Mixing
Does the curried version of a bifunctor contain more information?
$begingroup$
For functions $f:Xtimes Yto Z$, the curried version $lambda f:Xto (Yto Z)$ is a different way of representing the exact same information.
For functors, I’m assuming that the same should hold, but intuitively I’m thinking that the curried version contains more information.
Consider the similar case for a bifunctor. Given categories $C,D,E$, consider the functor $F:Ctimes Dto E$. We can curry this using the exponential category, and get the functor $lambda F : C to E^D$.
With the base functor $F$, we can EITHER input an object $(c,d)$ OR a morphism $(f,g)$ where $c,d$ are objects and $f,g$ morphisms of $C$ and $D$ respectively. In these two cases we get objects and morphisms in $E$ out of them.
But with the curried functor, we have more options: We can first input an object $c$, and then input an object $d$, which gives us the same as inputting $(c,d)$. We can also input first $c$ and then a morphism $gin D$. Alternatively, we can immediately input a morphism $fin C$ into $lambda F$.
Doesn’t this increased amount of uses of $lambda F$ mean that it has more information than $F$?
category-theory
asked Jan 30 at 14:15
user56834user56834
3,38621253
$endgroup$
add a comment |
$begingroup$
For functions $f:Xtimes Yto Z$, the curried version $lambda f:Xto (Yto Z)$ is a different way of representing the exact same information.
For functors, I’m assuming that the same should hold, but intuitively I’m thinking that the curried version contains more information.
Consider the similar case for a bifunctor. Given categories $C,D,E$, consider the functor $F:Ctimes Dto E$. We can curry this using the exponential category, and get the functor $lambda F : C to E^D$.
With the base functor $F$, we can EITHER input an object $(c,d)$ OR a morphism $(f,g)$ where $c,d$ are objects and $f,g$ morphisms of $C$ and $D$ respectively. In these two cases we get objects and morphisms in $E$ out of them.
But with the curried functor, we have more options: We can first input an object $c$, and then input an object $d$, which gives us the same as inputting $(c,d)$. We can also input first $c$ and then a morphism $gin D$. Alternatively, we can immediately input a morphism $fin C$ into $lambda F$.
Doesn’t this increased amount of uses of $lambda F$ mean that it has more information than $F$?
category-theory
asked Jan 30 at 14:15
user56834user56834
3,38621253
$endgroup$
add a comment |
$begingroup$
For functions $f:Xtimes Yto Z$, the curried version $lambda f:Xto (Yto Z)$ is a different way of representing the exact same information.
For functors, I’m assuming that the same should hold, but intuitively I’m thinking that the curried version contains more information.
Consider the similar case for a bifunctor. Given categories $C,D,E$, consider the functor $F:Ctimes Dto E$. We can curry this using the exponential category, and get the functor $lambda F : C to E^D$.
With the base functor $F$, we can EITHER input an object $(c,d)$ OR a morphism $(f,g)$ where $c,d$ are objects and $f,g$ morphisms of $C$ and $D$ respectively. In these two cases we get objects and morphisms in $E$ out of them.
But with the curried functor, we have more options: We can first input an object $c$, and then input an object $d$, which gives us the same as inputting $(c,d)$. We can also input first $c$ and then a morphism $gin D$. Alternatively, we can immediately input a morphism $fin C$ into $lambda F$.
Doesn’t this increased amount of uses of $lambda F$ mean that it has more information than $F$?
category-theory
asked Jan 30 at 14:15
user56834user56834
3,38621253
$endgroup$
For functions $f:Xtimes Yto Z$, the curried version $lambda f:Xto (Yto Z)$ is a different way of representing the exact same information.
For functors, I’m assuming that the same should hold, but intuitively I’m thinking that the curried version contains more information.
Consider the similar case for a bifunctor. Given categories $C,D,E$, consider the functor $F:Ctimes Dto E$. We can curry this using the exponential category, and get the functor $lambda F : C to E^D$.
With the base functor $F$, we can EITHER input an object $(c,d)$ OR a morphism $(f,g)$ where $c,d$ are objects and $f,g$ morphisms of $C$ and $D$ respectively. In these two cases we get objects and morphisms in $E$ out of them.
