Does $sigma$ commute with pullback.
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
asked 3 hours ago
caffeinemachine
6,48421250
add a comment |
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
asked 3 hours ago
caffeinemachine
6,48421250
add a comment |
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
asked 3 hours ago
caffeinemachine
6,48421250
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
measure-theory elementary-set-theory
asked 3 hours ago
caffeinemachine
6,48421250
asked 3 hours ago
caffeinemachine
6,48421250
asked 3 hours ago
caffeinemachine
6,48421250
asked 3 hours ago
caffeinemachine
6,48421250
asked 3 hours ago
caffeinemachine
6,48421250
6,48421250
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
answered 3 hours ago
Kavi Rama Murthy
49k31854
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
add a comment |
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%2f3051925%2fdoes-sigma-commute-with-pullback%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
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
answered 3 hours ago
Kavi Rama Murthy
49k31854
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
add a comment |
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
answered 3 hours ago
Kavi Rama Murthy
49k31854
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
add a comment |
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
answered 3 hours ago
Kavi Rama Murthy
49k31854
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
answered 3 hours ago
Kavi Rama Murthy
49k31854
edited 3 hours ago
answered 3 hours ago
Kavi Rama Murthy
49k31854
answered 3 hours ago
Kavi Rama Murthy
49k31854
answered 3 hours ago
Kavi Rama Murthy
49k31854
49k31854
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
add a comment |
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
+1 very nice...
– mathworker21
3 hours ago
+1 very nice...
– mathworker21
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
So we don't even need $T$ to be surjective? That is very nice!
– caffeinemachine
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
– Kavi Rama Murthy
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
– caffeinemachine
3 hours ago
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3051925%2fdoes-sigma-commute-with-pullback%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
