Showing a map $f:E rightarrow Bbb R^infty$ is continuous.












1












$begingroup$



Suppose for $i in Bbb N$, we have continuous maps $g_i:E rightarrow Bbb R^n$ satisfying local finiteness proeprty i.e. for each $xin E$ exists nhood $U_x$ such that only finitely many $g_i$ are nonzero.



Define $f:E rightarrow Bbb R^infty$, $Bbb R^infty$ being direct limit of inclusions $Bbb R^i$, as the composition,



$$ E xrightarrow{f'} bigsqcup Bbb R^n xrightarrow{q} Bbb R^infty$$
where $f':x mapsto (H_i(x))_{i=1}^infty$, and $bigsqcup Bbb R^n$ is coproduct in category of topological space, and $q$ is the quotient map.




I have two questions, (i) is the finiteness property necessary, and (ii) is it true that this map is continuous?




My argument:



It suffices to show $f'$ is continuous.
Let $U subseteq bigsqcup Bbb R^n$ be open nhood of $f'(x)$ and $U_x$ a local finite nhood of $x$ in $E$.



So it suffices to work locally, Let $U_i:= U cap Bbb R^i$. Then, $$f'^{-1}(U) cap U_x= bigcap_{i=1}^infty (g_i^{-1}(U_i) cap U_x) = bigcap_{i=1}^{k} g_{n_i}^{-1}(U_{n_i}) cap U_x$$



is non empty open .




This question is from Theorem 2.10, pg 15, line 7-10 of proof.






general-topology algebraic-topology differential-topology vector-bundles







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    Suppose for $i in Bbb N$, we have continuous maps $g_i:E rightarrow Bbb R^n$ satisfying local finiteness proeprty i.e. for each $xin E$ exists nhood $U_x$ such that only finitely many $g_i$ are nonzero.



    Define $f:E rightarrow Bbb R^infty$, $Bbb R^infty$ being direct limit of inclusions $Bbb R^i$, as the composition,



    $$ E xrightarrow{f'} bigsqcup Bbb R^n xrightarrow{q} Bbb R^infty$$
    where $f':x mapsto (H_i(x))_{i=1}^infty$, and $bigsqcup Bbb R^n$ is coproduct in category of topological space, and $q$ is the quotient map.




    I have two questions, (i) is the finiteness property necessary, and (ii) is it true that this map is continuous?




    My argument:



    It suffices to show $f'$ is continuous.
    Let $U subseteq bigsqcup Bbb R^n$ be open nhood of $f'(x)$ and $U_x$ a local finite nhood of $x$ in $E$.



    So it suffices to work locally, Let $U_i:= U cap Bbb R^i$. Then, $$f'^{-1}(U) cap U_x= bigcap_{i=1}^infty (g_i^{-1}(U_i) cap U_x) = bigcap_{i=1}^{k} g_{n_i}^{-1}(U_{n_i}) cap U_x$$



    is non empty open .




    This question is from Theorem 2.10, pg 15, line 7-10 of proof.






    general-topology algebraic-topology differential-topology vector-bundles







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Suppose for $i in Bbb N$, we have continuous maps $g_i:E rightarrow Bbb R^n$ satisfying local finiteness proeprty i.e. for each $xin E$ exists nhood $U_x$ such that only finitely many $g_i$ are nonzero.



      Define $f:E rightarrow Bbb R^infty$, $Bbb R^infty$ being direct limit of inclusions $Bbb R^i$, as the composition,



      $$ E xrightarrow{f'} bigsqcup Bbb R^n xrightarrow{q} Bbb R^infty$$
      where $f':x mapsto (H_i(x))_{i=1}^infty$, and $bigsqcup Bbb R^n$ is coproduct in category of topological space, and $q$ is the quotient map.




      I have two questions, (i) is the finiteness property necessary, and (ii) is it true that this map is continuous?




      My argument:



      It suffices to show $f'$ is continuous.
      Let $U subseteq bigsqcup Bbb R^n$ be open nhood of $f'(x)$ and $U_x$ a local finite nhood of $x$ in $E$.



      So it suffices to work locally, Let $U_i:= U cap Bbb R^i$. Then, $$f'^{-1}(U) cap U_x= bigcap_{i=1}^infty (g_i^{-1}(U_i) cap U_x) = bigcap_{i=1}^{k} g_{n_i}^{-1}(U_{n_i}) cap U_x$$



      is non empty open .




