Mixing
Can I define a functor F and a “ΔF” of sorts, which will uniquely determine a new functor?
$begingroup$
Let $F: mathcal{C} rightarrow mathcal{D}$ be a functor.
Natural transformation between $F$ and some other functor is defined as an assignment of a morphism in $mathcal{D}$ to each object in $mathcal{C}$, such that certain coherence conditions are satisfied.
However, given only such an assignment, it seems to me that the resulting functor isn't uniquely determined, as it doesn't specify what the morphisms are of a new functor. (as mentioned in this relevant question).
Is there a construction which will, given a functor $F$ and a "natural-transformation-like" construction $alpha': F Rightarrow ?$ map a functor to a new one?
In other words, can I define a functor $F$ and a "$Delta F$" of sorts, which will uniquely determine a new functor?
category-theory functors natural-transformations
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
$endgroup$
add a comment |
$begingroup$
Let $F: mathcal{C} rightarrow mathcal{D}$ be a functor.
Natural transformation between $F$ and some other functor is defined as an assignment of a morphism in $mathcal{D}$ to each object in $mathcal{C}$, such that certain coherence conditions are satisfied.
However, given only such an assignment, it seems to me that the resulting functor isn't uniquely determined, as it doesn't specify what the morphisms are of a new functor. (as mentioned in this relevant question).
Is there a construction which will, given a functor $F$ and a "natural-transformation-like" construction $alpha': F Rightarrow ?$ map a functor to a new one?
In other words, can I define a functor $F$ and a "$Delta F$" of sorts, which will uniquely determine a new functor?
category-theory functors natural-transformations
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
$endgroup$
$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55
add a comment |
$begingroup$
Let $F: mathcal{C} rightarrow mathcal{D}$ be a functor.
Natural transformation between $F$ and some other functor is defined as an assignment of a morphism in $mathcal{D}$ to each object in $mathcal{C}$, such that certain coherence conditions are satisfied.
However, given only such an assignment, it seems to me that the resulting functor isn't uniquely determined, as it doesn't specify what the morphisms are of a new functor. (as mentioned in this relevant question).
Is there a construction which will, given a functor $F$ and a "natural-transformation-like" construction $alpha': F Rightarrow ?$ map a functor to a new one?
In other words, can I define a functor $F$ and a "$Delta F$" of sorts, which will uniquely determine a new functor?
category-theory functors natural-transformations
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
$endgroup$
Let $F: mathcal{C} rightarrow mathcal{D}$ be a functor.
Natural transformation between $F$ and some other functor is defined as an assignment of a morphism in $mathcal{D}$ to each object in $mathcal{C}$, such that certain coherence conditions are satisfied.
However, given only such an assignment, it seems to me that the resulting functor isn't uniquely determined, as it doesn't specify what the morphisms are of a new functor. (as mentioned in this relevant question).
Is there a construction which will, given a functor $F$ and a "natural-transformation-like" construction $alpha': F Rightarrow ?$ map a functor to a new one?
In other words, can I define a functor $F$ and a "$Delta F$" of sorts, which will uniquely determine a new functor?
category-theory functors natural-transformations
category-theory functors natural-transformations
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
asked Jan 20 at 11:52
Bruno GavranovicBruno Gavranovic
434
434
$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55
add a comment |
$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55
$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm not sure if this is exactly what you are looking for, but there is an example of fundamental construction in category theory, when "natural transformation is given, but codomain functor is not".
Let $mathcal{C}$ and $mathcal{D}$ be categories, $mathcal{F}colonmathcal{C}tomathcal{D}$ be a functor. Suppose that we are interested in its left adjoint functor $mathcal{G}colonmathcal{D}tomathcal{C}$, we know the action of $mathcal{G}$ on objects and we know the unit morphisms $eta_dcolon dtomathcal{F}(mathcal{G}(d))$, but we don't know the action of $mathcal{G}$ on morphisms. In this case we can think about $eta$ as a "natural transformation without codomain functor" or "natural transformation with codomain functor defined only on objects" $etacolon mathcal{I}_{mathcal{D}}tomathcal{F}circmathcal{G}$. By this data, we can uniquely determine the action of $mathcal{G}$ on morphisms by the requirement that $eta$ is natural. Indeed, by the universal property of unit morphisms, for every $fcolon d_1to d_2$ there exists a unique morphism $g_fcolonmathcal{G}(d_1)tomathcal{G}(d_2)$, such that $eta_{d_2}circ f=mathcal{F}(g_f)circeta_{d_1}$, so we can only put $mathcal{G}(f)=g_f$.
