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What happens with the inverse limit if we relax the definition of the inverse system?
The Next CEO of Stack OverflowInverse limit of an inverse system of topological spacesInverse limit of small categoriesProving $varprojlim S_i cong varprojlim S_j $ where $J subseteq I$ cofinal.About inverse and inductive limits in category theoryGluing sheaves as direct limit. What is the ordering of this directed system.Direct limit of a directed systemQuestion about the definition of the inverse limit in category theory.An easy example of an inverse system which is not the constant inverse systemUnderstanding the inverse limit and universal property of topological spacesHow does a map between inverse systems induce the inverse limit of its components?
$begingroup$
An inverse system is a tuple $(X_i,varphi_ij,I)$ where
$(I,preceq)$ is a directed poset,
$X_i_i in I$ is a collection of topological spaces,
$varphi_ij : X_i to X_j$ is a continuous map whenever $i succeq j$
such that $varphi_jk varphi_ij = varphi_ik$ whenever $i succeq j succeq k$.
We know that each inverse system has an inverse limit $(X,varphi_i)$ where
$X$ is a topological space,
$varphi_i: X to X_i$ are continuous,
satisfying the universal property. Furthermore, the inverse limit is unique up to homeomorphism.
Question: What would happen with the inverse limit if we omit the condition $varphi_jk varphi_ij = varphi_ik$ for $i succeq j succeq k$?
- Would the inverse limit still exist?
- Would the inverse limit still be unique (up to homeomorphism)?
I was curious why we needed this condition in the first place and could not see when we used it explicitly. Could you please help me with this question?
Thank you!
general-topology category-theory order-theory limits-colimits
$endgroup$
add a comment |
$begingroup$
An inverse system is a tuple $(X_i,varphi_ij,I)$ where
$(I,preceq)$ is a directed poset,
$X_i_i in I$ is a collection of topological spaces,
$varphi_ij : X_i to X_j$ is a continuous map whenever $i succeq j$
such that $varphi_jk varphi_ij = varphi_ik$ whenever $i succeq j succeq k$.
We know that each inverse system has an inverse limit $(X,varphi_i)$ where
$X$ is a topological space,
$varphi_i: X to X_i$ are continuous,
satisfying the universal property. Furthermore, the inverse limit is unique up to homeomorphism.
Question: What would happen with the inverse limit if we omit the condition $varphi_jk varphi_ij = varphi_ik$ for $i succeq j succeq k$?
- Would the inverse limit still exist?
- Would the inverse limit still be unique (up to homeomorphism)?
I was curious why we needed this condition in the first place and could not see when we used it explicitly. Could you please help me with this question?
Thank you!
general-topology category-theory order-theory limits-colimits
$endgroup$
add a comment |
$begingroup$
An inverse system is a tuple $(X_i,varphi_ij,I)$ where
$(I,preceq)$ is a directed poset,
$X_i_i in I$ is a collection of topological spaces,
$varphi_ij : X_i to X_j$ is a continuous map whenever $i succeq j$
such that $varphi_jk varphi_ij = varphi_ik$ whenever $i succeq j succeq k$.
We know that each inverse system has an inverse limit $(X,varphi_i)$ where
$X$ is a topological space,
$varphi_i: X to X_i$ are continuous,
satisfying the universal property. Furthermore, the inverse limit is unique up to homeomorphism.
Question: What would happen with the inverse limit if we omit the condition $varphi_jk varphi_ij = varphi_ik$ for $i succeq j succeq k$?
- Would the inverse limit still exist?
- Would the inverse limit still be unique (up to homeomorphism)?
I was curious why we needed this condition in the first place and could not see when we used it explicitly. Could you please help me with this question?
Thank you!
general-topology category-theory order-theory limits-colimits
$endgroup$
An inverse system is a tuple $(X_i,varphi_ij,I)$ where
$(I,preceq)$ is a directed poset,
$X_i_i in I$ is a collection of topological spaces,
$varphi_ij : X_i to X_j$ is a continuous map whenever $i succeq j$
such that $varphi_jk varphi_ij = varphi_ik$ whenever $i succeq j succeq k$.
We know that each inverse system has an inverse limit $(X,varphi_i)$ where
$X$ is a topological space,
$varphi_i: X to X_i$ are continuous,
satisfying the universal property. Furthermore, the inverse limit is unique up to homeomorphism.
Question: What would happen with the inverse limit if we omit the condition $varphi_jk varphi_ij = varphi_ik$ for $i succeq j succeq k$?
