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Is the category of pointed sets cartesian closed?
Clash Royale CLAN TAG#URR8PPP
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Let's call this category $mathbfPt$ (if there's a standard name for it then I don't know it). I have been stuck for a few days on this problem. There is a terminal object $(bullet,bullet)$ and products can be taken to be $(A,a)times (B,b)=(Atimes B, (a,b))$, so one might expect $mathbfPt$ to be cartesian closed. However, I strongly suspect that it's not. Intuitively, it seems to me that the functor $U:mathbfPttomathbfSet$ sending objects $(A,a)$ to the set $A$ and sending arrows to themselves (i.e. the functor which "forgets" about the specified points) might preserve exponentials. I've solved the problem if this were true, but I don't know how to go about showing it (I don't even know if it's actually correct, honestly).
Any hints would be great.
category-theory
asked Jul 18 at 18:18
D. Brogan
463311
add a comment |Â
up vote
2
down vote
favorite
Let's call this category $mathbfPt$ (if there's a standard name for it then I don't know it). I have been stuck for a few days on this problem. There is a terminal object $(bullet,bullet)$ and products can be taken to be $(A,a)times (B,b)=(Atimes B, (a,b))$, so one might expect $mathbfPt$ to be cartesian closed. However, I strongly suspect that it's not. Intuitively, it seems to me that the functor $U:mathbfPttomathbfSet$ sending objects $(A,a)$ to the set $A$ and sending arrows to themselves (i.e. the functor which "forgets" about the specified points) might preserve exponentials. I've solved the problem if this were true, but I don't know how to go about showing it (I don't even know if it's actually correct, honestly).
Any hints would be great.
category-theory
asked Jul 18 at 18:18
D. Brogan
463311
2
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let's call this category $mathbfPt$ (if there's a standard name for it then I don't know it). I have been stuck for a few days on this problem. There is a terminal object $(bullet,bullet)$ and products can be taken to be $(A,a)times (B,b)=(Atimes B, (a,b))$, so one might expect $mathbfPt$ to be cartesian closed. However, I strongly suspect that it's not. Intuitively, it seems to me that the functor $U:mathbfPttomathbfSet$ sending objects $(A,a)$ to the set $A$ and sending arrows to themselves (i.e. the functor which "forgets" about the specified points) might preserve exponentials. I've solved the problem if this were true, but I don't know how to go about showing it (I don't even know if it's actually correct, honestly).
Any hints would be great.
category-theory
asked Jul 18 at 18:18
D. Brogan
463311
Let's call this category $mathbfPt$ (if there's a standard name for it then I don't know it). I have been stuck for a few days on this problem. There is a terminal object $(bullet,bullet)$ and products can be taken to be $(A,a)times (B,b)=(Atimes B, (a,b))$, so one might expect $mathbfPt$ to be cartesian closed. However, I strongly suspect that it's not. Intuitively, it seems to me that the functor $U:mathbfPttomathbfSet$ sending objects $(A,a)$ to the set $A$ and sending arrows to themselves (i.e. the functor which "forgets" about the specified points) might preserve exponentials. I've solved the problem if this were true, but I don't know how to go about showing it (I don't even know if it's actually correct, honestly).
Any hints would be great.
category-theory
asked Jul 18 at 18:18
D. Brogan
463311
asked Jul 18 at 18:18
D. Brogan
463311
asked Jul 18 at 18:18
D. Brogan
463311
asked Jul 18 at 18:18
D. Brogan
463311
463311
2
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37
add a comment |Â
2
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37
2
2
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37
add a comment |Â
1 Answer
1
active
oldest
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up vote
6
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accepted
One way to define a cartesian closed category is that it's a category with finite products where every product functor $(-) times X$ has a right adjoint, namely the exponential functor $(-)^X$. A necessary condition for this to be the case is that $(-) times X$ preserves colimits; in other words, that products distribute over colimits, and in particular coproducts, when they exist.
But products in pointed sets don't distribute over coproducts. For finite pointed sets $A, B, C$ the coproduct satisfies $|A sqcup B| = |A| + |B| - 1$ and you can use this to show that $(A sqcup B) times C$ and $(A times C) sqcup (B times C)$ have different cardinalities in general.
Alternatively, you can use the fact that pointed sets have a zero object. Here's a fun exercise: the only cartesian closed category with a zero object is the terminal category.
What is true is that pointed sets have a closed monoidal structure. The closed structure is given by defining the pointed hom-object $[A, B]$ to consist of all functions $A to B$, with basepoint given by the function which has constant value the basepoint of $B$. The monoidal structure is given by defining $A otimes B$ to be the smash product, which is the quotient of $A times B$ given by identifying all pairs $(a, b)$ such that either $a$ or $b$ is the basepoint down to a single basepoint.
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
One way to define a cartesian closed category is that it's a category with finite products where every product functor $(-) times X$ has a right adjoint, namely the exponential functor $(-)^X$. A necessary condition for this to be the case is that $(-) times X$ preserves colimits; in other words, that products distribute over colimits, and in particular coproducts, when they exist.
But products in pointed sets don't distribute over coproducts. For finite pointed sets $A, B, C$ the coproduct satisfies $|A sqcup B| = |A| + |B| - 1$ and you can use this to show that $(A sqcup B) times C$ and $(A times C) sqcup (B times C)$ have different cardinalities in general.
