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












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












$begingroup$



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

146k8147298




146k8147298










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
1






active

oldest

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f922348%2fprove-p-k-circ-f-continuous-implies-f-is-continuous%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown
























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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












        1








        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

        106k347114




        106k347114






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f922348%2fprove-p-k-circ-f-continuous-implies-f-is-continuous%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more