Mixing
A new category $C^*$ from a given category $C$
$begingroup$
For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $uto v$ are all finite sequences $(a_1,dots,a_n)$ of morphisms $a_iin Mor(C)$ such that $a_1:uto x$ and $a_n:yto v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,dots,a_m)(b_1,dots,b_n)=(a_1,dots,a_mb_1,dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?
Update:
1) Motivation: the construction $Cmapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C cup (Ctimes C)$ modulo the relations which hold in $C$.
2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:
a) type $w$: they tell me I should compose certain morphisms of $C$ the result of which is the morphism $a_1$, say, and thereby I run through the underlying graph $Gamma$ of $C$
b) type $w^{mathfrak m}$: they say I should jump elsewhere in the graph $Gamma$
The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $dots a_i,a_{i+1}dots$ and $dots a_ia_{i+1}dots$.
category-theory terminology
asked Jan 15 at 19:09
user 59363user 59363
1,249410
$endgroup$
|
show 3 more comments
$begingroup$
For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $uto v$ are all finite sequences $(a_1,dots,a_n)$ of morphisms $a_iin Mor(C)$ such that $a_1:uto x$ and $a_n:yto v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,dots,a_m)(b_1,dots,b_n)=(a_1,dots,a_mb_1,dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?
Update:
1) Motivation: the construction $Cmapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C cup (Ctimes C)$ modulo the relations which hold in $C$.
2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:
a) type $w$: they tell me I should compose certain morphisms of $C$ the result of which is the morphism $a_1$, say, and thereby I run through the underlying graph $Gamma$ of $C$
b) type $w^{mathfrak m}$: they say I should jump elsewhere in the graph $Gamma$
The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $dots a_i,a_{i+1}dots$ and $dots a_ia_{i+1}dots$.
category-theory terminology
asked Jan 15 at 19:09
user 59363user 59363
1,249410
$endgroup$
$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
$begingroup$
@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
$endgroup$
– user 59363
Jan 15 at 19:23
$begingroup$
And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
$endgroup$
– Dog_69
Jan 15 at 20:17
$begingroup$
Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
$endgroup$
– Arnaud D.
Jan 15 at 21:45
$begingroup$
@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
$endgroup$
– user 59363
Jan 16 at 10:42
|
show 3 more comments
$begingroup$
For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $uto v$ are all finite sequences $(a_1,dots,a_n)$ of morphisms $a_iin Mor(C)$ such that $a_1:uto x$ and $a_n:yto v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,dots,a_m)(b_1,dots,b_n)=(a_1,dots,a_mb_1,dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?
Update:
1) Motivation: the construction $Cmapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C cup (Ctimes C)$ modulo the relations which hold in $C$.
2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:
a) type $w$: they tell me I should compose certain morphisms of $C$ the result of which is the morphism $a_1$, say, and thereby I run through the underlying graph $Gamma$ of $C$
b) type $w^{mathfrak m}$: they say I should jump elsewhere in the graph $Gamma$
The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $dots a_i,a_{i+1}dots$ and $dots a_ia_{i+1}dots$.
category-theory terminology
asked Jan 15 at 19:09
user 59363user 59363
1,249410
$endgroup$
For a given category $C$ define the category $C^*$ as follows: the objects of $C^*$ are those of $C$; for given objects $u,v$, the $C^*$-morphisms $uto v$ are all finite sequences $(a_1,dots,a_n)$ of morphisms $a_iin Mor(C)$ such that $a_1:uto x$ and $a_n:yto v$ where $x$ and $y$ are arbitrary objects (no other restrictions). The composition of two composable morphisms is $$(a_1,dots,a_m)(b_1,dots,b_n)=(a_1,dots,a_mb_1,dots,b_n)$$ (where $a_mb_1$ are composed in $C$). My question: is there an established name for the category $C^*$?
Update:
1) Motivation: the construction $Cmapsto C^*$ appears as a tool of proof in my research and I wanted to know if, where and for which purpose this construction appears in literature, in order to insert some references. Intuitively I would say this is somehow the free category generated by $C cup (Ctimes C)$ modulo the relations which hold in $C$.
2) Intuitive background: I have algebraic terms of a certain type which can be interpreted as instructions what to do in a certain category $C$. There are two sorts of instructions:
a) type $w$: they tell me I should compose certain morphisms of $C$ the result of which is the morphism $a_1$, say, and thereby I run through the underlying graph $Gamma$ of $C$
b) type $w^{mathfrak m}$: they say I should jump elsewhere in the graph $Gamma$
The application of such instructions alternatingly ends up with a tuple $(a_1,a_2,dots,a_n)$ as in the question. The category $C^*$ seems to model exactly this behaviour. For technical reasons, I allow in type b) "empty jumps", that is, even if two consecutive morphisms $a_i$ and $a_{i+1}$ are composable in $C$ I would like to distinguish between $dots a_i,a_{i+1}dots$ and $dots a_ia_{i+1}dots$.
