understanding natural transformations that are not natural isomorphisms
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What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.
This means that the ordinary natural transformation law
$$ eta_Y circ F(f) = G(f) circ eta_x $$
can be written as
$$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$
or, using $triangleright$ as reverse composition ...
$$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$
In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.
However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.
If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?
category-theory natural-transformations
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What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.
This means that the ordinary natural transformation law
$$ eta_Y circ F(f) = G(f) circ eta_x $$
can be written as
$$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$
or, using $triangleright$ as reverse composition ...
$$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$
In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.
However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.
If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?
category-theory natural-transformations
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.
This means that the ordinary natural transformation law
$$ eta_Y circ F(f) = G(f) circ eta_x $$
can be written as
$$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$
or, using $triangleright$ as reverse composition ...
$$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$
In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.
However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.
If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?
category-theory natural-transformations
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
A natural equivalence or natural isomorphism is a natural transformation $eta : F stackrel{nt}{leftrightarrow} G$ where all the arrows in the "image" (what's the actual term?) of $eta$ are isomorphisms. So, for every $X$ in the object of the domain of our functors $F : C stackrel{ftr}{to} D$ and $G : C stackrel{ftr}{to} D$, $eta_X : D_{text{arr}}$ and $eta_X^{-1}$ both exist.
This means that the ordinary natural transformation law
$$ eta_Y circ F(f) = G(f) circ eta_x $$
can be written as
$$ F(f) = eta_Y^{-1} circ G(f) circ eta_X $$
or, using $triangleright$ as reverse composition ...
$$ F(f) = eta_X triangleright G(f) triangleright eta_Y^{-1} $$
In my opinion, writing it this way makes it clear just how strong a claim the existence of $eta$ is making. We can apply the functor $F$ to an arrow by applying $G$ instead and then "correcting it" by adding $eta_X$ in front and $eta_Y^{-1}$ behind it.
However, $eta_X$, crucially, does not depend on $f$ at all. It isn't enough that an arrow exists from the source of $F(f)$ to the source of $G(f)$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $f$'s homset.
If $eta$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $F$ and $G$ ?
category-theory natural-transformations
category-theory natural-transformations
edited Nov 19 at 0:31
Erick Wong
20.1k22666
20.1k22666
asked Nov 17 at 3:54
Gregory Nisbet
381211
381211
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1 Answer
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What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.
A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.
A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.
add a comment |
up vote
4
down vote
accepted
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.
A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.
A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.
What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?
No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $A$ is an abelian category and $F, G : C to A$ are two functors, then there is always a natural transformation $eta : F to G$, namely the zero natural transformation, which tells you nothing.
A nice concrete example here is if $A = text{Vect}$ and $C = BG$ is the one-object category with automorphisms a group $G$. Then the functor category $[BG, text{Vect}]$ is the category of linear representations of $G$, with natural transformations given by $G$-equivariant maps.
answered Nov 17 at 4:14
Qiaochu Yuan
275k32578914
275k32578914
add a comment |
add a comment |
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