      This question is from Theorem 2.10, pg 15, line 7-10 of proof.






      general-topology algebraic-topology differential-topology vector-bundles







      share|cite|improve this question









      $endgroup$





      Suppose for $i in Bbb N$, we have continuous maps $g_i:E rightarrow Bbb R^n$ satisfying local finiteness proeprty i.e. for each $xin E$ exists nhood $U_x$ such that only finitely many $g_i$ are nonzero.



      Define $f:E rightarrow Bbb R^infty$, $Bbb R^infty$ being direct limit of inclusions $Bbb R^i$, as the composition,



      $$ E xrightarrow{f'} bigsqcup Bbb R^n xrightarrow{q} Bbb R^infty$$
      where $f':x mapsto (H_i(x))_{i=1}^infty$, and $bigsqcup Bbb R^n$ is coproduct in category of topological space, and $q$ is the quotient map.




      I have two questions, (i) is the finiteness property necessary, and (ii) is it true that this map is continuous?




      My argument:



      It suffices to show $f'$ is continuous.
      Let $U subseteq bigsqcup Bbb R^n$ be open nhood of $f'(x)$ and $U_x$ a local finite nhood of $x$ in $E$.



      So it suffices to work locally, Let $U_i:= U cap Bbb R^i$. Then, $$f'^{-1}(U) cap U_x= bigcap_{i=1}^infty (g_i^{-1}(U_i) cap U_x) = bigcap_{i=1}^{k} g_{n_i}^{-1}(U_{n_i}) cap U_x$$



      is non empty open .




      This question is from Theorem 2.10, pg 15, line 7-10 of proof.






      general-topology algebraic-topology differential-topology vector-bundles




      general-topology algebraic-topology differential-topology vector-bundles






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 11 at 8:05









      CL.CL.

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          $begingroup$

          The author starts with a countable locally finite open cover ${ U_i }$ and constructs continuous maps $H_i : E to mathbb{R}^n$ such that $H_i(x) = 0$ for $x notin p^{-1}(U_i)$. Then he defines
          $$f : E to oplus_{i=1}^infty mathbb{R}^n = mathbb{R}^infty, f(x) = (H_i(x)) .$$
          $oplus_{i=1}^infty mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $oplus_{i=1}^infty mathbb{R}^n = { (y_i) in prod_{i=1}^infty mathbb{R}^n mid y_i ne 0 text{ only for finitely many } i }$. It receives the subspace topology from the infinite product $ prod_{i=1}^infty mathbb{R}^n$ (which has of course the product topology).



          Note that we may identify $oplus_{i=1}^infty mathbb{R}^n$ with $oplus_{i=1}^infty mathbb{R} = { (t_i) in prod_{i=1}^infty mathbb{R} mid t_i ne 0 text{ only for finitely many } i }$. The latter is the standard definition of $mathbb{R}^infty$.



          Regarding $f$ as a map into $prod_{i=1}^infty mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) subset mathbb{R}^infty$. This means that for each $x$ only finitely many $H_i(x) ne 0$. But this is clear because ${ U_i }$ is locally finite. Given $x in E$, there are only finitely many $i$ such that $p(x) in U_i$. For all other $i$ we have $p(x) notin U_i$, i.e. $x notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.






          share|cite|improve this answer









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            active

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            active

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            $begingroup$

            The author starts with a countable locally finite open cover ${ U_i }$ and constructs continuous maps $H_i : E to mathbb{R}^n$ such that $H_i(x) = 0$ for $x notin p^{-1}(U_i)$. Then he defines
            $$f : E to oplus_{i=1}^infty mathbb{R}^n = mathbb{R}^infty, f(x) = (H_i(x)) .$$
            $oplus_{i=1}^infty mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $oplus_{i=1}^infty mathbb{R}^n = { (y_i) in prod_{i=1}^infty mathbb{R}^n mid y_i ne 0 text{ only for finitely many } i }$. It receives the subspace topology from the infinite product $ prod_{i=1}^infty mathbb{R}^n$ (which has of course the product topology).



            Note that we may identify $oplus_{i=1}^infty mathbb{R}^n$ with $oplus_{i=1}^infty mathbb{R} = { (t_i) in prod_{i=1}^infty mathbb{R} mid t_i ne 0 text{ only for finitely many } i }$. The latter is the standard definition of $mathbb{R}^infty$.



