Define exponential object without reference to cartesian product?
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In category theory, the exponential object $X^Y$ is defined, roughly, by saying that if we take some other object $Z$ and the product between $Z$ and $X$, then any morphism from that product to $Y$ has an equivalent morphism that “goes through” $X^Y$.
However, what I find weird about this construction (note that category theory in general is weird to me), is that we need a third object $Z$ and the assumption that a cartesian product exists between them, to define the exponential object $X^Y$.
Is it not possible to define it more directly somehow? Intuitely it seems like we don’t need to talk about $Z$ or about cartesian products, to talk about functions from $X$ to $Y$ (I know that category theory is not only about functions, but this is how I intuitively think about it).
category-theory
asked Nov 18 at 9:57
user56834
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1
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In category theory, the exponential object $X^Y$ is defined, roughly, by saying that if we take some other object $Z$ and the product between $Z$ and $X$, then any morphism from that product to $Y$ has an equivalent morphism that “goes through” $X^Y$.
However, what I find weird about this construction (note that category theory in general is weird to me), is that we need a third object $Z$ and the assumption that a cartesian product exists between them, to define the exponential object $X^Y$.
Is it not possible to define it more directly somehow? Intuitely it seems like we don’t need to talk about $Z$ or about cartesian products, to talk about functions from $X$ to $Y$ (I know that category theory is not only about functions, but this is how I intuitively think about it).
category-theory
asked Nov 18 at 9:57
user56834
3,13821149
What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In category theory, the exponential object $X^Y$ is defined, roughly, by saying that if we take some other object $Z$ and the product between $Z$ and $X$, then any morphism from that product to $Y$ has an equivalent morphism that “goes through” $X^Y$.
However, what I find weird about this construction (note that category theory in general is weird to me), is that we need a third object $Z$ and the assumption that a cartesian product exists between them, to define the exponential object $X^Y$.
Is it not possible to define it more directly somehow? Intuitely it seems like we don’t need to talk about $Z$ or about cartesian products, to talk about functions from $X$ to $Y$ (I know that category theory is not only about functions, but this is how I intuitively think about it).
category-theory
asked Nov 18 at 9:57
user56834
3,13821149
In category theory, the exponential object $X^Y$ is defined, roughly, by saying that if we take some other object $Z$ and the product between $Z$ and $X$, then any morphism from that product to $Y$ has an equivalent morphism that “goes through” $X^Y$.
However, what I find weird about this construction (note that category theory in general is weird to me), is that we need a third object $Z$ and the assumption that a cartesian product exists between them, to define the exponential object $X^Y$.
Is it not possible to define it more directly somehow? Intuitely it seems like we don’t need to talk about $Z$ or about cartesian products, to talk about functions from $X$ to $Y$ (I know that category theory is not only about functions, but this is how I intuitively think about it).
category-theory
category-theory
asked Nov 18 at 9:57
user56834
3,13821149
asked Nov 18 at 9:57
user56834
3,13821149
asked Nov 18 at 9:57
user56834
3,13821149
asked Nov 18 at 9:57
user56834
3,13821149
asked Nov 18 at 9:57
user56834
3,13821149
3,13821149
What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17
add a comment |
What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17
What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17
What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:
$T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X to T$.
$C$ is the product of $A$ and $B$ if there are maps $C to A$ and $C to B$ and for any other object $D$ with maps $f: D to A$ and $g: D to B$, there is a unique morphism $D to C$ which $f,g$ factor through.
The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.
In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.
answered Nov 18 at 13:21
B. Mehta
11.8k22144
add a comment |
up vote
3
down vote
The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Yto (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $Xto X^Y$ and another family corresponding to diagonals.
Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:
$T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X to T$.
$C$ is the product of $A$ and $B$ if there are maps $C to A$ and $C to B$ and for any other object $D$ with maps $f: D to A$ and $g: D to B$, there is a unique morphism $D to C$ which $f,g$ factor through.
