Mixing
Are the homsets of Set themselves objects in the category?
So, the category $text{Set}$ has internal homsets. The set of total functions with domain $A$ and codomain $B$ is indeed a set.
I think this means the collection of arrows $hom_{C}(A,B)$ is always "isomorphic" or "equivalent" to some object in $C$ , written $[A,B]$ .
Are the homsets literally objects in $C$ ? For instance, is it possible to construct an infinite descending chain or other pathological object if the homsets are themselves objects instead of corresponding in some natural way to objects in the category? Assuming they aren't objects in the category, what is the relationship between a homset and the object it corresponds to in a category with internal homsets?
category-theory
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
add a comment |
So, the category $text{Set}$ has internal homsets. The set of total functions with domain $A$ and codomain $B$ is indeed a set.
I think this means the collection of arrows $hom_{C}(A,B)$ is always "isomorphic" or "equivalent" to some object in $C$ , written $[A,B]$ .
Are the homsets literally objects in $C$ ? For instance, is it possible to construct an infinite descending chain or other pathological object if the homsets are themselves objects instead of corresponding in some natural way to objects in the category? Assuming they aren't objects in the category, what is the relationship between a homset and the object it corresponds to in a category with internal homsets?
category-theory
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
3
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18
add a comment |
So, the category $text{Set}$ has internal homsets. The set of total functions with domain $A$ and codomain $B$ is indeed a set.
I think this means the collection of arrows $hom_{C}(A,B)$ is always "isomorphic" or "equivalent" to some object in $C$ , written $[A,B]$ .
Are the homsets literally objects in $C$ ? For instance, is it possible to construct an infinite descending chain or other pathological object if the homsets are themselves objects instead of corresponding in some natural way to objects in the category? Assuming they aren't objects in the category, what is the relationship between a homset and the object it corresponds to in a category with internal homsets?
category-theory
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
So, the category $text{Set}$ has internal homsets. The set of total functions with domain $A$ and codomain $B$ is indeed a set.
I think this means the collection of arrows $hom_{C}(A,B)$ is always "isomorphic" or "equivalent" to some object in $C$ , written $[A,B]$ .
Are the homsets literally objects in $C$ ? For instance, is it possible to construct an infinite descending chain or other pathological object if the homsets are themselves objects instead of corresponding in some natural way to objects in the category? Assuming they aren't objects in the category, what is the relationship between a homset and the object it corresponds to in a category with internal homsets?
category-theory
category-theory
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
edited Nov 25 '18 at 1:14
Gregory Nisbet
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
asked Nov 22 '18 at 6:20
Gregory NisbetGregory Nisbet
541312
541312
3
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18
add a comment |
3
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18
3
3
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18
add a comment |
2 Answers
2
active
oldest
votes
The Internal Hom-Sets, $[X,Y]$ as you denote them, are specified by
$$
X × Y → Z
quad≅quad
X → [Y, Z]
$$
Now a “point of $X$” is defined to be an arrow from the terminal object, 𝑰,
so we have the following relationship.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{equiv}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
begin{calc}
mathsf{points}, [A, B]
stepWith{≅}{ Definition of points }
𝑰 → [A, B]
stepWith{≅}{ Characterisation of internal-hom }
𝑰 × A → B
stepWith{≅}{ Terminal object is identity of × }
A → B
end{calc}
Hence the internal hom and external hom are ‘equivalent’ in that:
$$ mathsf{points}, [A,B] ;≅; (A → B) $$
As you have observed, in the category Set we have
$mathsf{points}, X ;≅; X$ and so the internal & external
homs are directly isomorphic.
Nice question :-)
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
add a comment |
I don't know if this will answer your question, but the first thing you made me think is the following situation widely studied in category theory. Let $(mathcal{C}, otimes, 1)$ be a symmetric monoidal model category, then it is said to admit internal homs if for every object $X$ of $mathcal{C}$ the functor $X otimes - colon mathcal{C} rightarrow mathcal{C}$ admits a right adjoint i.e. we have a functor $F(X,-)colon mathcal{C}rightarrow mathcal{C}$ right adjoint to $X otimes -$.
If the monoidal structure is not symmetric we have to be careful to differentiate between $X otimes - $ and $- otimes X$ and we have to ask if they have separately their own right adjoints, but for now lets take the easy case and consider symmetric monoidal structures.
From the definitions alone and Yoneda lemma you can deduce that there is a natural isomorphism $F(X otimes Y, Z) cong F(X, F(Y,Z))$, which is the version in the category $mathcal{C}$ of the exponential law $Z^{X times Y}cong (Z^{Y})^X$. So we usually consider $F(X,Y)$ as an object of $mathcal{C}$ which represents the set of morphisms from $X$ to $Y$.
You probably saw a lot of such situations, only not with this formulation.
If you consider $mathcal{C}$ to be the category of compactly generated topological spaces with $otimes = times$ the usual product then $F(X,Y)=Y^X$ is the space of continuous functions from $X$ to $Y$ with the compact-open topology.
Or if you fix $R$ a commutative ring, you can take $mathcal{C}$ to be the category of left $R$-modules and as $otimes$ the tensor product over $R$, then $F(X,Y)$ is the set of morphisms of $R$-modules from $X$ to $Y$ wich can be made in an $R$-module by taking $(r.phi)(x)=r.phi(x)$.
The example you presented is $mathcal{C}=Sets$ the category of sets with $otimes =times$ the usual product, then clearly $F(X,Y)$ is just the set of functions from $X$ to $Y$.
I think your question can be rephrased as follows: given any category $mathcal{C}$ does such structure exists? The answer is a clear no: there are easy counterexamples.
A non trivial counterexample is the category of topological spaces (even non-compactly generated!): here you have again a symmetric monoidal structure given by the usual product $times$ but for general spaces $X, Y, Z$ the exponential law $Z^{X times Y} cong (Z^{X})^Y$ does not hold. Therefore $X times -$ does not have a right adjoint.
Now an easy counterexample created ad hoc to deny the fact that the sets of morphisms are always internal hom objects in the sense I defined above. Consider $G$ a non-trivial group, then we can form $BG$ the category with just one object, call it $ast$, and as hom sets we take $text{Hom}(ast, ast)=G$ where the composition law is given by the product on $G$. Then for any symmetric monoidal structure $otimes$ on this category we have $ast otimes ast= ast$, hence $ast otimes -$ is just the identity functor on the category. Thus its right adjoint must be still the identity and we get $F(ast, ast)= ast$. But in $BG$ the hom set $text{Hom}(ast, ast)$ is $G$, not only a point.
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
The Internal Hom-Sets, $[X,Y]$ as you denote them, are specified by
$$
X × Y → Z
quad≅quad
X → [Y, Z]
$$
Now a “point of $X$” is defined to be an arrow from the terminal object, 𝑰,
so we have the following relationship.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{equiv}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
begin{calc}
mathsf{points}, [A, B]
stepWith{≅}{ Definition of points }
𝑰 → [A, B]
stepWith{≅}{ Characterisation of internal-hom }
𝑰 × A → B
stepWith{≅}{ Terminal object is identity of × }
A → B
end{calc}
Hence the internal hom and external hom are ‘equivalent’ in that:
$$ mathsf{points}, [A,B] ;≅; (A → B) $$
As you have observed, in the category Set we have
$mathsf{points}, X ;≅; X$ and so the internal & external
homs are directly isomorphic.
