Commuting diagrams and injectivity
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Let $C,D$ be a categories, $G$ a group and $g:GtotextAut( C)$, $h:GtotextAut(D)$ group homomorphisms (into the group of classes of autoequivalences). Further assume $g$ to be injective and let $F:Cto D$ be an equivalence such that the following diagram commutes for all $ain G$:
$$Fcirc g(a)=h(a)circ F.$$
Is it true that $h$ is injective as well?
abstract-algebra group-theory category-theory
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Let $C,D$ be a categories, $G$ a group and $g:GtotextAut( C)$, $h:GtotextAut(D)$ group homomorphisms (into the group of classes of autoequivalences). Further assume $g$ to be injective and let $F:Cto D$ be an equivalence such that the following diagram commutes for all $ain G$:
$$Fcirc g(a)=h(a)circ F.$$
Is it true that $h$ is injective as well?
abstract-algebra group-theory category-theory
do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
I do allow this :)
– user591938
Sep 10 at 20:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $C,D$ be a categories, $G$ a group and $g:GtotextAut( C)$, $h:GtotextAut(D)$ group homomorphisms (into the group of classes of autoequivalences). Further assume $g$ to be injective and let $F:Cto D$ be an equivalence such that the following diagram commutes for all $ain G$:
$$Fcirc g(a)=h(a)circ F.$$
Is it true that $h$ is injective as well?
abstract-algebra group-theory category-theory
Let $C,D$ be a categories, $G$ a group and $g:GtotextAut( C)$, $h:GtotextAut(D)$ group homomorphisms (into the group of classes of autoequivalences). Further assume $g$ to be injective and let $F:Cto D$ be an equivalence such that the following diagram commutes for all $ain G$:
$$Fcirc g(a)=h(a)circ F.$$
Is it true that $h$ is injective as well?
abstract-algebra group-theory category-theory
abstract-algebra group-theory category-theory
asked Sep 10 at 20:02
user591938
do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
I do allow this :)
– user591938
Sep 10 at 20:20
add a comment |Â
do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
I do allow this :)
– user591938
Sep 10 at 20:20
do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
I do allow this :)
– user591938
Sep 10 at 20:20
I do allow this :)
– user591938
Sep 10 at 20:20
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Assume $h(a) simeq mathrmid_D$. Then $h(a)circ F simeq F simeq Fcirc g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)simeq mathrmid_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)simeq g(b) implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)simeq mathbbZ/2Z$, $Aut(D)simeq 1$.
Put $G= mathbbZ/2Z$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:Cto D$ be the equivalence. Then $Fcirc g(a) = h(a)circ F$ for $ain G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.
answered Sep 10 at 20:30
Max
11k11037
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assume $h(a) simeq mathrmid_D$. Then $h(a)circ F simeq F simeq Fcirc g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)simeq mathrmid_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)simeq g(b) implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)simeq mathbbZ/2Z$, $Aut(D)simeq 1$.
Put $G= mathbbZ/2Z$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:Cto D$ be the equivalence. Then $Fcirc g(a) = h(a)circ F$ for $ain G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.
answered Sep 10 at 20:30
Max
11k11037
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
add a comment |Â
up vote
1
down vote
accepted
Assume $h(a) simeq mathrmid_D$. Then $h(a)circ F simeq F simeq Fcirc g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)simeq mathrmid_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)simeq g(b) implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)simeq mathbbZ/2Z$, $Aut(D)simeq 1$.
Put $G= mathbbZ/2Z$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:Cto D$ be the equivalence. Then $Fcirc g(a) = h(a)circ F$ for $ain G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.
answered Sep 10 at 20:30
Max
11k11037
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assume $h(a) simeq mathrmid_D$. Then $h(a)circ F simeq F simeq Fcirc g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)simeq mathrmid_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)simeq g(b) implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)simeq mathbbZ/2Z$, $Aut(D)simeq 1$.
Put $G= mathbbZ/2Z$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:Cto D$ be the equivalence. Then $Fcirc g(a) = h(a)circ F$ for $ain G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.
answered Sep 10 at 20:30
Max
11k11037
Assume $h(a) simeq mathrmid_D$. Then $h(a)circ F simeq F simeq Fcirc g(a)$.
Compose with a quasi-inverse $H$ of $F$ to get $g(a)simeq mathrmid_C$.
Thus if your "homomorphisms" are weak homomorphisms in the sense that $g(ab)simeq g(a)g(b)$, and $g$ is "strongly injective" in the sense that $g(a)simeq g(b) implies a=b$, then this implies $a=$ the neutral element; so indeed $h$ is strongly injective as well.
However, if you stick with "normal" injectivity (i.e. $g(a)=g(b) implies a=b$) then you can't prove that $h$ is injective too.
Consider indeed the trivial connected groupoïd with two objects $C$ and $D$ the trivial group. Then $Aut(C)simeq mathbbZ/2Z$, $Aut(D)simeq 1$.
Put $G= mathbbZ/2Z$, $g$ the unique isomorphism with $Aut(C)$, and $h$ the unique morphism to $Aut(D)$. Let $F:Cto D$ be the equivalence. Then $Fcirc g(a) = h(a)circ F$ for $ain G$, but clearly $h$ is not injective.
So from "strong injectivity" of $g$ you can get "strong injectivity" of $h$, but you can't get injectivity of $h$ from injectivity of $g$.
answered Sep 10 at 20:30
Max
11k11037
answered Sep 10 at 20:30
Max
11k11037
answered Sep 10 at 20:30
Max
11k11037
answered Sep 10 at 20:30
Max
11k11037
11k11037
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
add a comment |Â
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
I don't understand the difference of weak and strong injectivity. $textAut(...)$ is defined as the group of isomorphism classes of autoequivalences. So if I have $g(ab)simeq g(a)g(b)$ then this automorphisms classes are equal and thus $g(ab)=g(a)g(b)$ in $textAut(...)$.
– user591938
Sep 10 at 20:37
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Oh you're seeing $Aut(C)$ as isomorphism classes ? I didn't understand that
– Max
Sep 10 at 21:22
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Correct :) how would that change your answer?
– user591938
Sep 10 at 21:23
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
Well then the first part of my answer is the answer : yes, it does imply that
– Max
Sep 10 at 21:42
add a comment |Â
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do you actually want $g,h$ to be homomorphisms, or do you allow, say, $g(ab) simeq g(a)g(b)$ ?
– Max
Sep 10 at 20:19
I do allow this :)
– user591938
Sep 10 at 20:20