Continuous bijection equivalent to homeomorphism under suitable hypotheses











up vote
0
down vote

favorite












My question is pretty simple to state, although I did not find any satisfying answer on the internet or books. I assume that the answer has to be well-known, but I can not figure any proof or counterexample.



Suppose that $B subset mathbb{R}^n$ is a connected topological space (with the relative topology) and we have a map $f:B to mathbb{R}$ which is assumed to be a continuous bijection. The question is if under these simple hypotheses it is enough to conclude that $f$ an homeomorphism.



Some remarks:



1) I am already aware about the invariance of domain theorem, that would give a YES answer for the case $B=mathbb{R}$.



2) I know that the statement would be false if $mathbb{R}$ was in the domain and $B$ in the codomain (classical counterexample where $B=mathbb{S}^1$).



My intuitive guess is that it is true.



One idea is to induce an order in $B$ from the order in $mathbb{R}$. Therefore, if there is some result that ensures me that $B$ is homeomorphic to a subset of $mathbb{R}$ it is everything done, because connectedness forces this homeomorphic set to $B$ to be an interval and it is immediate to conclude that has to be an open one (and from there is obvious). Any ideas?










share|cite|improve this question




















  • 1




    What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
    – user120527
    Nov 22 at 10:59










  • It is OK for the general case, but not enough if we assume B to be connected.
    – DCao
    Nov 22 at 11:01










  • I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
    – DCao
    Nov 22 at 11:03















up vote
0
down vote

favorite












My question is pretty simple to state, although I did not find any satisfying answer on the internet or books. I assume that the answer has to be well-known, but I can not figure any proof or counterexample.



Suppose that $B subset mathbb{R}^n$ is a connected topological space (with the relative topology) and we have a map $f:B to mathbb{R}$ which is assumed to be a continuous bijection. The question is if under these simple hypotheses it is enough to conclude that $f$ an homeomorphism.



Some remarks:



1) I am already aware about the invariance of domain theorem, that would give a YES answer for the case $B=mathbb{R}$.



2) I know that the statement would be false if $mathbb{R}$ was in the domain and $B$ in the codomain (classical counterexample where $B=mathbb{S}^1$).



My intuitive guess is that it is true.



One idea is to induce an order in $B$ from the order in $mathbb{R}$. Therefore, if there is some result that ensures me that $B$ is homeomorphic to a subset of $mathbb{R}$ it is everything done, because connectedness forces this homeomorphic set to $B$ to be an interval and it is immediate to conclude that has to be an open one (and from there is obvious). Any ideas?










share|cite|improve this question




















  • 1




    What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
    – user120527
    Nov 22 at 10:59










  • It is OK for the general case, but not enough if we assume B to be connected.
    – DCao
    Nov 22 at 11:01










  • I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
    – DCao
    Nov 22 at 11:03













up vote
0
down vote

favorite









up vote
0
down vote

favorite











My question is pretty simple to state, although I did not find any satisfying answer on the internet or books. I assume that the answer has to be well-known, but I can not figure any proof or counterexample.



Suppose that $B subset mathbb{R}^n$ is a connected topological space (with the relative topology) and we have a map $f:B to mathbb{R}$ which is assumed to be a continuous bijection. The question is if under these simple hypotheses it is enough to conclude that $f$ an homeomorphism.



Some remarks:



1) I am already aware about the invariance of domain theorem, that would give a YES answer for the case $B=mathbb{R}$.



2) I know that the statement would be false if $mathbb{R}$ was in the domain and $B$ in the codomain (classical counterexample where $B=mathbb{S}^1$).



My intuitive guess is that it is true.



One idea is to induce an order in $B$ from the order in $mathbb{R}$. Therefore, if there is some result that ensures me that $B$ is homeomorphic to a subset of $mathbb{R}$ it is everything done, because connectedness forces this homeomorphic set to $B$ to be an interval and it is immediate to conclude that has to be an open one (and from there is obvious). Any ideas?










share|cite|improve this question















My question is pretty simple to state, although I did not find any satisfying answer on the internet or books. I assume that the answer has to be well-known, but I can not figure any proof or counterexample.



Suppose that $B subset mathbb{R}^n$ is a connected topological space (with the relative topology) and we have a map $f:B to mathbb{R}$ which is assumed to be a continuous bijection. The question is if under these simple hypotheses it is enough to conclude that $f$ an homeomorphism.



Some remarks:



1) I am already aware about the invariance of domain theorem, that would give a YES answer for the case $B=mathbb{R}$.



2) I know that the statement would be false if $mathbb{R}$ was in the domain and $B$ in the codomain (classical counterexample where $B=mathbb{S}^1$).



My intuitive guess is that it is true.