There are some similar constructions, connected with adjoint functors and limits, when we can uniquely determine domain or codomain functor of transformation by naturality requirements. In general case, as it was mentioned in answers by the link, we can't.
answered Jan 21 at 16:46
OskarOskar
3,1731819
$endgroup$
add a comment |
$begingroup$
We can start observing that natural transformations, despite their name, do not transform functors into functors, they are not functions between functors.
They are data that somehow relate functors.
Also we can observe that if we do not known the functors the natural transformations relate we cannot state the naturalness conditions, which I suppose you are reffering as coherence conditions.
What sometimes happens is what Oskar's answer describes. Sometimes we can find for a functor $mathcal F colon mathbf C to mathbf D$ a family of morphism of the form $tau colon F(c) to G(c)$ parametrized by objects of $mathbf C$ that satisfies a universal property which allows to complete the object function $G$, defined by the family, to a functor in such a way that the family of morphisms is a natural transformation between $F$ and this new functor.
Examples of this phenomena are give by adjoint functors where using the units and counits of the adjunction one can define the left/right adjoint to a given functor.
Another example is given by families of isomorphisms: if $F$ is a functor then every family of isomorphisms of the form
$(tau_c colon F(c) to G(c))_c$ defines a functor $G colon mathbf D to mathbf C$ whose action on morphisms is given by the equations
$$G(f) = tau_{c'} circ F(f) circ tau_c^{-1}$$
where $f in mathbf C(c,c')$.
I hope this helps.
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
add a comment |
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$begingroup$
I'm not sure if this is exactly what you are looking for, but there is an example of fundamental construction in category theory, when "natural transformation is given, but codomain functor is not".
Let $mathcal{C}$ and $mathcal{D}$ be categories, $mathcal{F}colonmathcal{C}tomathcal{D}$ be a functor. Suppose that we are interested in its left adjoint functor $mathcal{G}colonmathcal{D}tomathcal{C}$, we know the action of $mathcal{G}$ on objects and we know the unit morphisms $eta_dcolon dtomathcal{F}(mathcal{G}(d))$, but we don't know the action of $mathcal{G}$ on morphisms. In this case we can think about $eta$ as a "natural transformation without codomain functor" or "natural transformation with codomain functor defined only on objects" $etacolon mathcal{I}_{mathcal{D}}tomathcal{F}circmathcal{G}$. By this data, we can uniquely determine the action of $mathcal{G}$ on morphisms by the requirement that $eta$ is natural. Indeed, by the universal property of unit morphisms, for every $fcolon d_1to d_2$ there exists a unique morphism $g_fcolonmathcal{G}(d_1)tomathcal{G}(d_2)$, such that $eta_{d_2}circ f=mathcal{F}(g_f)circeta_{d_1}$, so we can only put $mathcal{G}(f)=g_f$.
There are some similar constructions, connected with adjoint functors and limits, when we can uniquely determine domain or codomain functor of transformation by naturality requirements. In general case, as it was mentioned in answers by the link, we can't.
answered Jan 21 at 16:46
OskarOskar
3,1731819
$endgroup$
add a comment |
$begingroup$
I'm not sure if this is exactly what you are looking for, but there is an example of fundamental construction in category theory, when "natural transformation is given, but codomain functor is not".
Let $mathcal{C}$ and $mathcal{D}$ be categories, $mathcal{F}colonmathcal{C}tomathcal{D}$ be a functor. Suppose that we are interested in its left adjoint functor $mathcal{G}colonmathcal{D}tomathcal{C}$, we know the action of $mathcal{G}$ on objects and we know the unit morphisms $eta_dcolon dtomathcal{F}(mathcal{G}(d))$, but we don't know the action of $mathcal{G}$ on morphisms. In this case we can think about $eta$ as a "natural transformation without codomain functor" or "natural transformation with codomain functor defined only on objects" $etacolon mathcal{I}_{mathcal{D}}tomathcal{F}circmathcal{G}$. By this data, we can uniquely determine the action of $mathcal{G}$ on morphisms by the requirement that $eta$ is natural. Indeed, by the universal property of unit morphisms, for every $fcolon d_1to d_2$ there exists a unique morphism $g_fcolonmathcal{G}(d_1)tomathcal{G}(d_2)$, such that $eta_{d_2}circ f=mathcal{F}(g_f)circeta_{d_1}$, so we can only put $mathcal{G}(f)=g_f$.
