Does $sigma$ commute with pullback.












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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







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    0















    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







    share|cite|improve this question

























      0












      0








      0








      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







      share|cite|improve this question














      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






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      asked 3 hours ago









      caffeinemachine

      6,48421250




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          1 Answer
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          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 $T$.






          share|cite|improve this answer























          • +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











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          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 $T$.






          share|cite|improve this answer























          • +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
















          2














          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 $T$.






          share|cite|improve this answer























          • +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














          2












          2








          2






          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 $T$.






          share|cite|improve this answer














          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 $T$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          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


















          • +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


















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