Prove $p_kcirc f$ continuous $implies$ f is continuous











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Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.




I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?






general-topology continuity







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  • Each open set in the product topology is the union of products of open sets. So you're close to the target.
    – Daniel Fischer♦
    Sep 7 '14 at 10:33










  • I added union before the product. Now that's correct?
    – user65985
    Sep 7 '14 at 10:36










  • You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
    – Daniel Fischer♦
    Sep 7 '14 at 11:48















up vote
1
down vote

favorite













Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.




I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?






general-topology continuity







share|cite|improve this question
























  • Each open set in the product topology is the union of products of open sets. So you're close to the target.
    – Daniel Fischer♦
    Sep 7 '14 at 10:33










  • I added union before the product. Now that's correct?
    – user65985
    Sep 7 '14 at 10:36










  • You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
    – Daniel Fischer♦
    Sep 7 '14 at 11:48













up vote
1
down vote

favorite









up vote
1
down vote

favorite












Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.




I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?






general-topology continuity







share|cite|improve this question
















Let $X_1,dots X_n$ topological space and $p_k:X_1timescdots X_nto X_k$ the projection to the kth component. Let $Y$ be topological space and
$f:Yto X_1timescdotstimes X_n$ function s.t $forall 1le kle nquad p_kcirc f$ is continuous. Prove that $f$ is continuous.




I thought that since $p_kcirc f$ is continuous $forall1le kle n$, $(p_kcirc f)^{-1}(O_k)intau_{Y}$ (where $O_kintau_k$ the topology of $(X_k,tau_k)$) and then each open set in $tau_{pi}$ is a product of open sets from these topologies, i.e $$forall Ointau_pi,O=bigcup_{iin I}prod_{k=1}^n O_{k,i}$$ (where $O_{k,i}$ is an open set in $tau_k$)$$Rightarrow f^{-1}(O)intau_Y$$ but that's seems fishy to me. Am I (even not completely) right?






general-topology continuity




general-topology continuity






share|cite|improve this question















share|cite|improve this question













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edited Sep 7 '14 at 12:13









Gerry Myerson

145k8147298




145k8147298










asked Sep 7 '14 at 10:29







user65985



















  • Each open set in the product topology is the union of products of open sets. So you're close to the target.
    – Daniel Fischer♦
    Sep 7 '14 at 10:33










  • I added union before the product. Now that's correct?
    – user65985
    Sep 7 '14 at 10:36










  • You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
    – Daniel Fischer♦
    Sep 7 '14 at 11:48


















  • Each open set in the product topology is the union of products of open sets. So you're close to the target.
    – Daniel Fischer♦
    Sep 7 '14 at 10:33










  • I added union before the product. Now that's correct?
    – user65985
    Sep 7 '14 at 10:36










  • You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
    – Daniel Fischer♦
    Sep 7 '14 at 11:48
















Each open set in the product topology is the union of products of open sets. So you're close to the target.
– Daniel Fischer♦
Sep 7 '14 at 10:33




Each open set in the product topology is the union of products of open sets. So you're close to the target.
– Daniel Fischer♦
Sep 7 '14 at 10:33












I added union before the product. Now that's correct?
– user65985
Sep 7 '14 at 10:36




I added union before the product. Now that's correct?
– user65985
Sep 7 '14 at 10:36












You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
– Daniel Fischer♦
Sep 7 '14 at 11:48




You have forgotten to mention the union in the text, but the bigger problem is that you haven't said why it follows that $f^{-1}(O)in tau_Y$.
– Daniel Fischer♦
Sep 7 '14 at 11:48










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Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.



Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$



Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.






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    1 Answer
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    up vote
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    Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.



    Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$



    Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.



      Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$



      Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
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        down vote



        accepted






        Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.



        Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$



        Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.






        share|cite|improve this answer














        Indeed $O = bigcup_{i in I}prod_{k=1}^n O_{k,i}$, where $O$ is open in $prod_{k=1}^n X_k$, $I$ is some index set, and for each $i$ we have open sets $O_{k,i} in tau_k$.



        Note (this you did not write down) that $prod_{k=1}^n O_{k,i} = cap_{k=1}^n p_k^{-1}[O_{k,i}]$, and so $$f^{-1}[O] = f^{-1}[bigcup_{i in I}prod_{k=1}^n O_{k,i}] = f^{-1} [bigcup_{i in I}bigcap_{k=1}^n (p_k)^{-1}[O_{k,i}]] = \ bigcup_{i in I} bigcap_{k=1}^n f^{-1}[p_k^{-1}[O_{k,i}]]$$ (we use $f^{-1}$ commutes with intersections and unions) which equals $$bigcup_{i in I}bigcap_{k=1}^n (p_k circ f)^{-1}[O_{k,i}]]text{.}$$



        Now the assumptions yield that indeed $f^{-1}[O]$ is open, using that finite intersections and arbitrary unions of open sets are open.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 at 9:51

























        answered Sep 7 '14 at 11:49









        Henno Brandsma

        103k345112




        103k345112






























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