One idea is to induce an order in $B$ from the order in $mathbb{R}$. Therefore, if there is some result that ensures me that $B$ is homeomorphic to a subset of $mathbb{R}$ it is everything done, because connectedness forces this homeomorphic set to $B$ to be an interval and it is immediate to conclude that has to be an open one (and from there is obvious). Any ideas?







general-topology continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 13:51

























asked Nov 22 at 10:53









DCao

251110




251110








  • 1




    What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
    – user120527
    Nov 22 at 10:59










  • It is OK for the general case, but not enough if we assume B to be connected.
    – DCao
    Nov 22 at 11:01










  • I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
    – DCao
    Nov 22 at 11:03














  • 1




    What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
    – user120527
    Nov 22 at 10:59










  • It is OK for the general case, but not enough if we assume B to be connected.
    – DCao
    Nov 22 at 11:01










  • I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
    – DCao
    Nov 22 at 11:03








1




1




What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
– user120527
Nov 22 at 10:59




What happens for instance if we take $B=mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ?
– user120527
Nov 22 at 10:59












It is OK for the general case, but not enough if we assume B to be connected.
– DCao
Nov 22 at 11:01




It is OK for the general case, but not enough if we assume B to be connected.
– DCao
Nov 22 at 11:01












I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
– DCao
Nov 22 at 11:03




I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/…
– DCao
Nov 22 at 11:03










1 Answer
1






active

oldest

votes

















up vote
0
down vote













You have the function $f:B to mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.



1) It is possible to induce a total order in $B$. Given $x,y in B$ just define $x>y$ iff $f(x)>f(y)$.



2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] subset B$ is closed since it is the continuous preimage of a closed set.



3) Since $B$ is path-connected there is a continuous map $h: [0,1] to B$, with $f(0)=x$ and $f(1)=y$. We have that $f circ h:[0,1] to mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.



4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-infty$ and $y$ for a sequence $y_n$ that goes to $+infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.



Remark: Point $4)$ holds for $mathbb{R}$ and not for $mathbb{S}^1$ (because you can not put $mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).






share|cite|improve this answer





















    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',
    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%2f3008998%2fcontinuous-bijection-equivalent-to-homeomorphism-under-suitable-hypotheses%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








    up vote
    0
    down vote













    You have the function $f:B to mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.



    1) It is possible to induce a total order in $B$. Given $x,y in B$ just define $x>y$ iff $f(x)>f(y)$.



    2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] subset B$ is closed since it is the continuous preimage of a closed set.



    3) Since $B$ is path-connected there is a continuous map $h: [0,1] to B$, with $f(0)=x$ and $f(1)=y$. We have that $f circ h:[0,1] to mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.



    4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-infty$ and $y$ for a sequence $y_n$ that goes to $+infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.



    Remark: Point $4)$ holds for $mathbb{R}$ and not for $mathbb{S}^1$ (because you can not put $mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).






    share|cite|improve this answer

























      up vote
      0
      down vote













      You have the function $f:B to mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.



      1) It is possible to induce a total order in $B$. Given $x,y in B$ just define $x>y$ iff $f(x)>f(y)$.



      2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] subset B$ is closed since it is the continuous preimage of a closed set.



      3) Since $B$ is path-connected there is a continuous map $h: [0,1] to B$, with $f(0)=x$ and $f(1)=y$. We have that $f circ h:[0,1] to mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.



      4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-infty$ and $y$ for a sequence $y_n$ that goes to $+infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.



      Remark: Point $4)$ holds for $mathbb{R}$ and not for $mathbb{S}^1$ (because you can not put $mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        You have the function $f:B to mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.



        1) It is possible to induce a total order in $B$. Given $x,y in B$ just define $x>y$ iff $f(x)>f(y)$.



        2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] subset B$ is closed since it is the continuous preimage of a closed set.



        3) Since $B$ is path-connected there is a continuous map $h: [0,1] to B$, with $f(0)=x$ and $f(1)=y$. We have that $f circ h:[0,1] to mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.



        4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-infty$ and $y$ for a sequence $y_n$ that goes to $+infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.



        Remark: Point $4)$ holds for $mathbb{R}$ and not for $mathbb{S}^1$ (because you can not put $mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).






        share|cite|improve this answer












        You have the function $f:B to mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.



        1) It is possible to induce a total order in $B$. Given $x,y in B$ just define $x>y$ iff $f(x)>f(y)$.



        2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] subset B$ is closed since it is the continuous preimage of a closed set.



        3) Since $B$ is path-connected there is a continuous map $h: [0,1] to B$, with $f(0)=x$ and $f(1)=y$. We have that $f circ h:[0,1] to mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.



        4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-infty$ and $y$ for a sequence $y_n$ that goes to $+infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.



        Remark: Point $4)$ holds for $mathbb{R}$ and not for $mathbb{S}^1$ (because you can not put $mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 20:56









        DCao

        251110




        251110






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3008998%2fcontinuous-bijection-equivalent-to-homeomorphism-under-suitable-hypotheses%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

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei