On a certain topology on the set of continuous functions between topological spaces

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Let $A,X,Y$ be topological spaces. Given a function $g: A times X to Y$, we can make a corresponding function $bar g: A to Y^X$ as $bar g(a) (x)=g(a,x), forall a in A, x in X$, and vice-versa. The "exponential topology" on $C(X,Y)$ (the set of all continuous functions from $X$ to $Y$) is a topology on it such that for every topological space $A$, $g in C(Atimes X,Y) iff bar g in C(A,C(X,Y))$ .



I can show that if an exponential topology on $C(X,Y)$ exists then it is unique.



My question is: Does there exists an exponential topology on $C(X,Y)$ when $X=Y=[0,1]$ ?







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    Let $A,X,Y$ be topological spaces. Given a function $g: A times X to Y$, we can make a corresponding function $bar g: A to Y^X$ as $bar g(a) (x)=g(a,x), forall a in A, x in X$, and vice-versa. The "exponential topology" on $C(X,Y)$ (the set of all continuous functions from $X$ to $Y$) is a topology on it such that for every topological space $A$, $g in C(Atimes X,Y) iff bar g in C(A,C(X,Y))$ .



    I can show that if an exponential topology on $C(X,Y)$ exists then it is unique.



    My question is: Does there exists an exponential topology on $C(X,Y)$ when $X=Y=[0,1]$ ?







    share|cite|improve this question





















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      Let $A,X,Y$ be topological spaces. Given a function $g: A times X to Y$, we can make a corresponding function $bar g: A to Y^X$ as $bar g(a) (x)=g(a,x), forall a in A, x in X$, and vice-versa. The "exponential topology" on $C(X,Y)$ (the set of all continuous functions from $X$ to $Y$) is a topology on it such that for every topological space $A$, $g in C(Atimes X,Y) iff bar g in C(A,C(X,Y))$ .



      I can show that if an exponential topology on $C(X,Y)$ exists then it is unique.



      My question is: Does there exists an exponential topology on $C(X,Y)$ when $X=Y=[0,1]$ ?







      share|cite|improve this question











      Let $A,X,Y$ be topological spaces. Given a function $g: A times X to Y$, we can make a corresponding function $bar g: A to Y^X$ as $bar g(a) (x)=g(a,x), forall a in A, x in X$, and vice-versa. The "exponential topology" on $C(X,Y)$ (the set of all continuous functions from $X$ to $Y$) is a topology on it such that for every topological space $A$, $g in C(Atimes X,Y) iff bar g in C(A,C(X,Y))$ .



      I can show that if an exponential topology on $C(X,Y)$ exists then it is unique.



      My question is: Does there exists an exponential topology on $C(X,Y)$ when $X=Y=[0,1]$ ?









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      asked Jul 28 at 21:50









      user521337

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          You should consult any book on general topology which covers function spaces, for example



          Engelking, Ryszard. "General topology." (1989)



          The correspondence $g mapsto overlineg$ is defined for all (not necessarily continuous) functions $g : A times X to Y$. It is called the exponential map $Lambda : mathcalF(A times X, Y) to mathcalF(A, mathcalF(X,Y))$. Here, $mathcalF(M,N)$ denotes the set of all functions $M to N$. If we restrict to continuous functions, we shall consider topologies $mathfrakT_X,Y$ on $C(X,Y)$. Let us define



          (1) $mathfrakT_X,Y$ is proper if $Lambda(C(A times X,Y) subset C(A, (C(X,Y),mathfrakT_X,Y))$ for all $A$.



          (2) $mathfrakT_X,Y$ is admissible if $Lambda^-1(C(A, (C(X,Y),mathfrakT_X,Y))) subset C(A times X,Y)$ for all $A$.



          (3) $mathfrakT_X,Y$ is acceptable if it is both proper and admissible (this is what you call exponential topology).



          It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is



          (a) proper for all spaces $X,Y$,



          (b) admissible for all locally compact spaces $X$ and all spaces $Y$.



          This answers your question in the affirmative.






          share|cite|improve this answer





















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            up vote
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            You should consult any book on general topology which covers function spaces, for example



            Engelking, Ryszard. "General topology." (1989)



            The correspondence $g mapsto overlineg$ is defined for all (not necessarily continuous) functions $g : A times X to Y$. It is called the exponential map $Lambda : mathcalF(A times X, Y) to mathcalF(A, mathcalF(X,Y))$. Here, $mathcalF(M,N)$ denotes the set of all functions $M to N$. If we restrict to continuous functions, we shall consider topologies $mathfrakT_X,Y$ on $C(X,Y)$. Let us define



            (1) $mathfrakT_X,Y$ is proper if $Lambda(C(A times X,Y) subset C(A, (C(X,Y),mathfrakT_X,Y))$ for all $A$.