Nice question :-)
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
add a comment |
The Internal Hom-Sets, $[X,Y]$ as you denote them, are specified by
$$
X × Y → Z
quad≅quad
X → [Y, Z]
$$
Now a “point of $X$” is defined to be an arrow from the terminal object, 𝑰,
so we have the following relationship.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{equiv}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
begin{calc}
mathsf{points}, [A, B]
stepWith{≅}{ Definition of points }
𝑰 → [A, B]
stepWith{≅}{ Characterisation of internal-hom }
𝑰 × A → B
stepWith{≅}{ Terminal object is identity of × }
A → B
end{calc}
Hence the internal hom and external hom are ‘equivalent’ in that:
$$ mathsf{points}, [A,B] ;≅; (A → B) $$
As you have observed, in the category Set we have
$mathsf{points}, X ;≅; X$ and so the internal & external
homs are directly isomorphic.
Nice question :-)
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
add a comment |
The Internal Hom-Sets, $[X,Y]$ as you denote them, are specified by
$$
X × Y → Z
quad≅quad
X → [Y, Z]
$$
Now a “point of $X$” is defined to be an arrow from the terminal object, 𝑰,
so we have the following relationship.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{equiv}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
begin{calc}
mathsf{points}, [A, B]
stepWith{≅}{ Definition of points }
𝑰 → [A, B]
stepWith{≅}{ Characterisation of internal-hom }
𝑰 × A → B
stepWith{≅}{ Terminal object is identity of × }
A → B
end{calc}
Hence the internal hom and external hom are ‘equivalent’ in that:
$$ mathsf{points}, [A,B] ;≅; (A → B) $$
As you have observed, in the category Set we have
$mathsf{points}, X ;≅; X$ and so the internal & external
homs are directly isomorphic.
Nice question :-)
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
The Internal Hom-Sets, $[X,Y]$ as you denote them, are specified by
$$
X × Y → Z
quad≅quad
X → [Y, Z]
$$
Now a “point of $X$” is defined to be an arrow from the terminal object, 𝑰,
so we have the following relationship.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{equiv}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
begin{calc}
mathsf{points}, [A, B]
stepWith{≅}{ Definition of points }
𝑰 → [A, B]
stepWith{≅}{ Characterisation of internal-hom }
𝑰 × A → B
stepWith{≅}{ Terminal object is identity of × }
A → B
end{calc}
Hence the internal hom and external hom are ‘equivalent’ in that:
$$ mathsf{points}, [A,B] ;≅; (A → B) $$
As you have observed, in the category Set we have
$mathsf{points}, X ;≅; X$ and so the internal & external
homs are directly isomorphic.
Nice question :-)
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
edited Nov 22 '18 at 12:54
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
answered Nov 22 '18 at 10:27
Musa Al-hassyMusa Al-hassy
1,3391711
1,3391711
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
add a comment |
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
2
2
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
$I$ is rather the terminal object.
– Berci
Nov 22 '18 at 11:29
1
1
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
Thanks for the fix -- typing before dawn seems to have led me to be silly ^_^
– Musa Al-hassy
Nov 22 '18 at 12:54
add a comment |
I don't know if this will answer your question, but the first thing you made me think is the following situation widely studied in category theory. Let $(mathcal{C}, otimes, 1)$ be a symmetric monoidal model category, then it is said to admit internal homs if for every object $X$ of $mathcal{C}$ the functor $X otimes - colon mathcal{C} rightarrow mathcal{C}$ admits a right adjoint i.e. we have a functor $F(X,-)colon mathcal{C}rightarrow mathcal{C}$ right adjoint to $X otimes -$.
If the monoidal structure is not symmetric we have to be careful to differentiate between $X otimes - $ and $- otimes X$ and we have to ask if they have separately their own right adjoints, but for now lets take the easy case and consider symmetric monoidal structures.
From the definitions alone and Yoneda lemma you can deduce that there is a natural isomorphism $F(X otimes Y, Z) cong F(X, F(Y,Z))$, which is the version in the category $mathcal{C}$ of the exponential law $Z^{X times Y}cong (Z^{Y})^X$. So we usually consider $F(X,Y)$ as an object of $mathcal{C}$ which represents the set of morphisms from $X$ to $Y$.
