f nonsurjective map between two topological spaces; continuity












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Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?



In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?










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$endgroup$












  • $begingroup$
    $f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
    $endgroup$
    – Paul Frost
    Jan 14 at 14:59












  • $begingroup$
    @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
    $endgroup$
    – metalder9
    Jan 14 at 17:56








  • 1




    $begingroup$
    Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
    $endgroup$
    – Paul Frost
    Jan 14 at 18:00
















0












$begingroup$


Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?



In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?










share|cite|improve this question











$endgroup$












  • $begingroup$
    $f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
    $endgroup$
    – Paul Frost
    Jan 14 at 14:59












  • $begingroup$
    @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
    $endgroup$
    – metalder9
    Jan 14 at 17:56








  • 1




    $begingroup$
    Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
    $endgroup$
    – Paul Frost
    Jan 14 at 18:00














0












0








0





$begingroup$


Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?



In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?










share|cite|improve this question











$endgroup$




Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?



In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?







general-topology continuity






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share|cite|improve this question













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edited Jan 14 at 8:31









José Carlos Santos

169k23132237




169k23132237










asked Jan 14 at 8:27









metalder9metalder9

647




647












  • $begingroup$
    $f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
    $endgroup$
    – Paul Frost
    Jan 14 at 14:59












  • $begingroup$
    @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
    $endgroup$
    – metalder9
    Jan 14 at 17:56








  • 1




    $begingroup$
    Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
    $endgroup$
    – Paul Frost
    Jan 14 at 18:00


















  • $begingroup$
    $f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
    $endgroup$
    – Paul Frost
    Jan 14 at 14:59












  • $begingroup$
    @PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
    $endgroup$
    – metalder9
    Jan 14 at 17:56








  • 1




    $begingroup$
    Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
    $endgroup$
    – Paul Frost
    Jan 14 at 18:00
















$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59






$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59














$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56






$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56






1




1




$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00




$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00










1 Answer
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$begingroup$

If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    $begingroup$

    If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.






    share|cite|improve this answer









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      3












      $begingroup$

      If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.






      share|cite|improve this answer









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        3












        3








        3





        $begingroup$

        If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.






        share|cite|improve this answer









        $endgroup$



        If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 8:29









        Kavi Rama MurthyKavi Rama Murthy

        69.1k53169




        69.1k53169






























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