            (2) $mathfrakT_X,Y$ is admissible if $Lambda^-1(C(A, (C(X,Y),mathfrakT_X,Y))) subset C(A times X,Y)$ for all $A$.



            (3) $mathfrakT_X,Y$ is acceptable if it is both proper and admissible (this is what you call exponential topology).



            It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is



            (a) proper for all spaces $X,Y$,



            (b) admissible for all locally compact spaces $X$ and all spaces $Y$.



            This answers your question in the affirmative.






            share|cite|improve this answer

























              up vote
              1
              down vote













              You should consult any book on general topology which covers function spaces, for example



              Engelking, Ryszard. "General topology." (1989)



              The correspondence $g mapsto overlineg$ is defined for all (not necessarily continuous) functions $g : A times X to Y$. It is called the exponential map $Lambda : mathcalF(A times X, Y) to mathcalF(A, mathcalF(X,Y))$. Here, $mathcalF(M,N)$ denotes the set of all functions $M to N$. If we restrict to continuous functions, we shall consider topologies $mathfrakT_X,Y$ on $C(X,Y)$. Let us define



              (1) $mathfrakT_X,Y$ is proper if $Lambda(C(A times X,Y) subset C(A, (C(X,Y),mathfrakT_X,Y))$ for all $A$.



              (2) $mathfrakT_X,Y$ is admissible if $Lambda^-1(C(A, (C(X,Y),mathfrakT_X,Y))) subset C(A times X,Y)$ for all $A$.



              (3) $mathfrakT_X,Y$ is acceptable if it is both proper and admissible (this is what you call exponential topology).



              It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is



              (a) proper for all spaces $X,Y$,



              (b) admissible for all locally compact spaces $X$ and all spaces $Y$.



              This answers your question in the affirmative.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                You should consult any book on general topology which covers function spaces, for example



                Engelking, Ryszard. "General topology." (1989)



                The correspondence $g mapsto overlineg$ is defined for all (not necessarily continuous) functions $g : A times X to Y$. It is called the exponential map $Lambda : mathcalF(A times X, Y) to mathcalF(A, mathcalF(X,Y))$. Here, $mathcalF(M,N)$ denotes the set of all functions $M to N$. If we restrict to continuous functions, we shall consider topologies $mathfrakT_X,Y$ on $C(X,Y)$. Let us define



                (1) $mathfrakT_X,Y$ is proper if $Lambda(C(A times X,Y) subset C(A, (C(X,Y),mathfrakT_X,Y))$ for all $A$.



                (2) $mathfrakT_X,Y$ is admissible if $Lambda^-1(C(A, (C(X,Y),mathfrakT_X,Y))) subset C(A times X,Y)$ for all $A$.



                (3) $mathfrakT_X,Y$ is acceptable if it is both proper and admissible (this is what you call exponential topology).



                It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is



                (a) proper for all spaces $X,Y$,



                (b) admissible for all locally compact spaces $X$ and all spaces $Y$.



                This answers your question in the affirmative.






                share|cite|improve this answer













                You should consult any book on general topology which covers function spaces, for example



                Engelking, Ryszard. "General topology." (1989)



                The correspondence $g mapsto overlineg$ is defined for all (not necessarily continuous) functions $g : A times X to Y$. It is called the exponential map $Lambda : mathcalF(A times X, Y) to mathcalF(A, mathcalF(X,Y))$. Here, $mathcalF(M,N)$ denotes the set of all functions $M to N$. If we restrict to continuous functions, we shall consider topologies $mathfrakT_X,Y$ on $C(X,Y)$. Let us define



                (1) $mathfrakT_X,Y$ is proper if $Lambda(C(A times X,Y) subset C(A, (C(X,Y),mathfrakT_X,Y))$ for all $A$.



                (2) $mathfrakT_X,Y$ is admissible if $Lambda^-1(C(A, (C(X,Y),mathfrakT_X,Y))) subset C(A times X,Y)$ for all $A$.



                (3) $mathfrakT_X,Y$ is acceptable if it is both proper and admissible (this is what you call exponential topology).



                It is a well-known fact that there exist at most one acceptable topology on $C(X,Y)$. Usually function spaces are endowed with the compact-open topology and it is well-known that the compact-open topology is



                (a) proper for all spaces $X,Y$,



                (b) admissible for all locally compact spaces $X$ and all spaces $Y$.



                This answers your question in the affirmative.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 29 at 0:07









                Paul Frost

                3,593420




                3,593420






















                     

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