What properties do natural isomorphisms between functors preserve?
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If $F,G: C rightarrow D$ are functors between regular categories and $F Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full? I believe this is true but would like some confirmation.
What about if $F$ is essentially surjective? Is it necessarily true that $G$ is essentially surjective?
Thanks.
category-theory
edited Nov 22 at 0:20
KReiser
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
add a comment |
up vote
1
down vote
favorite
If $F,G: C rightarrow D$ are functors between regular categories and $F Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full? I believe this is true but would like some confirmation.
What about if $F$ is essentially surjective? Is it necessarily true that $G$ is essentially surjective?
Thanks.
category-theory
edited Nov 22 at 0:20
KReiser
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $F,G: C rightarrow D$ are functors between regular categories and $F Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full? I believe this is true but would like some confirmation.
What about if $F$ is essentially surjective? Is it necessarily true that $G$ is essentially surjective?
Thanks.
category-theory
edited Nov 22 at 0:20
KReiser
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
If $F,G: C rightarrow D$ are functors between regular categories and $F Rightarrow G$ is a natural isomorphism, is it true that if $F$ is faithful {resp. full) then $G$ is full? I believe this is true but would like some confirmation.
What about if $F$ is essentially surjective? Is it necessarily true that $G$ is essentially surjective?
Thanks.
category-theory
category-theory
edited Nov 22 at 0:20
KReiser
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
edited Nov 22 at 0:20
KReiser
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
edited Nov 22 at 0:20
KReiser
9,21911335
edited Nov 22 at 0:20
KReiser
9,21911335
edited Nov 22 at 0:20
KReiser
9,21911335
9,21911335
asked Nov 26 '13 at 5:42
kjata
483
asked Nov 26 '13 at 5:42
kjata
483
asked Nov 26 '13 at 5:42
kjata
483
483
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let $C$ and $D$ be categories (arbitrary), $F,Gcolon Cto D$ functors, $alphacolon Fto G$ a natural isomorphism. Then:
- If $F$ is faithful, then $G$ is faithful.
Proof: Let $c_1,c_2in C$ be objects and $f_1,f_2:c_1to c_2$ morphisms in $C$. Then, by naturality:
$$
G(f_i)circ alpha(c_1)=alpha(c_2)circ F(f_i),quad i=1,2.
$$
We know that $alpha$ is an isomorphism, so we have:
$$
G(f_i)=alpha(c_2)circ F(f_i)circ(alpha(c_1))^{-1},quad i=1,2.
$$
Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.
- If $F$ is full, then $G$ is full.
Proof: Let $c_1,c_2in C$ be objects and $gcolon G(c_1)to G(c_2)$ a morphism in $D$. Let's define the morphism $kcolon F(c_1)to F(c_2)$ as follows:
$$
k=(alpha(c_2))^{-1}circ gcircalpha(c_1).
$$
$F$ is full, therefore there exists a morphism $fcolon c_1to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.
- If $F$ is essentially surjective, then $G$ is essentially surjective.
Proof: Let $din D$ be object, then there exists such $cin C$, that $dcong F(c)$. But $F(c)cong G(c)$ by $alpha$, thus, by transitivity, $dcong G(c)$.
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $C$ and $D$ be categories (arbitrary), $F,Gcolon Cto D$ functors, $alphacolon Fto G$ a natural isomorphism. Then:
- If $F$ is faithful, then $G$ is faithful.
Proof: Let $c_1,c_2in C$ be objects and $f_1,f_2:c_1to c_2$ morphisms in $C$. Then, by naturality:
$$
G(f_i)circ alpha(c_1)=alpha(c_2)circ F(f_i),quad i=1,2.
$$
We know that $alpha$ is an isomorphism, so we have:
$$
G(f_i)=alpha(c_2)circ F(f_i)circ(alpha(c_1))^{-1},quad i=1,2.
$$
Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.
- If $F$ is full, then $G$ is full.
Proof: Let $c_1,c_2in C$ be objects and $gcolon G(c_1)to G(c_2)$ a morphism in $D$. Let's define the morphism $kcolon F(c_1)to F(c_2)$ as follows:
$$
k=(alpha(c_2))^{-1}circ gcircalpha(c_1).
$$
$F$ is full, therefore there exists a morphism $fcolon c_1to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.
- If $F$ is essentially surjective, then $G$ is essentially surjective.
Proof: Let $din D$ be object, then there exists such $cin C$, that $dcong F(c)$. But $F(c)cong G(c)$ by $alpha$, thus, by transitivity, $dcong G(c)$.
