Inverse limit of epics in abelian category
$begingroup$
Does the AB4 axiom or AB5 axiom of abelian categories by grothendieck give the fact that cofiltered limits of epics are epic? Specifically, suppose $A_n$ maps surjectively to $A_{n-1}$ for each $n in mathbb{Z}$ does the limit map surjectively onto each $A_n$?
Motivation: trying to prove acyclic assembly lemma over abelian categories.
homological-algebra abelian-categories
asked Dec 1 '18 at 14:54
davikdavik
558317
$endgroup$
add a comment |
$begingroup$
Does the AB4 axiom or AB5 axiom of abelian categories by grothendieck give the fact that cofiltered limits of epics are epic? Specifically, suppose $A_n$ maps surjectively to $A_{n-1}$ for each $n in mathbb{Z}$ does the limit map surjectively onto each $A_n$?
Motivation: trying to prove acyclic assembly lemma over abelian categories.
homological-algebra abelian-categories
asked Dec 1 '18 at 14:54
davikdavik
558317
$endgroup$
add a comment |
$begingroup$
Does the AB4 axiom or AB5 axiom of abelian categories by grothendieck give the fact that cofiltered limits of epics are epic? Specifically, suppose $A_n$ maps surjectively to $A_{n-1}$ for each $n in mathbb{Z}$ does the limit map surjectively onto each $A_n$?
Motivation: trying to prove acyclic assembly lemma over abelian categories.
homological-algebra abelian-categories
asked Dec 1 '18 at 14:54
davikdavik
558317
$endgroup$
Does the AB4 axiom or AB5 axiom of abelian categories by grothendieck give the fact that cofiltered limits of epics are epic? Specifically, suppose $A_n$ maps surjectively to $A_{n-1}$ for each $n in mathbb{Z}$ does the limit map surjectively onto each $A_n$?
Motivation: trying to prove acyclic assembly lemma over abelian categories.
homological-algebra abelian-categories
homological-algebra abelian-categories
asked Dec 1 '18 at 14:54
davikdavik
558317
asked Dec 1 '18 at 14:54
davikdavik
558317
asked Dec 1 '18 at 14:54
davikdavik
558317
asked Dec 1 '18 at 14:54
davikdavik
558317
asked Dec 1 '18 at 14:54
davikdavik
558317
558317
add a comment |
add a comment |
1 Answer
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Let $mathsf{Ab}$ be the category of abelian groups, and $mathsf{TAb}$ the full subcategory of torsion abelian groups.
Then $mathsf{TAb}$ is an AB5 abelian subcategory of $mathsf{Ab}$, and the inclusion reflects colimits (i.e., the colimit in $mathsf{Ab}$ of any diagram of torsion abelian groups is torsion, and is also the colimit in $mathsf{TAb}$.
$mathsf{TAb}$ also has limits, but these are not the same as in $mathsf{Ab}$, since a limit of torsion groups may not be a torsion group. To construct a limit in $mathsf{TAb}$ you take the limit in $mathsf{Ab}$ and take the torsion subgroup of that.
Now consider the diagram
$$dotstomathbb{Z}/p^3mathbb{Z}tomathbb{Z}/p^2mathbb{Z}tomathbb{Z}/pmathbb{Z}$$
for some prime $p$, with all the maps surjective.
The limit in $mathsf{Ab}$ is the group $mathbb{Z}_p$ of $p$-adic integers, which is torsion free, and so its torsion subgroup (the limit in $mathsf{TAb}$) is zero, so the natural maps to the groups $mathbb{Z}/p^nmathbb{Z}$ are certainly not epimorphisms.
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
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1 Answer
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$begingroup$
Let $mathsf{Ab}$ be the category of abelian groups, and $mathsf{TAb}$ the full subcategory of torsion abelian groups.
Then $mathsf{TAb}$ is an AB5 abelian subcategory of $mathsf{Ab}$, and the inclusion reflects colimits (i.e., the colimit in $mathsf{Ab}$ of any diagram of torsion abelian groups is torsion, and is also the colimit in $mathsf{TAb}$.
$mathsf{TAb}$ also has limits, but these are not the same as in $mathsf{Ab}$, since a limit of torsion groups may not be a torsion group. To construct a limit in $mathsf{TAb}$ you take the limit in $mathsf{Ab}$ and take the torsion subgroup of that.