You probably saw a lot of such situations, only not with this formulation.
If you consider $mathcal{C}$ to be the category of compactly generated topological spaces with $otimes = times$ the usual product then $F(X,Y)=Y^X$ is the space of continuous functions from $X$ to $Y$ with the compact-open topology.
Or if you fix $R$ a commutative ring, you can take $mathcal{C}$ to be the category of left $R$-modules and as $otimes$ the tensor product over $R$, then $F(X,Y)$ is the set of morphisms of $R$-modules from $X$ to $Y$ wich can be made in an $R$-module by taking $(r.phi)(x)=r.phi(x)$.
The example you presented is $mathcal{C}=Sets$ the category of sets with $otimes =times$ the usual product, then clearly $F(X,Y)$ is just the set of functions from $X$ to $Y$.
I think your question can be rephrased as follows: given any category $mathcal{C}$ does such structure exists? The answer is a clear no: there are easy counterexamples.
A non trivial counterexample is the category of topological spaces (even non-compactly generated!): here you have again a symmetric monoidal structure given by the usual product $times$ but for general spaces $X, Y, Z$ the exponential law $Z^{X times Y} cong (Z^{X})^Y$ does not hold. Therefore $X times -$ does not have a right adjoint.
Now an easy counterexample created ad hoc to deny the fact that the sets of morphisms are always internal hom objects in the sense I defined above. Consider $G$ a non-trivial group, then we can form $BG$ the category with just one object, call it $ast$, and as hom sets we take $text{Hom}(ast, ast)=G$ where the composition law is given by the product on $G$. Then for any symmetric monoidal structure $otimes$ on this category we have $ast otimes ast= ast$, hence $ast otimes -$ is just the identity functor on the category. Thus its right adjoint must be still the identity and we get $F(ast, ast)= ast$. But in $BG$ the hom set $text{Hom}(ast, ast)$ is $G$, not only a point.
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
add a comment |
I don't know if this will answer your question, but the first thing you made me think is the following situation widely studied in category theory. Let $(mathcal{C}, otimes, 1)$ be a symmetric monoidal model category, then it is said to admit internal homs if for every object $X$ of $mathcal{C}$ the functor $X otimes - colon mathcal{C} rightarrow mathcal{C}$ admits a right adjoint i.e. we have a functor $F(X,-)colon mathcal{C}rightarrow mathcal{C}$ right adjoint to $X otimes -$.
If the monoidal structure is not symmetric we have to be careful to differentiate between $X otimes - $ and $- otimes X$ and we have to ask if they have separately their own right adjoints, but for now lets take the easy case and consider symmetric monoidal structures.
From the definitions alone and Yoneda lemma you can deduce that there is a natural isomorphism $F(X otimes Y, Z) cong F(X, F(Y,Z))$, which is the version in the category $mathcal{C}$ of the exponential law $Z^{X times Y}cong (Z^{Y})^X$. So we usually consider $F(X,Y)$ as an object of $mathcal{C}$ which represents the set of morphisms from $X$ to $Y$.
You probably saw a lot of such situations, only not with this formulation.
If you consider $mathcal{C}$ to be the category of compactly generated topological spaces with $otimes = times$ the usual product then $F(X,Y)=Y^X$ is the space of continuous functions from $X$ to $Y$ with the compact-open topology.
Or if you fix $R$ a commutative ring, you can take $mathcal{C}$ to be the category of left $R$-modules and as $otimes$ the tensor product over $R$, then $F(X,Y)$ is the set of morphisms of $R$-modules from $X$ to $Y$ wich can be made in an $R$-module by taking $(r.phi)(x)=r.phi(x)$.
The example you presented is $mathcal{C}=Sets$ the category of sets with $otimes =times$ the usual product, then clearly $F(X,Y)$ is just the set of functions from $X$ to $Y$.
I think your question can be rephrased as follows: given any category $mathcal{C}$ does such structure exists? The answer is a clear no: there are easy counterexamples.