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
add a comment |
up vote
4
down vote
accepted
Let $C$ and $D$ be categories (arbitrary), $F,Gcolon Cto D$ functors, $alphacolon Fto G$ a natural isomorphism. Then:
- If $F$ is faithful, then $G$ is faithful.
Proof: Let $c_1,c_2in C$ be objects and $f_1,f_2:c_1to c_2$ morphisms in $C$. Then, by naturality:
$$
G(f_i)circ alpha(c_1)=alpha(c_2)circ F(f_i),quad i=1,2.
$$
We know that $alpha$ is an isomorphism, so we have:
$$
G(f_i)=alpha(c_2)circ F(f_i)circ(alpha(c_1))^{-1},quad i=1,2.
$$
Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.
- If $F$ is full, then $G$ is full.
Proof: Let $c_1,c_2in C$ be objects and $gcolon G(c_1)to G(c_2)$ a morphism in $D$. Let's define the morphism $kcolon F(c_1)to F(c_2)$ as follows:
$$
k=(alpha(c_2))^{-1}circ gcircalpha(c_1).
$$
$F$ is full, therefore there exists a morphism $fcolon c_1to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.
- If $F$ is essentially surjective, then $G$ is essentially surjective.
Proof: Let $din D$ be object, then there exists such $cin C$, that $dcong F(c)$. But $F(c)cong G(c)$ by $alpha$, thus, by transitivity, $dcong G(c)$.
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $C$ and $D$ be categories (arbitrary), $F,Gcolon Cto D$ functors, $alphacolon Fto G$ a natural isomorphism. Then:
- If $F$ is faithful, then $G$ is faithful.
Proof: Let $c_1,c_2in C$ be objects and $f_1,f_2:c_1to c_2$ morphisms in $C$. Then, by naturality:
$$
G(f_i)circ alpha(c_1)=alpha(c_2)circ F(f_i),quad i=1,2.
$$
We know that $alpha$ is an isomorphism, so we have:
$$
G(f_i)=alpha(c_2)circ F(f_i)circ(alpha(c_1))^{-1},quad i=1,2.
$$
Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.
- If $F$ is full, then $G$ is full.
Proof: Let $c_1,c_2in C$ be objects and $gcolon G(c_1)to G(c_2)$ a morphism in $D$. Let's define the morphism $kcolon F(c_1)to F(c_2)$ as follows:
$$
k=(alpha(c_2))^{-1}circ gcircalpha(c_1).
$$
$F$ is full, therefore there exists a morphism $fcolon c_1to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.
- If $F$ is essentially surjective, then $G$ is essentially surjective.
Proof: Let $din D$ be object, then there exists such $cin C$, that $dcong F(c)$. But $F(c)cong G(c)$ by $alpha$, thus, by transitivity, $dcong G(c)$.
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
Let $C$ and $D$ be categories (arbitrary), $F,Gcolon Cto D$ functors, $alphacolon Fto G$ a natural isomorphism. Then:
- If $F$ is faithful, then $G$ is faithful.
Proof: Let $c_1,c_2in C$ be objects and $f_1,f_2:c_1to c_2$ morphisms in $C$. Then, by naturality:
$$
G(f_i)circ alpha(c_1)=alpha(c_2)circ F(f_i),quad i=1,2.
$$
We know that $alpha$ is an isomorphism, so we have:
$$
G(f_i)=alpha(c_2)circ F(f_i)circ(alpha(c_1))^{-1},quad i=1,2.
$$
Therefore, if $G(f_1)=G(f_2)$, then $F(f_1)=F(f_2)$, and finally $f_1=f_2$.
- If $F$ is full, then $G$ is full.
Proof: Let $c_1,c_2in C$ be objects and $gcolon G(c_1)to G(c_2)$ a morphism in $D$. Let's define the morphism $kcolon F(c_1)to F(c_2)$ as follows:
$$
k=(alpha(c_2))^{-1}circ gcircalpha(c_1).
$$
$F$ is full, therefore there exists a morphism $fcolon c_1to c_2$ in $C$, such that $F(f)=k$. It's easy to check that $G(f)=g$.
- If $F$ is essentially surjective, then $G$ is essentially surjective.
Proof: Let $din D$ be object, then there exists such $cin C$, that $dcong F(c)$. But $F(c)cong G(c)$ by $alpha$, thus, by transitivity, $dcong G(c)$.
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
edited Nov 26 '13 at 14:06
Stefan Hamcke
21.5k42877
21.5k42877
answered Nov 26 '13 at 7:17
Oskar
2,6961718
answered Nov 26 '13 at 7:17
Oskar
2,6961718
answered Nov 26 '13 at 7:17
Oskar
2,6961718
2,6961718
add a comment |
add a comment |
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