            Regarding $f$ as a map into $prod_{i=1}^infty mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) subset mathbb{R}^infty$. This means that for each $x$ only finitely many $H_i(x) ne 0$. But this is clear because ${ U_i }$ is locally finite. Given $x in E$, there are only finitely many $i$ such that $p(x) in U_i$. For all other $i$ we have $p(x) notin U_i$, i.e. $x notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The author starts with a countable locally finite open cover ${ U_i }$ and constructs continuous maps $H_i : E to mathbb{R}^n$ such that $H_i(x) = 0$ for $x notin p^{-1}(U_i)$. Then he defines
              $$f : E to oplus_{i=1}^infty mathbb{R}^n = mathbb{R}^infty, f(x) = (H_i(x)) .$$
              $oplus_{i=1}^infty mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $oplus_{i=1}^infty mathbb{R}^n = { (y_i) in prod_{i=1}^infty mathbb{R}^n mid y_i ne 0 text{ only for finitely many } i }$. It receives the subspace topology from the infinite product $ prod_{i=1}^infty mathbb{R}^n$ (which has of course the product topology).



              Note that we may identify $oplus_{i=1}^infty mathbb{R}^n$ with $oplus_{i=1}^infty mathbb{R} = { (t_i) in prod_{i=1}^infty mathbb{R} mid t_i ne 0 text{ only for finitely many } i }$. The latter is the standard definition of $mathbb{R}^infty$.



              Regarding $f$ as a map into $prod_{i=1}^infty mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) subset mathbb{R}^infty$. This means that for each $x$ only finitely many $H_i(x) ne 0$. But this is clear because ${ U_i }$ is locally finite. Given $x in E$, there are only finitely many $i$ such that $p(x) in U_i$. For all other $i$ we have $p(x) notin U_i$, i.e. $x notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The author starts with a countable locally finite open cover ${ U_i }$ and constructs continuous maps $H_i : E to mathbb{R}^n$ such that $H_i(x) = 0$ for $x notin p^{-1}(U_i)$. Then he defines
                $$f : E to oplus_{i=1}^infty mathbb{R}^n = mathbb{R}^infty, f(x) = (H_i(x)) .$$
                $oplus_{i=1}^infty mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $oplus_{i=1}^infty mathbb{R}^n = { (y_i) in prod_{i=1}^infty mathbb{R}^n mid y_i ne 0 text{ only for finitely many } i }$. It receives the subspace topology from the infinite product $ prod_{i=1}^infty mathbb{R}^n$ (which has of course the product topology).



                Note that we may identify $oplus_{i=1}^infty mathbb{R}^n$ with $oplus_{i=1}^infty mathbb{R} = { (t_i) in prod_{i=1}^infty mathbb{R} mid t_i ne 0 text{ only for finitely many } i }$. The latter is the standard definition of $mathbb{R}^infty$.



                Regarding $f$ as a map into $prod_{i=1}^infty mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) subset mathbb{R}^infty$. This means that for each $x$ only finitely many $H_i(x) ne 0$. But this is clear because ${ U_i }$ is locally finite. Given $x in E$, there are only finitely many $i$ such that $p(x) in U_i$. For all other $i$ we have $p(x) notin U_i$, i.e. $x notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.






                share|cite|improve this answer









                $endgroup$



                The author starts with a countable locally finite open cover ${ U_i }$ and constructs continuous maps $H_i : E to mathbb{R}^n$ such that $H_i(x) = 0$ for $x notin p^{-1}(U_i)$. Then he defines
                $$f : E to oplus_{i=1}^infty mathbb{R}^n = mathbb{R}^infty, f(x) = (H_i(x)) .$$
                $oplus_{i=1}^infty mathbb{R}^n$ is not the coproduct in category of topological spaces, it is the direct sum of vector spaces. That is, $oplus_{i=1}^infty mathbb{R}^n = { (y_i) in prod_{i=1}^infty mathbb{R}^n mid y_i ne 0 text{ only for finitely many } i }$. It receives the subspace topology from the infinite product $ prod_{i=1}^infty mathbb{R}^n$ (which has of course the product topology).



                Note that we may identify $oplus_{i=1}^infty mathbb{R}^n$ with $oplus_{i=1}^infty mathbb{R} = { (t_i) in prod_{i=1}^infty mathbb{R} mid t_i ne 0 text{ only for finitely many } i }$. The latter is the standard definition of $mathbb{R}^infty$.



                Regarding $f$ as a map into $prod_{i=1}^infty mathbb{R}^n$, continuity is obvious. It remains to show that $f(E) subset mathbb{R}^infty$. This means that for each $x$ only finitely many $H_i(x) ne 0$. But this is clear because ${ U_i }$ is locally finite. Given $x in E$, there are only finitely many $i$ such that $p(x) in U_i$. For all other $i$ we have $p(x) notin U_i$, i.e. $x notin p^{-1}(U_i)$ and hence $H_i(x) = 0$.







                share|cite|improve this answer












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                answered Jan 24 at 18:26









                Paul FrostPaul Frost

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