Do continuous injections preserve open sets?












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$begingroup$


Do continuous injections preserve open sets?



I'm pretty sure that's true in euclidean space.



If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?



If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?



Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbb{R}^m$ to $mathbb{R}^n$" in my title deleted it. Should I start a new one?






general-topology real-numbers open-map







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  • 3




    $begingroup$
    A bijective map is open iff its inverse is continuous.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    @KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
    $endgroup$
    – user3146
    2 days ago
















1












$begingroup$


Do continuous injections preserve open sets?



I'm pretty sure that's true in euclidean space.



If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?



If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?



Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbb{R}^m$ to $mathbb{R}^n$" in my title deleted it. Should I start a new one?






general-topology real-numbers open-map







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    A bijective map is open iff its inverse is continuous.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    @KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
    $endgroup$
    – user3146
    2 days ago














1












1








1





$begingroup$


Do continuous injections preserve open sets?



I'm pretty sure that's true in euclidean space.



If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?



If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?



Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbb{R}^m$ to $mathbb{R}^n$" in my title deleted it. Should I start a new one?






general-topology real-numbers open-map







share|cite|improve this question











$endgroup$




Do continuous injections preserve open sets?



I'm pretty sure that's true in euclidean space.



If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?



If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?



Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbb{R}^m$ to $mathbb{R}^n$" in my title deleted it. Should I start a new one?






general-topology real-numbers open-map




general-topology real-numbers open-map






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







user3146

















asked 2 days ago









user3146user3146

746




746








  • 3




    $begingroup$
    A bijective map is open iff its inverse is continuous.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    @KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
    $endgroup$
    – user3146
    2 days ago














  • 3




    $begingroup$
    A bijective map is open iff its inverse is continuous.
    $endgroup$
    – Kavi Rama Murthy
    2 days ago










  • $begingroup$
    @KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
    $endgroup$
    – user3146
    2 days ago








3




3




$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago




$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago












$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago




$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago










3 Answers
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11












$begingroup$

No they do not. It's not true in euclidean space.



The function $$f(x)=begin{cases}x& xleq 0\ x+1 & x>0end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.



      I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).






      share|cite|improve this answer











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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

        oldest

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        active

        oldest

        votes









        11












        $begingroup$

        No they do not. It's not true in euclidean space.



        The function $$f(x)=begin{cases}x& xleq 0\ x+1 & x>0end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.






        share|cite|improve this answer









        $endgroup$


















          11












          $begingroup$

          No they do not. It's not true in euclidean space.



          The function $$f(x)=begin{cases}x& xleq 0\ x+1 & x>0end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.






          share|cite|improve this answer









          $endgroup$
















            11












            11








            11





            $begingroup$

            No they do not. It's not true in euclidean space.



            The function $$f(x)=begin{cases}x& xleq 0\ x+1 & x>0end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.






            share|cite|improve this answer









            $endgroup$



            No they do not. It's not true in euclidean space.



            The function $$f(x)=begin{cases}x& xleq 0\ x+1 & x>0end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            5xum5xum

            92k394162




            92k394162























                2












                $begingroup$

                Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.






                    share|cite|improve this answer









                    $endgroup$



                    Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    JosuéJosué

                    3,51242672




                    3,51242672























                        1












                        $begingroup$

                        Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.



                        I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.



                          I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.



                            I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).






                            share|cite|improve this answer











                            $endgroup$



                            Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.



                            I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited yesterday

























                            answered 2 days ago









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