Effective epimorphisms in $G$-set
up vote
0
down vote
favorite
$newcommand{gset}{Gtext{-}mathsf{set}}$
Let $G$ be a group and $gset$ the category of $G$-sets, whose morphisms are $G$-maps (i.e. set maps where the map commutes with $G$-action).
What are the effective epimorphisms in $gset$? Recall that in a category $mathsf{C}$ an epimorphism $f:Ato B$ is a morphisms such that:
$$hom(B,Z)hookrightarrow hom(A,Z),$$
is injective for every $Zintext{ob}(mathsf{C})$, and it's an effective epimorphism if:
$$hom(B,Z)to hom(A,Z)stackrel{longrightarrow}to hom(Atimes_BA,Z),$$
is exact, where the two arrows are $pr_1^*,pr_2^*$, and by exact I mean $hom(B,Z)$ is the equalizer of this diagram.
In my previous question I make a claim what epimorphisms and the fibred products are. Apparently the effective epimorphisms in $gset$ are simply the surjective maps, but this seems false. I could take $Ato {1}$ where ${1}$ is just some singleton, and $Atimes_{{1}}A=Aprod A$ in set, with 'product action' $$rho(g)(a,a')=(rho(g)a,rho(g)a').$$
Certainly I can't obtain map $v:Ato Z$ such that $vcirc pr_1 = vcirc pr_2$ from $v'circ f$, where $v'$ is just a map ${1}to Z$?
$$require{AMScd}
begin{CD}
Atimes_{{1}}A@>>> Z;\
@V{pr_1,pr_2}VV@V{id}VV \
A @>{v}>> Z ;\
@V{f}VV @V{id}VV\
{1}@>{v'}>>Z
end{CD}$$
(Sorry about the diagram, not sure how to make commuting triangles)
abstract-algebra category-theory group-actions
asked Nov 15 at 14:45
user616128
111
add a comment |
up vote
0
down vote
favorite
$newcommand{gset}{Gtext{-}mathsf{set}}$
Let $G$ be a group and $gset$ the category of $G$-sets, whose morphisms are $G$-maps (i.e. set maps where the map commutes with $G$-action).
What are the effective epimorphisms in $gset$? Recall that in a category $mathsf{C}$ an epimorphism $f:Ato B$ is a morphisms such that:
$$hom(B,Z)hookrightarrow hom(A,Z),$$
is injective for every $Zintext{ob}(mathsf{C})$, and it's an effective epimorphism if:
$$hom(B,Z)to hom(A,Z)stackrel{longrightarrow}to hom(Atimes_BA,Z),$$
is exact, where the two arrows are $pr_1^*,pr_2^*$, and by exact I mean $hom(B,Z)$ is the equalizer of this diagram.
In my previous question I make a claim what epimorphisms and the fibred products are. Apparently the effective epimorphisms in $gset$ are simply the surjective maps, but this seems false. I could take $Ato {1}$ where ${1}$ is just some singleton, and $Atimes_{{1}}A=Aprod A$ in set, with 'product action' $$rho(g)(a,a')=(rho(g)a,rho(g)a').$$
Certainly I can't obtain map $v:Ato Z$ such that $vcirc pr_1 = vcirc pr_2$ from $v'circ f$, where $v'$ is just a map ${1}to Z$?
$$require{AMScd}
begin{CD}
Atimes_{{1}}A@>>> Z;\
@V{pr_1,pr_2}VV@V{id}VV \
A @>{v}>> Z ;\
@V{f}VV @V{id}VV\
{1}@>{v'}>>Z
end{CD}$$
(Sorry about the diagram, not sure how to make commuting triangles)
abstract-algebra category-theory group-actions
asked Nov 15 at 14:45
user616128
111
2
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$newcommand{gset}{Gtext{-}mathsf{set}}$
Let $G$ be a group and $gset$ the category of $G$-sets, whose morphisms are $G$-maps (i.e. set maps where the map commutes with $G$-action).
