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Let $C,D$ be a categories, $F,G:Ctimes Cto C$ functors with an natural isomorphism $eta_X$ between them, and $E:Cto D$ an equivalence.
Is it true that $E(eta_X)$ is a natural isomorphism as well?
abstract-algebra category-theory functors
asked Jul 30 at 10:15
user372565
1958
add a comment |Â
up vote
0
down vote
favorite
Let $C,D$ be a categories, $F,G:Ctimes Cto C$ functors with an natural isomorphism $eta_X$ between them, and $E:Cto D$ an equivalence.
Is it true that $E(eta_X)$ is a natural isomorphism as well?
abstract-algebra category-theory functors
asked Jul 30 at 10:15
user372565
1958
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $C,D$ be a categories, $F,G:Ctimes Cto C$ functors with an natural isomorphism $eta_X$ between them, and $E:Cto D$ an equivalence.
Is it true that $E(eta_X)$ is a natural isomorphism as well?
abstract-algebra category-theory functors
asked Jul 30 at 10:15
user372565
1958
Let $C,D$ be a categories, $F,G:Ctimes Cto C$ functors with an natural isomorphism $eta_X$ between them, and $E:Cto D$ an equivalence.
Is it true that $E(eta_X)$ is a natural isomorphism as well?
abstract-algebra category-theory functors
asked Jul 30 at 10:15
user372565
1958
asked Jul 30 at 10:15
user372565
1958
asked Jul 30 at 10:15
user372565
1958
asked Jul 30 at 10:15
user372565
1958
1958
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Given any morphism $f: (A,B) to (A',B')$ in $C times C$, then with your natural transformation $eta_(A,B): F(A,B) to G(A,B)$ you have
$$G(f) circ eta_(A,B) = eta_(A',B') circ F(f)$$
Applying $E$ on both sides, you get
$$EG(f) circ E(eta_(A,B)) = E(eta_(A',B')) circ EF(f)$$
so $E(eta_(A,B))_(A,B)$ is a natural transformation between $EF$ and $EG$.
Note that $eta$ is an isomorphism iff $eta_(A,B)$ is an isomorphism for all objects $(A,B)$ in $C times C$. Since functors map isomorphisms to isomorphisms we find that $E(eta)$ is a natural isomorphism if $eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C to D$.
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
add a comment |Â
up vote
0
down vote
Using juxtaposition for horizontal composition and $cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)
Let $eta : F to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,
$$(E eta) cdot (E eta^-1) = (E cdot E) (eta cdot eta^-1) = E G $$
$$(E eta^-1) cdot (E eta) = (E cdot E) (eta^-1 cdot eta) = E F $$
Both vertical composites are the respective identity natural transformations, and thus $E eta^-1$ is the inverse of $E eta$.
answered Jul 30 at 12:30
Hurkyl
107k9112253
add a comment |Â
up vote
0
down vote
I think as a matter of notation, it's worth asking what $E(eta_X)$ really means—it's just a shorthand for the horizontal composition $textid_Ecirc eta_Xin textHom_D^Ctimes C(Ecirc F,Ecirc G)$. And seeing as $textid_E$, $eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $textid_E^-1circ eta_X^-1$).
answered Jul 30 at 14:15
Rafay A.
1164
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Given any morphism $f: (A,B) to (A',B')$ in $C times C$, then with your natural transformation $eta_(A,B): F(A,B) to G(A,B)$ you have
$$G(f) circ eta_(A,B) = eta_(A',B') circ F(f)$$
Applying $E$ on both sides, you get
$$EG(f) circ E(eta_(A,B)) = E(eta_(A',B')) circ EF(f)$$
so $E(eta_(A,B))_(A,B)$ is a natural transformation between $EF$ and $EG$.
Note that $eta$ is an isomorphism iff $eta_(A,B)$ is an isomorphism for all objects $(A,B)$ in $C times C$. Since functors map isomorphisms to isomorphisms we find that $E(eta)$ is a natural isomorphism if $eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C to D$.
