Trying to understand a proof from math overflow regarding direct limit
$begingroup$
Direct limit of $mathbb {C^*}$ behaves well with quotients.
The following solution is from mathoverflow:
Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $phi_n: A_n to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $tilde{phi}_n: A_n/I_n to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $tilde{phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.
I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?
functional-analysis operator-algebras limits-colimits
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
$endgroup$
add a comment |
$begingroup$
Direct limit of $mathbb {C^*}$ behaves well with quotients.
The following solution is from mathoverflow:
Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $phi_n: A_n to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $tilde{phi}_n: A_n/I_n to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $tilde{phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.
I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?
functional-analysis operator-algebras limits-colimits
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
$endgroup$
add a comment |
$begingroup$
Direct limit of $mathbb {C^*}$ behaves well with quotients.
The following solution is from mathoverflow:
Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $phi_n: A_n to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $tilde{phi}_n: A_n/I_n to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $tilde{phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.
I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?
functional-analysis operator-algebras limits-colimits
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
$endgroup$
Direct limit of $mathbb {C^*}$ behaves well with quotients.
The following solution is from mathoverflow:
Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $phi_n: A_n to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $tilde{phi}_n: A_n/I_n to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $tilde{phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.
I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?
functional-analysis operator-algebras limits-colimits
functional-analysis operator-algebras limits-colimits
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
edited Dec 6 '18 at 13:08
Arpit Kansal
6,87911135
6,87911135
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
asked Dec 6 '18 at 10:45
Math LoverMath Lover
958315
958315
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
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%2f3028332%2ftrying-to-understand-a-proof-from-math-overflow-regarding-direct-limit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3028332%2ftrying-to-understand-a-proof-from-math-overflow-regarding-direct-limit%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