But with the curried functor, we have more options: We can first input an object $c$, and then input an object $d$, which gives us the same as inputting $(c,d)$. We can also input first $c$ and then a morphism $gin D$. Alternatively, we can immediately input a morphism $fin C$ into $lambda F$.
Doesn’t this increased amount of uses of $lambda F$ mean that it has more information than $F$?
category-theory
category-theory
asked Jan 30 at 14:15
user56834user56834
3,38621253
asked Jan 30 at 14:15
user56834user56834
3,38621253
asked Jan 30 at 14:15
user56834user56834
3,38621253
asked Jan 30 at 14:15
user56834user56834
3,38621253
asked Jan 30 at 14:15
user56834user56834
3,38621253
3,38621253
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Immediate intuitive answer: it can't, because things ought to be well-behaved, which means that currying ought to be an equivalence of the appropriate categories (in this case: $mathsf{Fun}(C times D, E)$ and $mathsf{Fun}(C, E^D)$) which means it ought to be reversible.
Let's chase down what happens in the curried version of the functor $lambda F$. If we put in an object $c in C$, this will get sent to what is informally referred to as $F(c, -)$. This functor takes an object $d in D$, and sends it to $F(c, d)$. So what does it do on morphisms? Given $g: d to d'$, we need to get a morphism $F(c, g): F(c, d) to F(c, d')$, and the only reasonable definition is to have $F((mathrm{id}_C, g))$, and you can check that this works. Thus: the information of what happens when you first put an object, and then a morphism, into $lambda F$ is already contained in $F$: you just use the identity morphism of the object.
For the other direction: if $f: c to c'$ is a morphism, $lambda F(f) = F(f, -)$ ought to be a morphism from $F(c, -) to F(c', -)$. That is, for each $d in D$ we should have a morphism $F(c, d) to F(c', d)$. Again our only real choice is to have this be $F((f, mathrm{id}_D))$. Again, you can check that this works.
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Immediate intuitive answer: it can't, because things ought to be well-behaved, which means that currying ought to be an equivalence of the appropriate categories (in this case: $mathsf{Fun}(C times D, E)$ and $mathsf{Fun}(C, E^D)$) which means it ought to be reversible.
Let's chase down what happens in the curried version of the functor $lambda F$. If we put in an object $c in C$, this will get sent to what is informally referred to as $F(c, -)$. This functor takes an object $d in D$, and sends it to $F(c, d)$. So what does it do on morphisms? Given $g: d to d'$, we need to get a morphism $F(c, g): F(c, d) to F(c, d')$, and the only reasonable definition is to have $F((mathrm{id}_C, g))$, and you can check that this works. Thus: the information of what happens when you first put an object, and then a morphism, into $lambda F$ is already contained in $F$: you just use the identity morphism of the object.
For the other direction: if $f: c to c'$ is a morphism, $lambda F(f) = F(f, -)$ ought to be a morphism from $F(c, -) to F(c', -)$. That is, for each $d in D$ we should have a morphism $F(c, d) to F(c', d)$. Again our only real choice is to have this be $F((f, mathrm{id}_D))$. Again, you can check that this works.
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
add a comment |
$begingroup$
Immediate intuitive answer: it can't, because things ought to be well-behaved, which means that currying ought to be an equivalence of the appropriate categories (in this case: $mathsf{Fun}(C times D, E)$ and $mathsf{Fun}(C, E^D)$) which means it ought to be reversible.
Let's chase down what happens in the curried version of the functor $lambda F$. If we put in an object $c in C$, this will get sent to what is informally referred to as $F(c, -)$. This functor takes an object $d in D$, and sends it to $F(c, d)$. So what does it do on morphisms? Given $g: d to d'$, we need to get a morphism $F(c, g): F(c, d) to F(c, d')$, and the only reasonable definition is to have $F((mathrm{id}_C, g))$, and you can check that this works. Thus: the information of what happens when you first put an object, and then a morphism, into $lambda F$ is already contained in $F$: you just use the identity morphism of the object.
For the other direction: if $f: c to c'$ is a morphism, $lambda F(f) = F(f, -)$ ought to be a morphism from $F(c, -) to F(c', -)$. That is, for each $d in D$ we should have a morphism $F(c, d) to F(c', d)$. Again our only real choice is to have this be $F((f, mathrm{id}_D))$. Again, you can check that this works.