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
add a comment |Â
up vote
3
down vote
accepted
Given any morphism $f: (A,B) to (A',B')$ in $C times C$, then with your natural transformation $eta_(A,B): F(A,B) to G(A,B)$ you have
$$G(f) circ eta_(A,B) = eta_(A',B') circ F(f)$$
Applying $E$ on both sides, you get
$$EG(f) circ E(eta_(A,B)) = E(eta_(A',B')) circ EF(f)$$
so $E(eta_(A,B))_(A,B)$ is a natural transformation between $EF$ and $EG$.
Note that $eta$ is an isomorphism iff $eta_(A,B)$ is an isomorphism for all objects $(A,B)$ in $C times C$. Since functors map isomorphisms to isomorphisms we find that $E(eta)$ is a natural isomorphism if $eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C to D$.
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Given any morphism $f: (A,B) to (A',B')$ in $C times C$, then with your natural transformation $eta_(A,B): F(A,B) to G(A,B)$ you have
$$G(f) circ eta_(A,B) = eta_(A',B') circ F(f)$$
Applying $E$ on both sides, you get
$$EG(f) circ E(eta_(A,B)) = E(eta_(A',B')) circ EF(f)$$
so $E(eta_(A,B))_(A,B)$ is a natural transformation between $EF$ and $EG$.
Note that $eta$ is an isomorphism iff $eta_(A,B)$ is an isomorphism for all objects $(A,B)$ in $C times C$. Since functors map isomorphisms to isomorphisms we find that $E(eta)$ is a natural isomorphism if $eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C to D$.
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
Given any morphism $f: (A,B) to (A',B')$ in $C times C$, then with your natural transformation $eta_(A,B): F(A,B) to G(A,B)$ you have
$$G(f) circ eta_(A,B) = eta_(A',B') circ F(f)$$
Applying $E$ on both sides, you get
$$EG(f) circ E(eta_(A,B)) = E(eta_(A',B')) circ EF(f)$$
so $E(eta_(A,B))_(A,B)$ is a natural transformation between $EF$ and $EG$.
Note that $eta$ is an isomorphism iff $eta_(A,B)$ is an isomorphism for all objects $(A,B)$ in $C times C$. Since functors map isomorphisms to isomorphisms we find that $E(eta)$ is a natural isomorphism if $eta$ is one. Note that we never used that $E$ is an equivalence, this works for arbitrary functors $C to D$.
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
answered Jul 30 at 10:27
Matthias Klupsch
6,0341127
6,0341127
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
add a comment |Â
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
Thank you very much! So if I have a braiding $c$ for a category $C$, which is a natural transformation, then $E(c)$ is a braiding of $D$? At least it is again a natural transformation?
– user372565
Jul 30 at 10:47
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
A braiding $c$ of $C$ is a natural transformation $c_x,y : x otimes y to y otimes x$, so we get a natural transformation $E(c): E(x otimes y) to E(y otimes x)$ but nothing more without further assumptions. To get a braiding you will have to have monoidal structures on $C$ and $D$ and $E$ to be a monoidal functor and an equivalence so that your braiding becomes $E(c_E^-1(a),E^-1(b)) : a otimes b cong E(E^-1(a) otimes E^-1(b)) to E(E^-1(b) otimes E^-1(a)) cong b otimes a$.
– Matthias Klupsch
Jul 30 at 11:05
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
Note that I am not completely rigorous here, since the isomorphisms left an right should also be included in the notation.
– Matthias Klupsch
Jul 30 at 11:11
add a comment |Â
up vote
0
down vote
Using juxtaposition for horizontal composition and $cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)
Let $eta : F to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,
$$(E eta) cdot (E eta^-1) = (E cdot E) (eta cdot eta^-1) = E G $$
$$(E eta^-1) cdot (E eta) = (E cdot E) (eta^-1 cdot eta) = E F $$
Both vertical composites are the respective identity natural transformations, and thus $E eta^-1$ is the inverse of $E eta$.
answered Jul 30 at 12:30
Hurkyl
107k9112253
add a comment |Â
up vote
0
down vote
Using juxtaposition for horizontal composition and $cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)
Let $eta : F to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,
$$(E eta) cdot (E eta^-1) = (E cdot E) (eta cdot eta^-1) = E G $$
$$(E eta^-1) cdot (E eta) = (E cdot E) (eta^-1 cdot eta) = E F $$
Both vertical composites are the respective identity natural transformations, and thus $E eta^-1$ is the inverse of $E eta$.
