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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
asked Aug 9 at 16:55
Niki Di Giano
819211
add a comment |
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
asked Aug 9 at 16:55
Niki Di Giano
819211
1
You have to assume $f$ is bijective in order for $f^{-1}$ to exist.
– Cheerful Parsnip
Aug 9 at 16:58
add a comment |
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
asked Aug 9 at 16:55
Niki Di Giano
819211
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
asked Aug 9 at 16:55
Niki Di Giano
819211
asked Aug 9 at 16:55
Niki Di Giano
819211
asked Aug 9 at 16:55
Niki Di Giano
819211
asked Aug 9 at 16:55
Niki Di Giano
819211
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
add a comment |
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
add a comment |
1 Answer
1
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.
answered Aug 9 at 16:59
José Carlos Santos
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
answered Aug 9 at 16:59
José Carlos Santos
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
add a comment |
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.
answered Aug 9 at 16:59
José Carlos Santos
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
add a comment |
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.
answered Aug 9 at 16:59
José Carlos Santos
139k18111203
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.
answered Aug 9 at 16:59
José Carlos Santos
139k18111203
edited yesterday
answered Aug 9 at 16:59
José Carlos Santos
139k18111203
answered Aug 9 at 16:59
José Carlos Santos
139k18111203
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
add a comment |
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
add a comment |
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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