The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.
In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.
answered Nov 18 at 13:21
B. Mehta
11.8k22144
add a comment |
up vote
3
down vote
First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:
$T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X to T$.
$C$ is the product of $A$ and $B$ if there are maps $C to A$ and $C to B$ and for any other object $D$ with maps $f: D to A$ and $g: D to B$, there is a unique morphism $D to C$ which $f,g$ factor through.
The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.
In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.
answered Nov 18 at 13:21
B. Mehta
11.8k22144
add a comment |
up vote
3
down vote
up vote
3
down vote
First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:
$T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X to T$.
$C$ is the product of $A$ and $B$ if there are maps $C to A$ and $C to B$ and for any other object $D$ with maps $f: D to A$ and $g: D to B$, there is a unique morphism $D to C$ which $f,g$ factor through.
The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.
In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.
answered Nov 18 at 13:21
B. Mehta
11.8k22144
First note that the idea of a universal property in category theory is virtually always defined in terms of some other objects, so the idea of adding $Z$ in the definition isn't special for exponentials:
$T$ is a terminal object in a category if, for any other object $X$, there is a unique morphism $X to T$.
$C$ is the product of $A$ and $B$ if there are maps $C to A$ and $C to B$ and for any other object $D$ with maps $f: D to A$ and $g: D to B$, there is a unique morphism $D to C$ which $f,g$ factor through.
The point is that to define an object, you can only refer to other objects in the category - so this is really the most direct way also to define the exponential object.
In terms of requiring binary products to exist, this may seem odd at first glance, but seemingly turns out to be the most sensible definition. Personally, I think of the exponential object $X^Y$ as the most general object (in some loose sense) for which giving a $Y$ combines with it to give an $X$.
answered Nov 18 at 13:21
B. Mehta
11.8k22144
answered Nov 18 at 13:21
B. Mehta
11.8k22144
answered Nov 18 at 13:21
B. Mehta
11.8k22144
answered Nov 18 at 13:21
B. Mehta
11.8k22144
11.8k22144
add a comment |
add a comment |
up vote
3
down vote
The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Yto (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $Xto X^Y$ and another family corresponding to diagonals.
Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
add a comment |
up vote
3
down vote
The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Yto (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $Xto X^Y$ and another family corresponding to diagonals.
Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
add a comment |
up vote
3
down vote
up vote
3
down vote
The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Yto (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $Xto X^Y$ and another family corresponding to diagonals.
Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
The closest thing you can do is to define a closed category. This is one for which there exist functorial choices of exponentials together with a unit object $I$ equipped with a map to every endomorphism object $X^X$ which is neutral for a given associative family of composition maps. Everything has to be said in curried form: for instance the composition map goes $X^Yto (Z^X)^{Y^Z}$. If you ask your exponentiation functors to be right adjoints, then you have a monoidal closed category, not in general a Cartesian closed one. There is a proposal on the nLab page for closed categories that there should be a characterization of the internal Hom structures whose left adjoints, if they exist, are Cartesian monoidal in terms of the existence of constant-morphism maps $Xto X^Y$ and another family corresponding to diagonals.
Closed categories are used relatively infrequently, in part for psychological reasons: contravariant homs create extra natural transformations, and people find these annoying. There is also the practical reason that closed categories which aren't monoidal are rather rare, and the theoretical reason that one can embed a closed category in a closed monoidal one.
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
edited Nov 18 at 18:01
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
answered Nov 18 at 17:49
Kevin Carlson
32.2k23270
32.2k23270
add a comment |
add a comment |
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What other property, that you can express in category theoretic terms, do you think an exponential should have? We can't generally talk about "members" of an object in an arbitrary category, and sometimes when we can it would be misleading (e.g. for any two abelian groups, there's an abelian group of homomorphisms between them, but it's no good at all as an exponential object).
– Malice Vidrine
Nov 18 at 19:17