 at 16:30
user3482749
1,709411
answered Nov 15 at 16:30
user3482749
1,709411
1,709411
add a comment |
add a comment |
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1
It's rarely true. Consider the category of sets.
– Randall
Nov 15 at 16:27
No -- or at least, not in a way that has anything to do with the category. Categories like $mathsf{Set}, mathsf{Top}, mathsf{Graph}$ form some counterexamples.
– Mees de Vries
Nov 15 at 16:28
In general you can at most say that it is a set.
– drhab
Nov 15 at 16:28
What would the group operation be ? If $X$ and $Y$ are two different objects, and $f$ and $g$ are two different morphisms from $X$ to $Y$, how would you "compose" $f$ and $g$ ?
– gandalf61
Nov 15 at 16:32
2
Such category is called $mathbf{Ab}$-enriched category.
– Seewoo Lee
Nov 15 at 16:38