There are some similar constructions, connected with adjoint functors and limits, when we can uniquely determine domain or codomain functor of transformation by naturality requirements. In general case, as it was mentioned in answers by the link, we can't.
answered Jan 21 at 16:46
OskarOskar
3,1731819
$endgroup$
add a comment |
$begingroup$
I'm not sure if this is exactly what you are looking for, but there is an example of fundamental construction in category theory, when "natural transformation is given, but codomain functor is not".
Let $mathcal{C}$ and $mathcal{D}$ be categories, $mathcal{F}colonmathcal{C}tomathcal{D}$ be a functor. Suppose that we are interested in its left adjoint functor $mathcal{G}colonmathcal{D}tomathcal{C}$, we know the action of $mathcal{G}$ on objects and we know the unit morphisms $eta_dcolon dtomathcal{F}(mathcal{G}(d))$, but we don't know the action of $mathcal{G}$ on morphisms. In this case we can think about $eta$ as a "natural transformation without codomain functor" or "natural transformation with codomain functor defined only on objects" $etacolon mathcal{I}_{mathcal{D}}tomathcal{F}circmathcal{G}$. By this data, we can uniquely determine the action of $mathcal{G}$ on morphisms by the requirement that $eta$ is natural. Indeed, by the universal property of unit morphisms, for every $fcolon d_1to d_2$ there exists a unique morphism $g_fcolonmathcal{G}(d_1)tomathcal{G}(d_2)$, such that $eta_{d_2}circ f=mathcal{F}(g_f)circeta_{d_1}$, so we can only put $mathcal{G}(f)=g_f$.
There are some similar constructions, connected with adjoint functors and limits, when we can uniquely determine domain or codomain functor of transformation by naturality requirements. In general case, as it was mentioned in answers by the link, we can't.
answered Jan 21 at 16:46
OskarOskar
3,1731819
$endgroup$
I'm not sure if this is exactly what you are looking for, but there is an example of fundamental construction in category theory, when "natural transformation is given, but codomain functor is not".
Let $mathcal{C}$ and $mathcal{D}$ be categories, $mathcal{F}colonmathcal{C}tomathcal{D}$ be a functor. Suppose that we are interested in its left adjoint functor $mathcal{G}colonmathcal{D}tomathcal{C}$, we know the action of $mathcal{G}$ on objects and we know the unit morphisms $eta_dcolon dtomathcal{F}(mathcal{G}(d))$, but we don't know the action of $mathcal{G}$ on morphisms. In this case we can think about $eta$ as a "natural transformation without codomain functor" or "natural transformation with codomain functor defined only on objects" $etacolon mathcal{I}_{mathcal{D}}tomathcal{F}circmathcal{G}$. By this data, we can uniquely determine the action of $mathcal{G}$ on morphisms by the requirement that $eta$ is natural. Indeed, by the universal property of unit morphisms, for every $fcolon d_1to d_2$ there exists a unique morphism $g_fcolonmathcal{G}(d_1)tomathcal{G}(d_2)$, such that $eta_{d_2}circ f=mathcal{F}(g_f)circeta_{d_1}$, so we can only put $mathcal{G}(f)=g_f$.
There are some similar constructions, connected with adjoint functors and limits, when we can uniquely determine domain or codomain functor of transformation by naturality requirements. In general case, as it was mentioned in answers by the link, we can't.
answered Jan 21 at 16:46
OskarOskar
3,1731819
answered Jan 21 at 16:46
OskarOskar
3,1731819
answered Jan 21 at 16:46
OskarOskar
3,1731819
answered Jan 21 at 16:46
OskarOskar
3,1731819
3,1731819
add a comment |
add a comment |
$begingroup$
We can start observing that natural transformations, despite their name, do not transform functors into functors, they are not functions between functors.
They are data that somehow relate functors.
Also we can observe that if we do not known the functors the natural transformations relate we cannot state the naturalness conditions, which I suppose you are reffering as coherence conditions.
What sometimes happens is what Oskar's answer describes. Sometimes we can find for a functor $mathcal F colon mathbf C to mathbf D$ a family of morphism of the form $tau colon F(c) to G(c)$ parametrized by objects of $mathbf C$ that satisfies a universal property which allows to complete the object function $G$, defined by the family, to a functor in such a way that the family of morphisms is a natural transformation between $F$ and this new functor.
Examples of this phenomena are give by adjoint functors where using the units and counits of the adjunction one can define the left/right adjoint to a given functor.
Another example is given by families of isomorphisms: if $F$ is a functor then every family of isomorphisms of the form
$(tau_c colon F(c) to G(c))_c$ defines a functor $G colon mathbf D to mathbf C$ whose action on morphisms is given by the equations
$$G(f) = tau_{c'} circ F(f) circ tau_c^{-1}$$
where $f in mathbf C(c,c')$.