answered Jul 30 at 12:30
Hurkyl
107k9112253
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using juxtaposition for horizontal composition and $cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)
Let $eta : F to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,
$$(E eta) cdot (E eta^-1) = (E cdot E) (eta cdot eta^-1) = E G $$
$$(E eta^-1) cdot (E eta) = (E cdot E) (eta^-1 cdot eta) = E F $$
Both vertical composites are the respective identity natural transformations, and thus $E eta^-1$ is the inverse of $E eta$.
answered Jul 30 at 12:30
Hurkyl
107k9112253
Using juxtaposition for horizontal composition and $cdot$ for vertical composition (and identifying functors with their corresponding identity natural transformations as appropriate)
Let $eta : F to G$ be any natural isomorphism and $E$ be any functor at all. Then by the interchange law,
$$(E eta) cdot (E eta^-1) = (E cdot E) (eta cdot eta^-1) = E G $$
$$(E eta^-1) cdot (E eta) = (E cdot E) (eta^-1 cdot eta) = E F $$
Both vertical composites are the respective identity natural transformations, and thus $E eta^-1$ is the inverse of $E eta$.
answered Jul 30 at 12:30
Hurkyl
107k9112253
answered Jul 30 at 12:30
Hurkyl
107k9112253
answered Jul 30 at 12:30
Hurkyl
107k9112253
answered Jul 30 at 12:30
Hurkyl
107k9112253
107k9112253
add a comment |Â
add a comment |Â
up vote
0
down vote
I think as a matter of notation, it's worth asking what $E(eta_X)$ really means—it's just a shorthand for the horizontal composition $textid_Ecirc eta_Xin textHom_D^Ctimes C(Ecirc F,Ecirc G)$. And seeing as $textid_E$, $eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $textid_E^-1circ eta_X^-1$).
answered Jul 30 at 14:15
Rafay A.
1164
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
add a comment |Â
up vote
0
down vote
I think as a matter of notation, it's worth asking what $E(eta_X)$ really means—it's just a shorthand for the horizontal composition $textid_Ecirc eta_Xin textHom_D^Ctimes C(Ecirc F,Ecirc G)$. And seeing as $textid_E$, $eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $textid_E^-1circ eta_X^-1$).
answered Jul 30 at 14:15
Rafay A.
1164
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think as a matter of notation, it's worth asking what $E(eta_X)$ really means—it's just a shorthand for the horizontal composition $textid_Ecirc eta_Xin textHom_D^Ctimes C(Ecirc F,Ecirc G)$. And seeing as $textid_E$, $eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $textid_E^-1circ eta_X^-1$).
answered Jul 30 at 14:15
Rafay A.
1164
I think as a matter of notation, it's worth asking what $E(eta_X)$ really means—it's just a shorthand for the horizontal composition $textid_Ecirc eta_Xin textHom_D^Ctimes C(Ecirc F,Ecirc G)$. And seeing as $textid_E$, $eta_X$ are both natural isomorphisms, their horizontal composition must also be a natural isomorphism (with explicit vertical inverse $textid_E^-1circ eta_X^-1$).
answered Jul 30 at 14:15
Rafay A.
1164
edited Jul 30 at 14:51
answered Jul 30 at 14:15
Rafay A.
1164
answered Jul 30 at 14:15
Rafay A.
1164
answered Jul 30 at 14:15
Rafay A.
1164
1164
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
add a comment |Â
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
I'm not sure this is really that helpful. Here "natural isomorphism" means isomorphism for the vertical composition, so the reason the composite is an iso has more to do with the functoriality of horizontal composition than it being a form of composition; and for this specific case it really boils down to the fact that $E$ is a functor, as the accepted answer already mentions.
– Arnaud D.
Jul 30 at 14:25
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
Fair enough, it seems to boil down to applying interchange implicitly vs. explicitly.
– Rafay A.
Jul 30 at 14:33
add a comment |Â
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