Inverse limits over coinitial sets.
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Given a directed set $A$, a subset $B subset A$ is said to be coinitial if for every $a in A$ there is some $b in B$ such that $b leq a$.
Now consider an inverse system of rings $(S_i, f_ij)$ indexed by $A$ and let $B subset A$ be coinitial.
Is it true that $varprojlim_i in AS_i cong varprojlim_i in B S_i$?
I am aware that the answer is true if $B$ is cofinal instead of coinitial. I as wondering if there is some sort of modification in the proof for cofinal sets that works as well for coinitial sets. Thank you in advance.
abstract-algebra category-theory limits-colimits
asked Aug 8 at 16:25
user313212
622519
 |Â
show 1 more comment
up vote
1
down vote
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Given a directed set $A$, a subset $B subset A$ is said to be coinitial if for every $a in A$ there is some $b in B$ such that $b leq a$.
Now consider an inverse system of rings $(S_i, f_ij)$ indexed by $A$ and let $B subset A$ be coinitial.
Is it true that $varprojlim_i in AS_i cong varprojlim_i in B S_i$?
I am aware that the answer is true if $B$ is cofinal instead of coinitial. I as wondering if there is some sort of modification in the proof for cofinal sets that works as well for coinitial sets. Thank you in advance.
abstract-algebra category-theory limits-colimits
asked Aug 8 at 16:25
user313212
622519
2
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
1
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a directed set $A$, a subset $B subset A$ is said to be coinitial if for every $a in A$ there is some $b in B$ such that $b leq a$.
Now consider an inverse system of rings $(S_i, f_ij)$ indexed by $A$ and let $B subset A$ be coinitial.
Is it true that $varprojlim_i in AS_i cong varprojlim_i in B S_i$?
I am aware that the answer is true if $B$ is cofinal instead of coinitial. I as wondering if there is some sort of modification in the proof for cofinal sets that works as well for coinitial sets. Thank you in advance.
abstract-algebra category-theory limits-colimits
asked Aug 8 at 16:25
user313212
622519
Given a directed set $A$, a subset $B subset A$ is said to be coinitial if for every $a in A$ there is some $b in B$ such that $b leq a$.
Now consider an inverse system of rings $(S_i, f_ij)$ indexed by $A$ and let $B subset A$ be coinitial.
Is it true that $varprojlim_i in AS_i cong varprojlim_i in B S_i$?
I am aware that the answer is true if $B$ is cofinal instead of coinitial. I as wondering if there is some sort of modification in the proof for cofinal sets that works as well for coinitial sets. Thank you in advance.
abstract-algebra category-theory limits-colimits
asked Aug 8 at 16:25
user313212
622519
asked Aug 8 at 16:25
user313212
622519
asked Aug 8 at 16:25
user313212
622519
asked Aug 8 at 16:25
user313212
622519
622519
2
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
1
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51
 |Â
show 1 more comment
2
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
1
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51
2
2
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
1
1
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51
 |Â
show 1 more comment
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2
To get an inverse system, shouldn't you ask for a codirected set $A$?
– Arnaud D.
Aug 8 at 16:36
@ArnaudD. Pardon my ignorance, but I don't know what a codirected set is...
– user313212
Aug 8 at 18:27
If the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $igeq j$, then you want a directed index set $A$, and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a cofinal subset $B$. If, on the other hand, the homomorphisms in your inverse system go from $S_i$ to $S_j$ with $ileq j$, then you want a co-directed index set $A$ (i.e., every two elements have a common lower bound), and the inverse limit will be unchanged (up to isomorphism) if you replace $A$ with a coinitial subset $B$.
– Andreas Blass
Aug 8 at 23:07
@AndreasBlass Thank you your comment. It seems more clear now. Can you point me to any reference (book or notes) where I can find the result you quote?
– user313212
Aug 9 at 8:55
1
@user313212 I'm confused : you say in your question that you know already know the result for $B$ cofinal. I guess this is in the first case mentioned by Andreas? Note that the second is really the same result, but with the order relation inverted.
– Arnaud D.
Aug 9 at 9:51