Is the inverse of a continuous function an open map?











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The title is pretty self-explanatory, but I'll state the full question.



Let $f: X rightarrow Y$ be a continuous function between topological spaces. Is $f^{-1}$ an open map?



By definition of continuous function, if $V$ is open in $Y$, $f^{-1}(V)$ is open in $X$. Since the image under $f^{-1}$ of an open set is open, $f^{-1}$ is an open map.



I'm not sure about the converse. Is there something I'm missing?






general-topology continuity open-map







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    You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
    – Cheerful Parsnip
    Aug 9 at 16:58















up vote
1
down vote

favorite












The title is pretty self-explanatory, but I'll state the full question.



Let $f: X rightarrow Y$ be a continuous function between topological spaces. Is $f^{-1}$ an open map?



By definition of continuous function, if $V$ is open in $Y$, $f^{-1}(V)$ is open in $X$. Since the image under $f^{-1}$ of an open set is open, $f^{-1}$ is an open map.



I'm not sure about the converse. Is there something I'm missing?






general-topology continuity open-map







share|cite|improve this question


















  • 1




    You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
    – Cheerful Parsnip
    Aug 9 at 16:58













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The title is pretty self-explanatory, but I'll state the full question.



Let $f: X rightarrow Y$ be a continuous function between topological spaces. Is $f^{-1}$ an open map?



By definition of continuous function, if $V$ is open in $Y$, $f^{-1}(V)$ is open in $X$. Since the image under $f^{-1}$ of an open set is open, $f^{-1}$ is an open map.



I'm not sure about the converse. Is there something I'm missing?






general-topology continuity open-map







share|cite|improve this question













The title is pretty self-explanatory, but I'll state the full question.



Let $f: X rightarrow Y$ be a continuous function between topological spaces. Is $f^{-1}$ an open map?



By definition of continuous function, if $V$ is open in $Y$, $f^{-1}(V)$ is open in $X$. Since the image under $f^{-1}$ of an open set is open, $f^{-1}$ is an open map.



I'm not sure about the converse. Is there something I'm missing?






general-topology continuity open-map




general-topology continuity open-map






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share|cite|improve this question











share|cite|improve this question




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asked Aug 9 at 16:55









Niki Di Giano

819211




819211








  • 1




    You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
    – Cheerful Parsnip
    Aug 9 at 16:58














  • 1




    You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
    – Cheerful Parsnip
    Aug 9 at 16:58








1




1




You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
– Cheerful Parsnip
Aug 9 at 16:58




You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
– Cheerful Parsnip
Aug 9 at 16:58










1 Answer
1






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oldest

votes

















up vote
3
down vote



accepted










In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherewise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.






share|cite|improve this answer























  • Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
    – Niki Di Giano
    Aug 9 at 17:16












  • @NikiDiGiano Yes, that's right.
    – José Carlos Santos
    Aug 9 at 17:16










  • All clear. Thank you.
    – Niki Di Giano
    Aug 9 at 17:18











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1 Answer
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1 Answer
1






active

oldest

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oldest

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active

oldest

votes








up vote
3
down vote



accepted










In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherewise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.






share|cite|improve this answer























  • Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
    – Niki Di Giano
    Aug 9 at 17:16












  • @NikiDiGiano Yes, that's right.
    – José Carlos Santos
    Aug 9 at 17:16










  • All clear. Thank you.
    – Niki Di Giano
    Aug 9 at 17:18















up vote
3
down vote



accepted










In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherewise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.






share|cite|improve this answer























  • Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
    – Niki Di Giano
    Aug 9 at 17:16












  • @NikiDiGiano Yes, that's right.
    – José Carlos Santos
    Aug 9 at 17:16










  • All clear. Thank you.
    – Niki Di Giano
    Aug 9 at 17:18













up vote
3
down vote



accepted







up vote
3
down vote



accepted






In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherewise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.






share|cite|improve this answer














In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherewise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered Aug 9 at 16:59









José Carlos Santos

139k18111203




139k18111203












  • Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
    – Niki Di Giano
    Aug 9 at 17:16












  • @NikiDiGiano Yes, that's right.
    – José Carlos Santos
    Aug 9 at 17:16










  • All clear. Thank you.
    – Niki Di Giano
    Aug 9 at 17:18


















  • Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
    – Niki Di Giano
    Aug 9 at 17:16












  • @NikiDiGiano Yes, that's right.
    – José Carlos Santos
    Aug 9 at 17:16










  • All clear. Thank you.
    – Niki Di Giano
    Aug 9 at 17:18
















Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
– Niki Di Giano
Aug 9 at 17:16






Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right?
– Niki Di Giano
Aug 9 at 17:16














@NikiDiGiano Yes, that's right.
– José Carlos Santos
Aug 9 at 17:16




@NikiDiGiano Yes, that's right.
– José Carlos Santos
Aug 9 at 17:16












All clear. Thank you.
– Niki Di Giano
Aug 9 at 17:18




All clear. Thank you.
– Niki Di Giano
Aug 9 at 17:18


















 

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