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
$endgroup$
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
add a comment |
$begingroup$
Immediate intuitive answer: it can't, because things ought to be well-behaved, which means that currying ought to be an equivalence of the appropriate categories (in this case: $mathsf{Fun}(C times D, E)$ and $mathsf{Fun}(C, E^D)$) which means it ought to be reversible.
Let's chase down what happens in the curried version of the functor $lambda F$. If we put in an object $c in C$, this will get sent to what is informally referred to as $F(c, -)$. This functor takes an object $d in D$, and sends it to $F(c, d)$. So what does it do on morphisms? Given $g: d to d'$, we need to get a morphism $F(c, g): F(c, d) to F(c, d')$, and the only reasonable definition is to have $F((mathrm{id}_C, g))$, and you can check that this works. Thus: the information of what happens when you first put an object, and then a morphism, into $lambda F$ is already contained in $F$: you just use the identity morphism of the object.
For the other direction: if $f: c to c'$ is a morphism, $lambda F(f) = F(f, -)$ ought to be a morphism from $F(c, -) to F(c', -)$. That is, for each $d in D$ we should have a morphism $F(c, d) to F(c', d)$. Again our only real choice is to have this be $F((f, mathrm{id}_D))$. Again, you can check that this works.
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
$endgroup$
Immediate intuitive answer: it can't, because things ought to be well-behaved, which means that currying ought to be an equivalence of the appropriate categories (in this case: $mathsf{Fun}(C times D, E)$ and $mathsf{Fun}(C, E^D)$) which means it ought to be reversible.
Let's chase down what happens in the curried version of the functor $lambda F$. If we put in an object $c in C$, this will get sent to what is informally referred to as $F(c, -)$. This functor takes an object $d in D$, and sends it to $F(c, d)$. So what does it do on morphisms? Given $g: d to d'$, we need to get a morphism $F(c, g): F(c, d) to F(c, d')$, and the only reasonable definition is to have $F((mathrm{id}_C, g))$, and you can check that this works. Thus: the information of what happens when you first put an object, and then a morphism, into $lambda F$ is already contained in $F$: you just use the identity morphism of the object.
For the other direction: if $f: c to c'$ is a morphism, $lambda F(f) = F(f, -)$ ought to be a morphism from $F(c, -) to F(c', -)$. That is, for each $d in D$ we should have a morphism $F(c, d) to F(c', d)$. Again our only real choice is to have this be $F((f, mathrm{id}_D))$. Again, you can check that this works.
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
answered Jan 30 at 14:35
Mees de VriesMees de Vries
17.6k13060
17.6k13060
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
add a comment |
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
What still confuses me about the last part is that $lambda F(f)$ is not a functor but a natural transformation (i.e. a morphism in $C$), and so I don't really see why it even makes sense to talk about $lambda F (f) = F(f,-)$. (this is different from the case $lambda F(c)$ where $c$ is an object, because $lambda F(c)$ really is a functor.)
$endgroup$
– user56834
Feb 4 at 17:06
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
If you don't like it you can ignore the notation $F(f, -)$ -- it's an abuse of notation that makes it easier for me to keep track of what's happening.
$endgroup$
– Mees de Vries
Feb 4 at 17:12
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
but even dropping that notation, I am unsure what $lambda F(f)$ is. It is a morphism, and therefore I don’t see how we can “apply” something to it. In other words, I see how we can curry the “object version” of $F$: $F(c,d) = $lambda F(c) (d)$, because $lambda F(c)$ is a functor again. But I can’t see what $lambda F(f) (g)$ even means, because $lambda F(f)$ is not a functor.
$endgroup$
– user56834
Feb 4 at 17:25
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
$begingroup$
You're right that you cannot "apply" $lambda F(f)$ to something in the same way that you might $lambda F(c)$. The exact way in which the information is encoded in $lambda F$ is different from the way it is encoded in $F$. In particular, $lambda F(f)(d)$ could be an abuse of notation for $lambda F(f)_d : lambda F(c)(d) to lambda F(c')(d)$. And you could abuse your notation further to make $lambda F(f)(g)$, with $g: d to d'$, be equal to either the composition $lambda F(f)(d') circ lambda F(c)(g)$ or $lambda F(c')(g) circ lambda F(f)(d)$ -- naturality says they are equal.
$endgroup$
– Mees de Vries
Feb 4 at 17:31
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