A non trivial counterexample is the category of topological spaces (even non-compactly generated!): here you have again a symmetric monoidal structure given by the usual product $times$ but for general spaces $X, Y, Z$ the exponential law $Z^{X times Y} cong (Z^{X})^Y$ does not hold. Therefore $X times -$ does not have a right adjoint.
Now an easy counterexample created ad hoc to deny the fact that the sets of morphisms are always internal hom objects in the sense I defined above. Consider $G$ a non-trivial group, then we can form $BG$ the category with just one object, call it $ast$, and as hom sets we take $text{Hom}(ast, ast)=G$ where the composition law is given by the product on $G$. Then for any symmetric monoidal structure $otimes$ on this category we have $ast otimes ast= ast$, hence $ast otimes -$ is just the identity functor on the category. Thus its right adjoint must be still the identity and we get $F(ast, ast)= ast$. But in $BG$ the hom set $text{Hom}(ast, ast)$ is $G$, not only a point.
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
add a comment |
I don't know if this will answer your question, but the first thing you made me think is the following situation widely studied in category theory. Let $(mathcal{C}, otimes, 1)$ be a symmetric monoidal model category, then it is said to admit internal homs if for every object $X$ of $mathcal{C}$ the functor $X otimes - colon mathcal{C} rightarrow mathcal{C}$ admits a right adjoint i.e. we have a functor $F(X,-)colon mathcal{C}rightarrow mathcal{C}$ right adjoint to $X otimes -$.
If the monoidal structure is not symmetric we have to be careful to differentiate between $X otimes - $ and $- otimes X$ and we have to ask if they have separately their own right adjoints, but for now lets take the easy case and consider symmetric monoidal structures.
From the definitions alone and Yoneda lemma you can deduce that there is a natural isomorphism $F(X otimes Y, Z) cong F(X, F(Y,Z))$, which is the version in the category $mathcal{C}$ of the exponential law $Z^{X times Y}cong (Z^{Y})^X$. So we usually consider $F(X,Y)$ as an object of $mathcal{C}$ which represents the set of morphisms from $X$ to $Y$.
You probably saw a lot of such situations, only not with this formulation.
If you consider $mathcal{C}$ to be the category of compactly generated topological spaces with $otimes = times$ the usual product then $F(X,Y)=Y^X$ is the space of continuous functions from $X$ to $Y$ with the compact-open topology.
Or if you fix $R$ a commutative ring, you can take $mathcal{C}$ to be the category of left $R$-modules and as $otimes$ the tensor product over $R$, then $F(X,Y)$ is the set of morphisms of $R$-modules from $X$ to $Y$ wich can be made in an $R$-module by taking $(r.phi)(x)=r.phi(x)$.
The example you presented is $mathcal{C}=Sets$ the category of sets with $otimes =times$ the usual product, then clearly $F(X,Y)$ is just the set of functions from $X$ to $Y$.
I think your question can be rephrased as follows: given any category $mathcal{C}$ does such structure exists? The answer is a clear no: there are easy counterexamples.
A non trivial counterexample is the category of topological spaces (even non-compactly generated!): here you have again a symmetric monoidal structure given by the usual product $times$ but for general spaces $X, Y, Z$ the exponential law $Z^{X times Y} cong (Z^{X})^Y$ does not hold. Therefore $X times -$ does not have a right adjoint.
Now an easy counterexample created ad hoc to deny the fact that the sets of morphisms are always internal hom objects in the sense I defined above. Consider $G$ a non-trivial group, then we can form $BG$ the category with just one object, call it $ast$, and as hom sets we take $text{Hom}(ast, ast)=G$ where the composition law is given by the product on $G$. Then for any symmetric monoidal structure $otimes$ on this category we have $ast otimes ast= ast$, hence $ast otimes -$ is just the identity functor on the category. Thus its right adjoint must be still the identity and we get $F(ast, ast)= ast$. But in $BG$ the hom set $text{Hom}(ast, ast)$ is $G$, not only a point.
