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Problem in understanding principle of duality in category theory.
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$mathbfdef:$ a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p^op$ iff $mathcalC^op$ has $p$
$mathbfquestion 1:$ is the above def is equivalent to the statement: a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p$ iff $mathcalC^op$ has $p^op$ . And why ?
$bullet$ Let $ Pi= iin I $ be a set of properties and let $ Pi^op= q_i^op $ be the set of dual properties. Let $p$
be a single property. Consider the statement
$1)$ If a category $mathcalC$ has $Pi$, then it also has $p$.
Since all categories have the form $mathcalC^op$ for some category $mathcalC$, this statement is logically equivalent to the
statement
$2)$ If a category $mathcalC^op$ has $Pi$, then it also has $p$.
and this is logically equivalent to
$3)$If a category $mathcalC$ has $Pi^op$ , then it also has $p^op$
$mathbfquestion 2:$ Why $2$ and $3$ are equivalent.please explain with as much detail as possible. Thanks
logic category-theory proof-explanation
asked Jul 17 at 4:57
bumba
246
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up vote
0
down vote
favorite
$mathbfdef:$ a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p^op$ iff $mathcalC^op$ has $p$
$mathbfquestion 1:$ is the above def is equivalent to the statement: a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p$ iff $mathcalC^op$ has $p^op$ . And why ?
$bullet$ Let $ Pi= iin I $ be a set of properties and let $ Pi^op= q_i^op $ be the set of dual properties. Let $p$
be a single property. Consider the statement
$1)$ If a category $mathcalC$ has $Pi$, then it also has $p$.
Since all categories have the form $mathcalC^op$ for some category $mathcalC$, this statement is logically equivalent to the
statement
$2)$ If a category $mathcalC^op$ has $Pi$, then it also has $p$.
and this is logically equivalent to
$3)$If a category $mathcalC$ has $Pi^op$ , then it also has $p^op$
$mathbfquestion 2:$ Why $2$ and $3$ are equivalent.please explain with as much detail as possible. Thanks
logic category-theory proof-explanation
asked Jul 17 at 4:57
bumba
246
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$mathbfdef:$ a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p^op$ iff $mathcalC^op$ has $p$
$mathbfquestion 1:$ is the above def is equivalent to the statement: a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p$ iff $mathcalC^op$ has $p^op$ . And why ?
$bullet$ Let $ Pi= iin I $ be a set of properties and let $ Pi^op= q_i^op $ be the set of dual properties. Let $p$
be a single property. Consider the statement
$1)$ If a category $mathcalC$ has $Pi$, then it also has $p$.
Since all categories have the form $mathcalC^op$ for some category $mathcalC$, this statement is logically equivalent to the
statement
$2)$ If a category $mathcalC^op$ has $Pi$, then it also has $p$.
and this is logically equivalent to
$3)$If a category $mathcalC$ has $Pi^op$ , then it also has $p^op$
$mathbfquestion 2:$ Why $2$ and $3$ are equivalent.please explain with as much detail as possible. Thanks
logic category-theory proof-explanation
asked Jul 17 at 4:57
bumba
246
$mathbfdef:$ a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p^op$ iff $mathcalC^op$ has $p$
$mathbfquestion 1:$ is the above def is equivalent to the statement: a property $p^op$ is a dual property to $p$ if for all categories $mathcalC$, $mathcalC$ has $p$ iff $mathcalC^op$ has $p^op$ . And why ?
$bullet$ Let $ Pi= iin I $ be a set of properties and let $ Pi^op= q_i^op $ be the set of dual properties. Let $p$
be a single property. Consider the statement
$1)$ If a category $mathcalC$ has $Pi$, then it also has $p$.
Since all categories have the form $mathcalC^op$ for some category $mathcalC$, this statement is logically equivalent to the
statement
$2)$ If a category $mathcalC^op$ has $Pi$, then it also has $p$.
and this is logically equivalent to
$3)$If a category $mathcalC$ has $Pi^op$ , then it also has $p^op$
$mathbfquestion 2:$ Why $2$ and $3$ are equivalent.please explain with as much detail as possible. Thanks
logic category-theory proof-explanation
asked Jul 17 at 4:57
bumba
246
asked Jul 17 at 4:57
bumba
246
asked Jul 17 at 4:57
bumba
246
asked Jul 17 at 4:57
bumba
246
246
add a comment |Â
add a comment |Â
1 Answer
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Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^op$; but also every category is the opposite of some category, because $(C^op)^op= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^op$'s and all $C^op$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $Pi^op$. Then $C^op$ has $Pi$ by definition of dual property. Thus $C^op$ has $p$ by 2. Thus $C$ has $p^op$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.
answered Jul 17 at 7:52
Max
10.3k1736
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^op$; but also every category is the opposite of some category, because $(C^op)^op= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^op$'s and all $C^op$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $Pi^op$. Then $C^op$ has $Pi$ by definition of dual property. Thus $C^op$ has $p$ by 2. Thus $C$ has $p^op$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.
answered Jul 17 at 7:52
Max
10.3k1736
add a comment |Â
up vote
2
down vote
Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^op$; but also every category is the opposite of some category, because $(C^op)^op= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^op$'s and all $C^op$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $Pi^op$. Then $C^op$ has $Pi$ by definition of dual property. Thus $C^op$ has $p$ by 2. Thus $C$ has $p^op$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.
answered Jul 17 at 7:52
Max
10.3k1736
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^op$; but also every category is the opposite of some category, because $(C^op)^op= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^op$'s and all $C^op$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $Pi^op$. Then $C^op$ has $Pi$ by definition of dual property. Thus $C^op$ has $p$ by 2. Thus $C$ has $p^op$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.
answered Jul 17 at 7:52
Max
10.3k1736
Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^op$; but also every category is the opposite of some category, because $(C^op)^op= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^op$'s and all $C^op$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $Pi^op$. Then $C^op$ has $Pi$ by definition of dual property. Thus $C^op$ has $p$ by 2. Thus $C$ has $p^op$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.
answered Jul 17 at 7:52
Max
10.3k1736
answered Jul 17 at 7:52
Max
10.3k1736
answered Jul 17 at 7:52
Max
10.3k1736
answered Jul 17 at 7:52
Max
10.3k1736
10.3k1736
add a comment |Â
add a comment |Â
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