Proving $f$ is continuous iff all the functions parameterized are continuous












2












$begingroup$



Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.




My proof:



$rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
Suppose $f$ is continuous then $f^{-1}(U)intau$



$f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.



As $f_i(U)$ is a composition of two continuous functions hence it is continuous.



$leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$



Supposing $f_i$ is continuous forall the $i=1,2...n$ then



$f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$



But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.



Question:



Is my proof right? If not. Why not?



Thanks in advance!






general-topology proof-verification







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$endgroup$

















    2












    $begingroup$



    Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.




    My proof:



    $rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
    Suppose $f$ is continuous then $f^{-1}(U)intau$



    $f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.



    As $f_i(U)$ is a composition of two continuous functions hence it is continuous.



    $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$



    Supposing $f_i$ is continuous forall the $i=1,2...n$ then



    $f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$



    But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.



    Question:



    Is my proof right? If not. Why not?



    Thanks in advance!






    general-topology proof-verification







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.




      My proof:



      $rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
      Suppose $f$ is continuous then $f^{-1}(U)intau$



      $f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.



      As $f_i(U)$ is a composition of two continuous functions hence it is continuous.



      $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$



      Supposing $f_i$ is continuous forall the $i=1,2...n$ then



      $f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$



      But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.



      Question:



      Is my proof right? If not. Why not?



      Thanks in advance!






      general-topology proof-verification







      share|cite|improve this question









      $endgroup$





      Let $(Y,tau)$ and $(X_i,tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,tau)$ into $(X_i,tau_i)$. Porve that the mapping $f:(Y,tau)toprod_limits{i=1}^{n}(X_i,tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.




      My proof:



      $rightarrow$ Let $U$ be an open set of $(prod_limits{i=1}^{n}X_i,tau_i)$
      Suppose $f$ is continuous then $f^{-1}(U)intau$



      $f_i(U)=p_icirc f(U)$ where $p_i$ is the projection $p_i:(prod_limits{i=1}^{n}X_i,tau_i)to (X_itau_i)$.



      As $f_i(U)$ is a composition of two continuous functions hence it is continuous.



      $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n$ be an open set in the subspace topology $f(Y),tau_{nY}$



      Supposing $f_i$ is continuous forall the $i=1,2...n$ then



      $f^{-1}_icirc p_i(U)=f^{-1}(U_i)intau_i$



      But $f^{-1}(U)=f^{-1}_icirc p_i(U)$ then $f$ is continuous.



      Question:



      Is my proof right? If not. Why not?



      Thanks in advance!






      general-topology proof-verification




      general-topology proof-verification






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      asked Dec 24 '18 at 15:10









      Pedro GomesPedro Gomes

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          I would recommend the following change



          $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.






          share|cite|improve this answer









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            1 Answer
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            1 Answer
            1






            active

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            active

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            active

            oldest

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            1












            $begingroup$

            I would recommend the following change



            $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I would recommend the following change



              $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I would recommend the following change



                $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.






                share|cite|improve this answer









                $endgroup$



                I would recommend the following change



                $leftarrow$ Let $U=U_1times U_2times...times U_itimes...times U_n subset prod_limits{i=1}^{n}(X_i,tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 16:08









                CopyPasteItCopyPasteIt

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