Ternary function in axiomatization of category theory
$begingroup$
I am interested in axiomatizing category theory (for example, the category of categories or category of sets just like Lawvere did). I went through some sources and everywhere I looked I saw that the composition was axiomatized instead of $circ(u,v)$ which would be binary function but using ternary function $Gamma(u,v,w)$ which is interpreted in metamathematics as "$u circ v = w$". Why is that so? I don't see why we choose this because I feel that $circ$ is a more natural choice and equality is already assumed while axiomatizing category theory.
category-theory first-order-logic
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
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add a comment |
$begingroup$
I am interested in axiomatizing category theory (for example, the category of categories or category of sets just like Lawvere did). I went through some sources and everywhere I looked I saw that the composition was axiomatized instead of $circ(u,v)$ which would be binary function but using ternary function $Gamma(u,v,w)$ which is interpreted in metamathematics as "$u circ v = w$". Why is that so? I don't see why we choose this because I feel that $circ$ is a more natural choice and equality is already assumed while axiomatizing category theory.
category-theory first-order-logic
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
$endgroup$
add a comment |
$begingroup$
I am interested in axiomatizing category theory (for example, the category of categories or category of sets just like Lawvere did). I went through some sources and everywhere I looked I saw that the composition was axiomatized instead of $circ(u,v)$ which would be binary function but using ternary function $Gamma(u,v,w)$ which is interpreted in metamathematics as "$u circ v = w$". Why is that so? I don't see why we choose this because I feel that $circ$ is a more natural choice and equality is already assumed while axiomatizing category theory.
category-theory first-order-logic
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
$endgroup$
I am interested in axiomatizing category theory (for example, the category of categories or category of sets just like Lawvere did). I went through some sources and everywhere I looked I saw that the composition was axiomatized instead of $circ(u,v)$ which would be binary function but using ternary function $Gamma(u,v,w)$ which is interpreted in metamathematics as "$u circ v = w$". Why is that so? I don't see why we choose this because I feel that $circ$ is a more natural choice and equality is already assumed while axiomatizing category theory.
category-theory first-order-logic
category-theory first-order-logic
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
asked Jan 10 at 6:05
Daniels KrimansDaniels Krimans
51739
51739
add a comment |
add a comment |
1 Answer
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In traditional first-order logic, functions must be total. So, if $circ$ were a binary operation, $circ(u,v)$ would need to be defined for any two morphisms $u$ and $v$. This is obviously not desirable, since composition should only be defined when the domain of one morphism matches the codomain of the other. So, since it is not a total function, you must encode $circ$ instead as a relation.
See Model theory: What is the signature of `Category theory` for some related discussion.
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
$endgroup$
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
add a comment |
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$begingroup$
In traditional first-order logic, functions must be total. So, if $circ$ were a binary operation, $circ(u,v)$ would need to be defined for any two morphisms $u$ and $v$. This is obviously not desirable, since composition should only be defined when the domain of one morphism matches the codomain of the other. So, since it is not a total function, you must encode $circ$ instead as a relation.
See Model theory: What is the signature of `Category theory` for some related discussion.
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
$endgroup$
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
add a comment |
$begingroup$
In traditional first-order logic, functions must be total. So, if $circ$ were a binary operation, $circ(u,v)$ would need to be defined for any two morphisms $u$ and $v$. This is obviously not desirable, since composition should only be defined when the domain of one morphism matches the codomain of the other. So, since it is not a total function, you must encode $circ$ instead as a relation.
See Model theory: What is the signature of `Category theory` for some related discussion.
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
$endgroup$
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
add a comment |
$begingroup$
In traditional first-order logic, functions must be total. So, if $circ$ were a binary operation, $circ(u,v)$ would need to be defined for any two morphisms $u$ and $v$. This is obviously not desirable, since composition should only be defined when the domain of one morphism matches the codomain of the other. So, since it is not a total function, you must encode $circ$ instead as a relation.
See Model theory: What is the signature of `Category theory` for some related discussion.
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
$endgroup$
In traditional first-order logic, functions must be total. So, if $circ$ were a binary operation, $circ(u,v)$ would need to be defined for any two morphisms $u$ and $v$. This is obviously not desirable, since composition should only be defined when the domain of one morphism matches the codomain of the other. So, since it is not a total function, you must encode $circ$ instead as a relation.
See Model theory: What is the signature of `Category theory` for some related discussion.
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
answered Jan 10 at 6:13
Eric WofseyEric Wofsey
182k12209337
182k12209337
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
add a comment |
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
1
1
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
$begingroup$
I am not sure I understand why this isn't desirable. Can't I just add as an axiom: $circ(u,v) = w implies c(u) = d(v)$, where $c$ is interpreted as codomain and $d$ as domain. Then, even though it is defined for all $u,v$ I can use it only when codomain and domain match.
$endgroup$
– Daniels Krimans
Jan 10 at 6:23
2
2
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
$begingroup$
Assuming you want the variables in that axiom to be universally quantified, that axiom implies that $c(u)=d(v)$ for all $u$ and $v$. Indeed, you can substitute the term $circ(u,v)$ for $w$ to make the antecedent of the implication true no matter what $u$ and $v$ are.
$endgroup$
– Eric Wofsey
Jan 10 at 6:24
1
1
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
$begingroup$
you are right. Thanks for the help! Can you maybe give some more references about first order axiomatization of categories?
$endgroup$
– Daniels Krimans
Jan 10 at 6:26
add a comment |
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