Does $sigma$ commute with pullback.












0












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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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$endgroup$

















    0












    $begingroup$



    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









    $endgroup$















      0












      0








      0





      $begingroup$



      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









      $endgroup$





      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






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 25 '18 at 7:48









      caffeinemachinecaffeinemachine

      6,63221355




      6,63221355






















          1 Answer
          1






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          $begingroup$

          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











          $endgroup$













          • $begingroup$
            +1 very nice...
            $endgroup$
            – mathworker21
            Dec 25 '18 at 8:06










          • $begingroup$
            So we don't even need $T$ to be surjective? That is very nice!
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:13










          • $begingroup$
            @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
            $endgroup$
            – Kavi Rama Murthy
            Dec 25 '18 at 8:18










          • $begingroup$
            @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.
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:29











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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

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          2












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            +1 very nice...
            $endgroup$
            – mathworker21
            Dec 25 '18 at 8:06










          • $begingroup$
            So we don't even need $T$ to be surjective? That is very nice!
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:13










          • $begingroup$
            @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
            $endgroup$
            – Kavi Rama Murthy
            Dec 25 '18 at 8:18










          • $begingroup$
            @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.
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:29
















          2












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            +1 very nice...
            $endgroup$
            – mathworker21
            Dec 25 '18 at 8:06










          • $begingroup$
            So we don't even need $T$ to be surjective? That is very nice!
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:13










          • $begingroup$
            @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
            $endgroup$
            – Kavi Rama Murthy
            Dec 25 '18 at 8:18










          • $begingroup$
            @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.
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:29














          2












          2








          2





          $begingroup$

          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











          $endgroup$



          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 Dec 25 '18 at 8:07

























          answered Dec 25 '18 at 7:55









          Kavi Rama MurthyKavi Rama Murthy

          66.9k53067




          66.9k53067












          • $begingroup$
            +1 very nice...
            $endgroup$
            – mathworker21
            Dec 25 '18 at 8:06










          • $begingroup$
            So we don't even need $T$ to be surjective? That is very nice!
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:13










          • $begingroup$
            @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
            $endgroup$
            – Kavi Rama Murthy
            Dec 25 '18 at 8:18










          • $begingroup$
            @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.
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:29


















          • $begingroup$
            +1 very nice...
            $endgroup$
            – mathworker21
            Dec 25 '18 at 8:06










          • $begingroup$
            So we don't even need $T$ to be surjective? That is very nice!
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:13










          • $begingroup$
            @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
            $endgroup$
            – Kavi Rama Murthy
            Dec 25 '18 at 8:18










          • $begingroup$
            @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.
            $endgroup$
            – caffeinemachine
            Dec 25 '18 at 8:29
















          $begingroup$
          +1 very nice...
          $endgroup$
          – mathworker21
          Dec 25 '18 at 8:06




          $begingroup$
          +1 very nice...
          $endgroup$
          – mathworker21
          Dec 25 '18 at 8:06












          $begingroup$
          So we don't even need $T$ to be surjective? That is very nice!
          $endgroup$
          – caffeinemachine
          Dec 25 '18 at 8:13




          $begingroup$
          So we don't even need $T$ to be surjective? That is very nice!
          $endgroup$
          – caffeinemachine
          Dec 25 '18 at 8:13












          $begingroup$
          @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
          $endgroup$
          – Kavi Rama Murthy
          Dec 25 '18 at 8:18




          $begingroup$
          @caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
          $endgroup$
          – Kavi Rama Murthy
          Dec 25 '18 at 8:18












          $begingroup$
          @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.
          $endgroup$
          – caffeinemachine
          Dec 25 '18 at 8:29




          $begingroup$
          @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.
          $endgroup$
          – caffeinemachine
          Dec 25 '18 at 8:29


















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