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
I don't know if this will answer your question, but the first thing you made me think is the following situation widely studied in category theory. Let $(mathcal{C}, otimes, 1)$ be a symmetric monoidal model category, then it is said to admit internal homs if for every object $X$ of $mathcal{C}$ the functor $X otimes - colon mathcal{C} rightarrow mathcal{C}$ admits a right adjoint i.e. we have a functor $F(X,-)colon mathcal{C}rightarrow mathcal{C}$ right adjoint to $X otimes -$.
If the monoidal structure is not symmetric we have to be careful to differentiate between $X otimes - $ and $- otimes X$ and we have to ask if they have separately their own right adjoints, but for now lets take the easy case and consider symmetric monoidal structures.
From the definitions alone and Yoneda lemma you can deduce that there is a natural isomorphism $F(X otimes Y, Z) cong F(X, F(Y,Z))$, which is the version in the category $mathcal{C}$ of the exponential law $Z^{X times Y}cong (Z^{Y})^X$. So we usually consider $F(X,Y)$ as an object of $mathcal{C}$ which represents the set of morphisms from $X$ to $Y$.
You probably saw a lot of such situations, only not with this formulation.
If you consider $mathcal{C}$ to be the category of compactly generated topological spaces with $otimes = times$ the usual product then $F(X,Y)=Y^X$ is the space of continuous functions from $X$ to $Y$ with the compact-open topology.
Or if you fix $R$ a commutative ring, you can take $mathcal{C}$ to be the category of left $R$-modules and as $otimes$ the tensor product over $R$, then $F(X,Y)$ is the set of morphisms of $R$-modules from $X$ to $Y$ wich can be made in an $R$-module by taking $(r.phi)(x)=r.phi(x)$.
The example you presented is $mathcal{C}=Sets$ the category of sets with $otimes =times$ the usual product, then clearly $F(X,Y)$ is just the set of functions from $X$ to $Y$.
I think your question can be rephrased as follows: given any category $mathcal{C}$ does such structure exists? The answer is a clear no: there are easy counterexamples.
A non trivial counterexample is the category of topological spaces (even non-compactly generated!): here you have again a symmetric monoidal structure given by the usual product $times$ but for general spaces $X, Y, Z$ the exponential law $Z^{X times Y} cong (Z^{X})^Y$ does not hold. Therefore $X times -$ does not have a right adjoint.
Now an easy counterexample created ad hoc to deny the fact that the sets of morphisms are always internal hom objects in the sense I defined above. Consider $G$ a non-trivial group, then we can form $BG$ the category with just one object, call it $ast$, and as hom sets we take $text{Hom}(ast, ast)=G$ where the composition law is given by the product on $G$. Then for any symmetric monoidal structure $otimes$ on this category we have $ast otimes ast= ast$, hence $ast otimes -$ is just the identity functor on the category. Thus its right adjoint must be still the identity and we get $F(ast, ast)= ast$. But in $BG$ the hom set $text{Hom}(ast, ast)$ is $G$, not only a point.
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
edited Nov 22 '18 at 11:16
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
answered Nov 22 '18 at 11:11
N.B.N.B.
689313
689313
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3
Are sets literally sets? That's a philosophical question that's far too difficult for me.
– Lord Shark the Unknown
Nov 22 '18 at 7:07
@LordSharktheUnknown ... Can you help me to rephrase the question? I didn't intend it to be trivial or meaningless. Changing the question slightly to illustrate, the homsets in $text{Set}$ are of course filled with arrows. I don't know if you can freely decide whether to treat an arrow as some kind of ur-element or as the graph of a function. If you go the ur-element route, is the homset still an object of $text{Set}$ ... do you get a contradiction somewhere down the line if you say yes?
– Gregory Nisbet
Nov 22 '18 at 7:18