Alternatively, you can use the fact that pointed sets have a zero object. Here's a fun exercise: the only cartesian closed category with a zero object is the terminal category.
What is true is that pointed sets have a closed monoidal structure. The closed structure is given by defining the pointed hom-object $[A, B]$ to consist of all functions $A to B$, with basepoint given by the function which has constant value the basepoint of $B$. The monoidal structure is given by defining $A otimes B$ to be the smash product, which is the quotient of $A times B$ given by identifying all pairs $(a, b)$ such that either $a$ or $b$ is the basepoint down to a single basepoint.
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
add a comment |Â
up vote
6
down vote
accepted
One way to define a cartesian closed category is that it's a category with finite products where every product functor $(-) times X$ has a right adjoint, namely the exponential functor $(-)^X$. A necessary condition for this to be the case is that $(-) times X$ preserves colimits; in other words, that products distribute over colimits, and in particular coproducts, when they exist.
But products in pointed sets don't distribute over coproducts. For finite pointed sets $A, B, C$ the coproduct satisfies $|A sqcup B| = |A| + |B| - 1$ and you can use this to show that $(A sqcup B) times C$ and $(A times C) sqcup (B times C)$ have different cardinalities in general.
Alternatively, you can use the fact that pointed sets have a zero object. Here's a fun exercise: the only cartesian closed category with a zero object is the terminal category.
What is true is that pointed sets have a closed monoidal structure. The closed structure is given by defining the pointed hom-object $[A, B]$ to consist of all functions $A to B$, with basepoint given by the function which has constant value the basepoint of $B$. The monoidal structure is given by defining $A otimes B$ to be the smash product, which is the quotient of $A times B$ given by identifying all pairs $(a, b)$ such that either $a$ or $b$ is the basepoint down to a single basepoint.
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
One way to define a cartesian closed category is that it's a category with finite products where every product functor $(-) times X$ has a right adjoint, namely the exponential functor $(-)^X$. A necessary condition for this to be the case is that $(-) times X$ preserves colimits; in other words, that products distribute over colimits, and in particular coproducts, when they exist.
But products in pointed sets don't distribute over coproducts. For finite pointed sets $A, B, C$ the coproduct satisfies $|A sqcup B| = |A| + |B| - 1$ and you can use this to show that $(A sqcup B) times C$ and $(A times C) sqcup (B times C)$ have different cardinalities in general.
Alternatively, you can use the fact that pointed sets have a zero object. Here's a fun exercise: the only cartesian closed category with a zero object is the terminal category.
What is true is that pointed sets have a closed monoidal structure. The closed structure is given by defining the pointed hom-object $[A, B]$ to consist of all functions $A to B$, with basepoint given by the function which has constant value the basepoint of $B$. The monoidal structure is given by defining $A otimes B$ to be the smash product, which is the quotient of $A times B$ given by identifying all pairs $(a, b)$ such that either $a$ or $b$ is the basepoint down to a single basepoint.
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
One way to define a cartesian closed category is that it's a category with finite products where every product functor $(-) times X$ has a right adjoint, namely the exponential functor $(-)^X$. A necessary condition for this to be the case is that $(-) times X$ preserves colimits; in other words, that products distribute over colimits, and in particular coproducts, when they exist.
But products in pointed sets don't distribute over coproducts. For finite pointed sets $A, B, C$ the coproduct satisfies $|A sqcup B| = |A| + |B| - 1$ and you can use this to show that $(A sqcup B) times C$ and $(A times C) sqcup (B times C)$ have different cardinalities in general.
Alternatively, you can use the fact that pointed sets have a zero object. Here's a fun exercise: the only cartesian closed category with a zero object is the terminal category.
What is true is that pointed sets have a closed monoidal structure. The closed structure is given by defining the pointed hom-object $[A, B]$ to consist of all functions $A to B$, with basepoint given by the function which has constant value the basepoint of $B$. The monoidal structure is given by defining $A otimes B$ to be the smash product, which is the quotient of $A times B$ given by identifying all pairs $(a, b)$ such that either $a$ or $b$ is the basepoint down to a single basepoint.
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
edited Jul 18 at 19:08
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
answered Jul 18 at 18:31
Qiaochu Yuan
269k32564900
269k32564900
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
add a comment |Â
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
Whoops, thanks.
– Qiaochu Yuan
Jul 18 at 19:07
add a comment |Â
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2
It’s not. You can verify that products don’t distribute over coproducts.
– Qiaochu Yuan
Jul 18 at 18:23
@QiaochuYuan Interesting. I never thought to look at it like that. I'll take a look at it. Thank you.
– D. Brogan
Jul 18 at 18:25
Another possible approach might start with: the forgetful functor $U$ is represented by $2$, the pointed set $( 0, 1 , 0)$. So, if the category were cartesian closed, the exponential would have to satisfy $U([A, B]) simeq operatornameHom_mathbfPt(2, [A, B]) simeq operatornameHom_mathbfPt(2 times A, B)$. An element of this would be equivalent to two functions $A to B$, one preserving base points and the other not necessarily preserving base points.
– Daniel Schepler
Jul 18 at 19:34
A standard notation is $1/Set$.
– Berci
Jul 18 at 21:37