Countable Completion is Isomorphic to Full Completion (Lang Algebra)
$begingroup$
This is from Lang's $Algebra$, revised third edition, page 52. I will state my understanding of the problem.
Suppose $G$ is a group, and $F$ is a family of normal subgroups of $G$, partially ordered by the subset relation. Then $F$ is directed, so we define the inverse limit: $lim_{H∈F}G/H$ by the canonical homomorphisms.
Suppose we have an $A⊆F$ and $A = {H_i}_{i∈mathbb{Z_+}}$, where for all i, we have $H_{i+1}⊆H_i$. Then we also similarly define the inverse limit $lim_{i∈mathbb{Z_+}}G/H_i$.
The problem is to show that these two limits are isomorphic if A is cofinal, that is, if for all $H∈F$ there exists $H_i∈A$ such that $H_i⊆H$. The problem seems to be counterintuitive and I do not know how to start. I would be thankful for hints as well as answers.
abstract-algebra group-theory category-theory
asked Jan 16 at 1:25
rr01rr01
1158
$endgroup$
add a comment |
$begingroup$
This is from Lang's $Algebra$, revised third edition, page 52. I will state my understanding of the problem.
Suppose $G$ is a group, and $F$ is a family of normal subgroups of $G$, partially ordered by the subset relation. Then $F$ is directed, so we define the inverse limit: $lim_{H∈F}G/H$ by the canonical homomorphisms.
Suppose we have an $A⊆F$ and $A = {H_i}_{i∈mathbb{Z_+}}$, where for all i, we have $H_{i+1}⊆H_i$. Then we also similarly define the inverse limit $lim_{i∈mathbb{Z_+}}G/H_i$.
The problem is to show that these two limits are isomorphic if A is cofinal, that is, if for all $H∈F$ there exists $H_i∈A$ such that $H_i⊆H$. The problem seems to be counterintuitive and I do not know how to start. I would be thankful for hints as well as answers.
abstract-algebra group-theory category-theory
asked Jan 16 at 1:25
rr01rr01
1158
$endgroup$
1
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15
add a comment |
$begingroup$
This is from Lang's $Algebra$, revised third edition, page 52. I will state my understanding of the problem.
Suppose $G$ is a group, and $F$ is a family of normal subgroups of $G$, partially ordered by the subset relation. Then $F$ is directed, so we define the inverse limit: $lim_{H∈F}G/H$ by the canonical homomorphisms.
Suppose we have an $A⊆F$ and $A = {H_i}_{i∈mathbb{Z_+}}$, where for all i, we have $H_{i+1}⊆H_i$. Then we also similarly define the inverse limit $lim_{i∈mathbb{Z_+}}G/H_i$.
The problem is to show that these two limits are isomorphic if A is cofinal, that is, if for all $H∈F$ there exists $H_i∈A$ such that $H_i⊆H$. The problem seems to be counterintuitive and I do not know how to start. I would be thankful for hints as well as answers.
abstract-algebra group-theory category-theory
asked Jan 16 at 1:25
rr01rr01
1158
$endgroup$
This is from Lang's $Algebra$, revised third edition, page 52. I will state my understanding of the problem.
Suppose $G$ is a group, and $F$ is a family of normal subgroups of $G$, partially ordered by the subset relation. Then $F$ is directed, so we define the inverse limit: $lim_{H∈F}G/H$ by the canonical homomorphisms.
Suppose we have an $A⊆F$ and $A = {H_i}_{i∈mathbb{Z_+}}$, where for all i, we have $H_{i+1}⊆H_i$. Then we also similarly define the inverse limit $lim_{i∈mathbb{Z_+}}G/H_i$.
The problem is to show that these two limits are isomorphic if A is cofinal, that is, if for all $H∈F$ there exists $H_i∈A$ such that $H_i⊆H$. The problem seems to be counterintuitive and I do not know how to start. I would be thankful for hints as well as answers.
abstract-algebra group-theory category-theory
abstract-algebra group-theory category-theory
asked Jan 16 at 1:25
rr01rr01
1158
asked Jan 16 at 1:25
rr01rr01
1158
edited Jan 17 at 1:41
rr01
asked Jan 16 at 1:25
rr01rr01
1158
asked Jan 16 at 1:25
rr01rr01
1158
asked Jan 16 at 1:25
rr01rr01
1158
1158
1
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15
add a comment |
1
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15
1
1
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
After user10354138's comment it is an easy task to arrive at the desired isomorphism.
As for the intuition, it follows directly from the proof; the $H$th entry of $x ∈ lim_{H∈F}G/H$ contains information about all larger $K$ encompassing $H$ and therefore the "collapsed" form of $x$ in the countable limit contains information about the original $x$ that is entirely recoverable.
answered Jan 17 at 8:04
rr01rr01
1158
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3075202%2fcountable-completion-is-isomorphic-to-full-completion-lang-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After user10354138's comment it is an easy task to arrive at the desired isomorphism.
As for the intuition, it follows directly from the proof; the $H$th entry of $x ∈ lim_{H∈F}G/H$ contains information about all larger $K$ encompassing $H$ and therefore the "collapsed" form of $x$ in the countable limit contains information about the original $x$ that is entirely recoverable.
answered Jan 17 at 8:04
rr01rr01
1158
$endgroup$
add a comment |
$begingroup$
After user10354138's comment it is an easy task to arrive at the desired isomorphism.
As for the intuition, it follows directly from the proof; the $H$th entry of $x ∈ lim_{H∈F}G/H$ contains information about all larger $K$ encompassing $H$ and therefore the "collapsed" form of $x$ in the countable limit contains information about the original $x$ that is entirely recoverable.
answered Jan 17 at 8:04
rr01rr01
1158
$endgroup$
add a comment |
$begingroup$
After user10354138's comment it is an easy task to arrive at the desired isomorphism.
As for the intuition, it follows directly from the proof; the $H$th entry of $x ∈ lim_{H∈F}G/H$ contains information about all larger $K$ encompassing $H$ and therefore the "collapsed" form of $x$ in the countable limit contains information about the original $x$ that is entirely recoverable.
answered Jan 17 at 8:04
rr01rr01
1158
$endgroup$
After user10354138's comment it is an easy task to arrive at the desired isomorphism.
As for the intuition, it follows directly from the proof; the $H$th entry of $x ∈ lim_{H∈F}G/H$ contains information about all larger $K$ encompassing $H$ and therefore the "collapsed" form of $x$ in the countable limit contains information about the original $x$ that is entirely recoverable.
answered Jan 17 at 8:04
rr01rr01
1158
answered Jan 17 at 8:04
rr01rr01
1158
answered Jan 17 at 8:04
rr01rr01
1158
answered Jan 17 at 8:04
rr01rr01
1158
1158
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3075202%2fcountable-completion-is-isomorphic-to-full-completion-lang-algebra%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
1
$begingroup$
Show that the identity on $projlim_H G/H$ factors through $projlim_i G/H_i$, i.e. for every $H'in F$ the canonical $projlim_H G/Hto G/H'$ factors through $G/H_i$ for large enough $i$ (and similarly for the other way round).
$endgroup$
– user10354138
Jan 16 at 5:15