I hope this helps.
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
add a comment |
$begingroup$
We can start observing that natural transformations, despite their name, do not transform functors into functors, they are not functions between functors.
They are data that somehow relate functors.
Also we can observe that if we do not known the functors the natural transformations relate we cannot state the naturalness conditions, which I suppose you are reffering as coherence conditions.
What sometimes happens is what Oskar's answer describes. Sometimes we can find for a functor $mathcal F colon mathbf C to mathbf D$ a family of morphism of the form $tau colon F(c) to G(c)$ parametrized by objects of $mathbf C$ that satisfies a universal property which allows to complete the object function $G$, defined by the family, to a functor in such a way that the family of morphisms is a natural transformation between $F$ and this new functor.
Examples of this phenomena are give by adjoint functors where using the units and counits of the adjunction one can define the left/right adjoint to a given functor.
Another example is given by families of isomorphisms: if $F$ is a functor then every family of isomorphisms of the form
$(tau_c colon F(c) to G(c))_c$ defines a functor $G colon mathbf D to mathbf C$ whose action on morphisms is given by the equations
$$G(f) = tau_{c'} circ F(f) circ tau_c^{-1}$$
where $f in mathbf C(c,c')$.
I hope this helps.
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
add a comment |
$begingroup$
We can start observing that natural transformations, despite their name, do not transform functors into functors, they are not functions between functors.
They are data that somehow relate functors.
Also we can observe that if we do not known the functors the natural transformations relate we cannot state the naturalness conditions, which I suppose you are reffering as coherence conditions.
What sometimes happens is what Oskar's answer describes. Sometimes we can find for a functor $mathcal F colon mathbf C to mathbf D$ a family of morphism of the form $tau colon F(c) to G(c)$ parametrized by objects of $mathbf C$ that satisfies a universal property which allows to complete the object function $G$, defined by the family, to a functor in such a way that the family of morphisms is a natural transformation between $F$ and this new functor.
Examples of this phenomena are give by adjoint functors where using the units and counits of the adjunction one can define the left/right adjoint to a given functor.
Another example is given by families of isomorphisms: if $F$ is a functor then every family of isomorphisms of the form
$(tau_c colon F(c) to G(c))_c$ defines a functor $G colon mathbf D to mathbf C$ whose action on morphisms is given by the equations
$$G(f) = tau_{c'} circ F(f) circ tau_c^{-1}$$
where $f in mathbf C(c,c')$.
I hope this helps.
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
We can start observing that natural transformations, despite their name, do not transform functors into functors, they are not functions between functors.
They are data that somehow relate functors.
Also we can observe that if we do not known the functors the natural transformations relate we cannot state the naturalness conditions, which I suppose you are reffering as coherence conditions.
What sometimes happens is what Oskar's answer describes. Sometimes we can find for a functor $mathcal F colon mathbf C to mathbf D$ a family of morphism of the form $tau colon F(c) to G(c)$ parametrized by objects of $mathbf C$ that satisfies a universal property which allows to complete the object function $G$, defined by the family, to a functor in such a way that the family of morphisms is a natural transformation between $F$ and this new functor.
Examples of this phenomena are give by adjoint functors where using the units and counits of the adjunction one can define the left/right adjoint to a given functor.
Another example is given by families of isomorphisms: if $F$ is a functor then every family of isomorphisms of the form
$(tau_c colon F(c) to G(c))_c$ defines a functor $G colon mathbf D to mathbf C$ whose action on morphisms is given by the equations
$$G(f) = tau_{c'} circ F(f) circ tau_c^{-1}$$
where $f in mathbf C(c,c')$.
I hope this helps.
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
edited Jan 22 at 8:38
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
answered Jan 22 at 1:12
Giorgio MossaGiorgio Mossa
14.2k11749
14.2k11749
add a comment |
add a comment |
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$begingroup$
What do you mean "resulting functor"? Natural transformations don't have "resulting functors".
$endgroup$
– Malice Vidrine
Jan 20 at 17:39
$begingroup$
@MaliceVidrine, that's correct, which is why I explain the context and restate the question below the sentence you're referring to in a different way. To make an analogy: given a number $2$, I can perform the operation $(+3)$ on it to obtain 5. Given a functor $F$, is there a $Delta F$ I can apply on it to obtain a new functor?
$endgroup$
– Bruno Gavranovic
Jan 20 at 17:55