category-theory terminology
category-theory terminology
asked Jan 15 at 19:09
user 59363user 59363
1,249410
asked Jan 15 at 19:09
user 59363user 59363
1,249410
edited Jan 16 at 10:41
user 59363
asked Jan 15 at 19:09
user 59363user 59363
1,249410
asked Jan 15 at 19:09
user 59363user 59363
1,249410
asked Jan 15 at 19:09
user 59363user 59363
1,249410
1,249410
$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
$begingroup$
@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
$endgroup$
– user 59363
Jan 15 at 19:23
$begingroup$
And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
$endgroup$
– Dog_69
Jan 15 at 20:17
$begingroup$
Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
$endgroup$
– Arnaud D.
Jan 15 at 21:45
$begingroup$
@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
$endgroup$
– user 59363
Jan 16 at 10:42
|
show 3 more comments
$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
$begingroup$
@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
$endgroup$
– user 59363
Jan 15 at 19:23
$begingroup$
And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
$endgroup$
– Dog_69
Jan 15 at 20:17
$begingroup$
Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
$endgroup$
– Arnaud D.
Jan 15 at 21:45
$begingroup$
@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
$endgroup$
– user 59363
Jan 16 at 10:42
$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
$begingroup$
@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
$endgroup$
– user 59363
Jan 15 at 19:23
$begingroup$
@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
$endgroup$
– user 59363
Jan 15 at 19:23
$begingroup$
And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
$endgroup$
– Dog_69
Jan 15 at 20:17
$begingroup$
And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
$endgroup$
– Dog_69
Jan 15 at 20:17
$begingroup$
Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
$endgroup$
– Arnaud D.
Jan 15 at 21:45
$begingroup$
Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
$endgroup$
– Arnaud D.
Jan 15 at 21:45
$begingroup$
@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
$endgroup$
– user 59363
Jan 16 at 10:42
$begingroup$
@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
$endgroup$
– user 59363
Jan 16 at 10:42
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.
answered Jan 15 at 19:50
magmamagma
3,36911124
$endgroup$
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
|
show 2 more comments
Your Answer
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$begingroup$
It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.
answered Jan 15 at 19:50
magmamagma
3,36911124
$endgroup$
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
|
show 2 more comments
$begingroup$
It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.
answered Jan 15 at 19:50
magmamagma
3,36911124
$endgroup$
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
|
show 2 more comments
$begingroup$
It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.
answered Jan 15 at 19:50
magmamagma
3,36911124
$endgroup$
It looks to me like the two categories are exactly same! The objects are the same so take 2 arbitrary objects A and B, then you can see that the Hom(A,B) in C and C* are in bijection.
answered Jan 15 at 19:50
magmamagma
3,36911124
answered Jan 15 at 19:50
magmamagma
3,36911124
answered Jan 15 at 19:50
magmamagma
3,36911124
answered Jan 15 at 19:50
magmamagma
3,36911124
3,36911124
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
|
show 2 more comments
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
This is not the case. For example, if $C$ is finite, $C^*$ will be infinite (if $C$ has at least two different morphisms).
$endgroup$
– user 59363
Jan 15 at 20:03
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
I don't see this. Among the morphisms in $C^*$ from $A$ to $B$, you have all the morphisms in $C$ from $A$ to $B$ (regarded as sequences of length 1), and you also have sequence of greater length. Note that a directed path n $C$ from $A$ to $B$ is a morphism in $C^*$ and has not been identified with the composite of its members. Furthermore, the terms $a_i$ in a $C^*$-morphism needn't even line up to form a path in $C$.
$endgroup$
– Andreas Blass
Jan 15 at 20:05
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
Yes my answer is wrong, I will fix it when I can. The empty sequence of morphisms should be admitted too, as identity, correct?
$endgroup$
– magma
Jan 26 at 6:10
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
@magma No, the empty sequence is not admitted.
$endgroup$
– user 59363
Jan 27 at 22:37
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
$begingroup$
So what sequence is the identity on A?
$endgroup$
– magma
Feb 2 at 8:02
|
show 2 more comments
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$begingroup$
Does the composition have $n+m-1$ terms?
$endgroup$
– Dog_69
Jan 15 at 19:16
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@Dog_69 Yes, the adjacent morphisms $a_m$ and $b_1$ are composed in $C$, so $a_mb_1$ is one entry of the sequence.
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– user 59363
Jan 15 at 19:23
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And I suppose that two morphisms if $m=n$ and they agree in each component, isn't it?
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– Dog_69
Jan 15 at 20:17
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Do you have a particular reason to think that this category may have an established name? The definition looks a bit arbitrary to me...
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– Arnaud D.
Jan 15 at 21:45
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@Dog_69 No, for every $n$ there can be many tuples of length $n$ (I'm not sure if I understood the question correctly).
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– user 59363
Jan 16 at 10:42