Now consider the diagram
$$dotstomathbb{Z}/p^3mathbb{Z}tomathbb{Z}/p^2mathbb{Z}tomathbb{Z}/pmathbb{Z}$$
for some prime $p$, with all the maps surjective.
The limit in $mathsf{Ab}$ is the group $mathbb{Z}_p$ of $p$-adic integers, which is torsion free, and so its torsion subgroup (the limit in $mathsf{TAb}$) is zero, so the natural maps to the groups $mathbb{Z}/p^nmathbb{Z}$ are certainly not epimorphisms.
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
add a comment |
$begingroup$
Let $mathsf{Ab}$ be the category of abelian groups, and $mathsf{TAb}$ the full subcategory of torsion abelian groups.
Then $mathsf{TAb}$ is an AB5 abelian subcategory of $mathsf{Ab}$, and the inclusion reflects colimits (i.e., the colimit in $mathsf{Ab}$ of any diagram of torsion abelian groups is torsion, and is also the colimit in $mathsf{TAb}$.
$mathsf{TAb}$ also has limits, but these are not the same as in $mathsf{Ab}$, since a limit of torsion groups may not be a torsion group. To construct a limit in $mathsf{TAb}$ you take the limit in $mathsf{Ab}$ and take the torsion subgroup of that.
Now consider the diagram
$$dotstomathbb{Z}/p^3mathbb{Z}tomathbb{Z}/p^2mathbb{Z}tomathbb{Z}/pmathbb{Z}$$
for some prime $p$, with all the maps surjective.
The limit in $mathsf{Ab}$ is the group $mathbb{Z}_p$ of $p$-adic integers, which is torsion free, and so its torsion subgroup (the limit in $mathsf{TAb}$) is zero, so the natural maps to the groups $mathbb{Z}/p^nmathbb{Z}$ are certainly not epimorphisms.
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
add a comment |
$begingroup$
Let $mathsf{Ab}$ be the category of abelian groups, and $mathsf{TAb}$ the full subcategory of torsion abelian groups.
Then $mathsf{TAb}$ is an AB5 abelian subcategory of $mathsf{Ab}$, and the inclusion reflects colimits (i.e., the colimit in $mathsf{Ab}$ of any diagram of torsion abelian groups is torsion, and is also the colimit in $mathsf{TAb}$.
$mathsf{TAb}$ also has limits, but these are not the same as in $mathsf{Ab}$, since a limit of torsion groups may not be a torsion group. To construct a limit in $mathsf{TAb}$ you take the limit in $mathsf{Ab}$ and take the torsion subgroup of that.
Now consider the diagram
$$dotstomathbb{Z}/p^3mathbb{Z}tomathbb{Z}/p^2mathbb{Z}tomathbb{Z}/pmathbb{Z}$$
for some prime $p$, with all the maps surjective.
The limit in $mathsf{Ab}$ is the group $mathbb{Z}_p$ of $p$-adic integers, which is torsion free, and so its torsion subgroup (the limit in $mathsf{TAb}$) is zero, so the natural maps to the groups $mathbb{Z}/p^nmathbb{Z}$ are certainly not epimorphisms.
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
Let $mathsf{Ab}$ be the category of abelian groups, and $mathsf{TAb}$ the full subcategory of torsion abelian groups.
Then $mathsf{TAb}$ is an AB5 abelian subcategory of $mathsf{Ab}$, and the inclusion reflects colimits (i.e., the colimit in $mathsf{Ab}$ of any diagram of torsion abelian groups is torsion, and is also the colimit in $mathsf{TAb}$.
$mathsf{TAb}$ also has limits, but these are not the same as in $mathsf{Ab}$, since a limit of torsion groups may not be a torsion group. To construct a limit in $mathsf{TAb}$ you take the limit in $mathsf{Ab}$ and take the torsion subgroup of that.
Now consider the diagram
$$dotstomathbb{Z}/p^3mathbb{Z}tomathbb{Z}/p^2mathbb{Z}tomathbb{Z}/pmathbb{Z}$$
for some prime $p$, with all the maps surjective.
The limit in $mathsf{Ab}$ is the group $mathbb{Z}_p$ of $p$-adic integers, which is torsion free, and so its torsion subgroup (the limit in $mathsf{TAb}$) is zero, so the natural maps to the groups $mathbb{Z}/p^nmathbb{Z}$ are certainly not epimorphisms.
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 17:30
Jeremy RickardJeremy Rickard
16.1k11643
16.1k11643
add a comment |
add a comment |
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