- Would the inverse limit still exist?
- Would the inverse limit still be unique (up to homeomorphism)?
I was curious why we needed this condition in the first place and could not see when we used it explicitly. Could you please help me with this question?
Thank you!
general-topology category-theory order-theory limits-colimits
general-topology category-theory order-theory limits-colimits
asked yesterday
DiglettDiglett
1,0281521
1,0281521
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $varphi_jk varphi_ij = varphi_ik$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $varphi_i:Xto X_i$ satisfying $varphi_ijvarphi_i=varphi_j$ whenever $isucceq j$, we have $$varphi_jk varphi_ijvarphi_i=varphi_jkvarphi_j=varphi_k=varphi_ikvarphi_i$$ whenever $isucceq jsucceq k$. In other words, the image of the map $varphi_i$ must be contained in the subset $Y_isubseteq X_i$ consisting of elements $x$ such that $varphi_jk varphi_ij(x) = varphi_ik(x)$ whenever $isucceq jsucceq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $varphi_ij(Y_i)subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $varphi_jk varphi_ij = varphi_ik$ are true.
To put it another way, the assumption that $varphi_jk varphi_ij = varphi_ik$ is essentially inherent in the condition $varphi_ijvarphi_i=varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $varphi_jk varphi_ij = varphi_ik$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $varphi_ii$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $varphi_jk varphi_ij = varphi_ik$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
$endgroup$
add a comment |
$begingroup$
EDIT: Doesn't answer the question that was asked
An inverse limit in this sense is a classical special case of a much more general concept, called a limit in category theory. You can certainly construct (inverse) limits, unique up to homeomorphism, of topological spaces indexed by any poset whatsoever. More generally, you could replace the poset with any small category. The construction is similar no matter what: the space of tuples of points from all the $X_i$ which respect all the given maps $X_ito X_j$.
There's this basically no point to focusing on inverse limits in your specialized sense. However, direct limits, which are the dual construction, are much easier for directed posets than for general posets or categories.
$endgroup$
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $varphi_jk varphi_ij = varphi_ik$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $varphi_i:Xto X_i$ satisfying $varphi_ijvarphi_i=varphi_j$ whenever $isucceq j$, we have $$varphi_jk varphi_ijvarphi_i=varphi_jkvarphi_j=varphi_k=varphi_ikvarphi_i$$ whenever $isucceq jsucceq k$. In other words, the image of the map $varphi_i$ must be contained in the subset $Y_isubseteq X_i$ consisting of elements $x$ such that $varphi_jk varphi_ij(x) = varphi_ik(x)$ whenever $isucceq jsucceq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $varphi_ij(Y_i)subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $varphi_jk varphi_ij = varphi_ik$ are true.
To put it another way, the assumption that $varphi_jk varphi_ij = varphi_ik$ is essentially inherent in the condition $varphi_ijvarphi_i=varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $varphi_jk varphi_ij = varphi_ik$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $varphi_ii$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $varphi_jk varphi_ij = varphi_ik$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
$endgroup$
add a comment |
$begingroup$
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $varphi_jk varphi_ij = varphi_ik$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $varphi_i:Xto X_i$ satisfying $varphi_ijvarphi_i=varphi_j$ whenever $isucceq j$, we have $$varphi_jk varphi_ijvarphi_i=varphi_jkvarphi_j=varphi_k=varphi_ikvarphi_i$$ whenever $isucceq jsucceq k$. In other words, the image of the map $varphi_i$ must be contained in the subset $Y_isubseteq X_i$ consisting of elements $x$ such that $varphi_jk varphi_ij(x) = varphi_ik(x)$ whenever $isucceq jsucceq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $varphi_ij(Y_i)subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $varphi_jk varphi_ij = varphi_ik$ are true.
To put it another way, the assumption that $varphi_jk varphi_ij = varphi_ik$ is essentially inherent in the condition $varphi_ijvarphi_i=varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $varphi_jk varphi_ij = varphi_ik$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $varphi_ii$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $varphi_jk varphi_ij = varphi_ik$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
$endgroup$
add a comment |
$begingroup$
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $varphi_jk varphi_ij = varphi_ik$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $varphi_i:Xto X_i$ satisfying $varphi_ijvarphi_i=varphi_j$ whenever $isucceq j$, we have $$varphi_jk varphi_ijvarphi_i=varphi_jkvarphi_j=varphi_k=varphi_ikvarphi_i$$ whenever $isucceq jsucceq k$. In other words, the image of the map $varphi_i$ must be contained in the subset $Y_isubseteq X_i$ consisting of elements $x$ such that $varphi_jk varphi_ij(x) = varphi_ik(x)$ whenever $isucceq jsucceq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $varphi_ij(Y_i)subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $varphi_jk varphi_ij = varphi_ik$ are true.