What are the effective epimorphisms in $gset$? Recall that in a category $mathsf{C}$ an epimorphism $f:Ato B$ is a morphisms such that:
$$hom(B,Z)hookrightarrow hom(A,Z),$$
is injective for every $Zintext{ob}(mathsf{C})$, and it's an effective epimorphism if:
$$hom(B,Z)to hom(A,Z)stackrel{longrightarrow}to hom(Atimes_BA,Z),$$
is exact, where the two arrows are $pr_1^*,pr_2^*$, and by exact I mean $hom(B,Z)$ is the equalizer of this diagram.
In my previous question I make a claim what epimorphisms and the fibred products are. Apparently the effective epimorphisms in $gset$ are simply the surjective maps, but this seems false. I could take $Ato {1}$ where ${1}$ is just some singleton, and $Atimes_{{1}}A=Aprod A$ in set, with 'product action' $$rho(g)(a,a')=(rho(g)a,rho(g)a').$$
Certainly I can't obtain map $v:Ato Z$ such that $vcirc pr_1 = vcirc pr_2$ from $v'circ f$, where $v'$ is just a map ${1}to Z$?
$$require{AMScd}
begin{CD}
Atimes_{{1}}A@>>> Z;\
@V{pr_1,pr_2}VV@V{id}VV \
A @>{v}>> Z ;\
@V{f}VV @V{id}VV\
{1}@>{v'}>>Z
end{CD}$$
(Sorry about the diagram, not sure how to make commuting triangles)
abstract-algebra category-theory group-actions
asked Nov 15 at 14:45
user616128
111
$newcommand{gset}{Gtext{-}mathsf{set}}$
Let $G$ be a group and $gset$ the category of $G$-sets, whose morphisms are $G$-maps (i.e. set maps where the map commutes with $G$-action).
What are the effective epimorphisms in $gset$? Recall that in a category $mathsf{C}$ an epimorphism $f:Ato B$ is a morphisms such that:
$$hom(B,Z)hookrightarrow hom(A,Z),$$
is injective for every $Zintext{ob}(mathsf{C})$, and it's an effective epimorphism if:
$$hom(B,Z)to hom(A,Z)stackrel{longrightarrow}to hom(Atimes_BA,Z),$$
is exact, where the two arrows are $pr_1^*,pr_2^*$, and by exact I mean $hom(B,Z)$ is the equalizer of this diagram.
In my previous question I make a claim what epimorphisms and the fibred products are. Apparently the effective epimorphisms in $gset$ are simply the surjective maps, but this seems false. I could take $Ato {1}$ where ${1}$ is just some singleton, and $Atimes_{{1}}A=Aprod A$ in set, with 'product action' $$rho(g)(a,a')=(rho(g)a,rho(g)a').$$
Certainly I can't obtain map $v:Ato Z$ such that $vcirc pr_1 = vcirc pr_2$ from $v'circ f$, where $v'$ is just a map ${1}to Z$?
$$require{AMScd}
begin{CD}
Atimes_{{1}}A@>>> Z;\
@V{pr_1,pr_2}VV@V{id}VV \
A @>{v}>> Z ;\
@V{f}VV @V{id}VV\
{1}@>{v'}>>Z
end{CD}$$
(Sorry about the diagram, not sure how to make commuting triangles)
abstract-algebra category-theory group-actions
abstract-algebra category-theory group-actions
asked Nov 15 at 14:45
user616128
111
asked Nov 15 at 14:45
user616128
111
asked Nov 15 at 14:45
user616128
111
asked Nov 15 at 14:45
user616128
111
asked Nov 15 at 14:45
user616128
111
111
2
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57
add a comment |
2
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57
2
2
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999789%2feffective-epimorphisms-in-g-set%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
The category of $G$-sets is a topos, so every epimorphism is effective.
– Malice Vidrine
Nov 15 at 17:54
Yes, you can obtain such a map: $v=v'circ f$ certainly works. If it helps clarify, note that $v'$ corresponds to a $G$-fixed point in $Z$.
– Kevin Carlson
Nov 15 at 17:57