Seeking examples of $f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$











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Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$










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    the preimage of the codomain is the domain itself by definition
    – Alvin Lepik
    Nov 24 at 8:26















up vote
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Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$










share|cite|improve this question




















  • 1




    the preimage of the codomain is the domain itself by definition
    – Alvin Lepik
    Nov 24 at 8:26













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$










share|cite|improve this question















Could you give me some examples for this statement?
$$f: (X,tau) rightarrow (Y, tau) ;text{continuous, but}; f^{-1} (Y) neq X$$







general-topology






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edited Nov 24 at 8:31









Blue

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47k870148










asked Nov 24 at 8:19









Minh

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  • 1




    the preimage of the codomain is the domain itself by definition
    – Alvin Lepik
    Nov 24 at 8:26














  • 1




    the preimage of the codomain is the domain itself by definition
    – Alvin Lepik
    Nov 24 at 8:26








1




1




the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26




the preimage of the codomain is the domain itself by definition
– Alvin Lepik
Nov 24 at 8:26










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By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.






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    By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.






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      3
      down vote













      By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.






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        up vote
        3
        down vote









        By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.






        share|cite|improve this answer














        By definition $f^{-1}[Y] = {x: f(x) in Y } = X$ as all points of $X$ map into $Y$ by $f$ being a map from $X$ to $Y$ by definition! Continuity of $f$ is irrelevant, this is just set theory and the definition of a function and of inverse images.







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        edited Nov 24 at 14:16

























        answered Nov 24 at 8:25









        Henno Brandsma

        102k345109




        102k345109






























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