To put it another way, the assumption that $varphi_jk varphi_ij = varphi_ik$ is essentially inherent in the condition $varphi_ijvarphi_i=varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $varphi_jk varphi_ij = varphi_ik$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $varphi_ii$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $varphi_jk varphi_ij = varphi_ik$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
$endgroup$
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $varphi_jk varphi_ij = varphi_ik$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $varphi_i:Xto X_i$ satisfying $varphi_ijvarphi_i=varphi_j$ whenever $isucceq j$, we have $$varphi_jk varphi_ijvarphi_i=varphi_jkvarphi_j=varphi_k=varphi_ikvarphi_i$$ whenever $isucceq jsucceq k$. In other words, the image of the map $varphi_i$ must be contained in the subset $Y_isubseteq X_i$ consisting of elements $x$ such that $varphi_jk varphi_ij(x) = varphi_ik(x)$ whenever $isucceq jsucceq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $varphi_ij(Y_i)subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $varphi_jk varphi_ij = varphi_ik$ are true.
To put it another way, the assumption that $varphi_jk varphi_ij = varphi_ik$ is essentially inherent in the condition $varphi_ijvarphi_i=varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $varphi_jk varphi_ij = varphi_ik$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $varphi_ii$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $varphi_jk varphi_ij = varphi_ik$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
edited yesterday
answered yesterday
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
EDIT: Doesn't answer the question that was asked
An inverse limit in this sense is a classical special case of a much more general concept, called a limit in category theory. You can certainly construct (inverse) limits, unique up to homeomorphism, of topological spaces indexed by any poset whatsoever. More generally, you could replace the poset with any small category. The construction is similar no matter what: the space of tuples of points from all the $X_i$ which respect all the given maps $X_ito X_j$.
There's this basically no point to focusing on inverse limits in your specialized sense. However, direct limits, which are the dual construction, are much easier for directed posets than for general posets or categories.
$endgroup$
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
add a comment |
$begingroup$
EDIT: Doesn't answer the question that was asked
An inverse limit in this sense is a classical special case of a much more general concept, called a limit in category theory. You can certainly construct (inverse) limits, unique up to homeomorphism, of topological spaces indexed by any poset whatsoever. More generally, you could replace the poset with any small category. The construction is similar no matter what: the space of tuples of points from all the $X_i$ which respect all the given maps $X_ito X_j$.
There's this basically no point to focusing on inverse limits in your specialized sense. However, direct limits, which are the dual construction, are much easier for directed posets than for general posets or categories.
$endgroup$
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
add a comment |
$begingroup$
EDIT: Doesn't answer the question that was asked
An inverse limit in this sense is a classical special case of a much more general concept, called a limit in category theory. You can certainly construct (inverse) limits, unique up to homeomorphism, of topological spaces indexed by any poset whatsoever. More generally, you could replace the poset with any small category. The construction is similar no matter what: the space of tuples of points from all the $X_i$ which respect all the given maps $X_ito X_j$.
There's this basically no point to focusing on inverse limits in your specialized sense. However, direct limits, which are the dual construction, are much easier for directed posets than for general posets or categories.
$endgroup$
EDIT: Doesn't answer the question that was asked
An inverse limit in this sense is a classical special case of a much more general concept, called a limit in category theory. You can certainly construct (inverse) limits, unique up to homeomorphism, of topological spaces indexed by any poset whatsoever. More generally, you could replace the poset with any small category. The construction is similar no matter what: the space of tuples of points from all the $X_i$ which respect all the given maps $X_ito X_j$.
There's this basically no point to focusing on inverse limits in your specialized sense. However, direct limits, which are the dual construction, are much easier for directed posets than for general posets or categories.
edited yesterday
answered yesterday
Kevin CarlsonKevin Carlson
33.8k23372
33.8k23372
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
add a comment |
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
The question is not about generalizing the index poset, though; it's about removing the assumption that the diagram respects composition.
$endgroup$
– Eric Wofsey
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
$begingroup$
@EricWofsey Woops, thanks, I guess I'll leave it up anyway.
$endgroup$
– Kevin Carlson
yesterday
add a comment |
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