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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$endgroup$












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










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










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
















1












$begingroup$



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











$endgroup$












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










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










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














1












1








1





$begingroup$



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











$endgroup$





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













share|cite|improve this question




share|cite|improve this question








edited Sep 7 '14 at 12:13









Gerry Myerson

146k8147299




146k8147299










asked Sep 7 '14 at 10:29







user65985



















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










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










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


















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










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










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
















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




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












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




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












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




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










1 Answer
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$begingroup$

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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    $begingroup$

    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











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      1












      $begingroup$

      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











      $endgroup$
















        1












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        1





        $begingroup$

        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











        $endgroup$



        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 '18 at 9:51

























        answered Sep 7 '14 at 11:49









        Henno BrandsmaHenno Brandsma

        108k347114




        108k347114






























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