Homeomorphism: can $f:Xmapsto X'$ be continuous, bijective without $f^{-1}:X'mapsto X$ being continuous?...
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Finding counterexamples: bijective continuous functions that are not homeomorphisms
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In our definition of homeomorphic topological spaces there must exist a bijective function $f:Xmapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.
Is it necessary to assure that $f^{-1}$ is continuous on $X'$?
general-topology continuity
asked Jan 20 at 10:00
John CataldoJohn Cataldo
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marked as duplicate by José Carlos Santos general-topology
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Jan 20 at 10:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Finding counterexamples: bijective continuous functions that are not homeomorphisms
3 answers
In our definition of homeomorphic topological spaces there must exist a bijective function $f:Xmapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.
Is it necessary to assure that $f^{-1}$ is continuous on $X'$?
general-topology continuity
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
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marked as duplicate by José Carlos Santos general-topology
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Jan 20 at 10:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
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– David Mitra
Jan 20 at 10:04
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$begingroup$
This question already has an answer here:
Finding counterexamples: bijective continuous functions that are not homeomorphisms
3 answers
In our definition of homeomorphic topological spaces there must exist a bijective function $f:Xmapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.
Is it necessary to assure that $f^{-1}$ is continuous on $X'$?
general-topology continuity
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
$endgroup$
This question already has an answer here:
Finding counterexamples: bijective continuous functions that are not homeomorphisms
3 answers
In our definition of homeomorphic topological spaces there must exist a bijective function $f:Xmapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.
Is it necessary to assure that $f^{-1}$ is continuous on $X'$?
This question already has an answer here:
Finding counterexamples: bijective continuous functions that are not homeomorphisms
3 answers
general-topology continuity
general-topology continuity
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
asked Jan 20 at 10:00
John CataldoJohn Cataldo
1,1831316
1,1831316
marked as duplicate by José Carlos Santos general-topology
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Jan 20 at 10:09
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Jan 20 at 10:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
$endgroup$
– David Mitra
Jan 20 at 10:04
add a comment |
1
$begingroup$
Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
$endgroup$
– David Mitra
Jan 20 at 10:04
1
1
$begingroup$
Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
$endgroup$
– David Mitra
Jan 20 at 10:04
$begingroup$
Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
$endgroup$
– David Mitra
Jan 20 at 10:04
add a comment |
2 Answers
2
active
oldest
votes
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The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 pi)rightarrow S_1$ (the unit circle in $mathbb{R}^2$ defined by
$f(phi )=(cos phi ,sin phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2pi )$ is not).
The function
$f^{-1}$ is not continuous at the point
$(1,0)$, because although
$f^{-1}$ maps
$(1,0)$ to
$0$, any neighbourhood of this point also includes points that the function maps close to
$2pi $, but the points it maps to numbers in between lie outside the neighbourhood.
answered Jan 20 at 10:08
user289143user289143
903313
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You have to assume this, it is not automatic:
If $X={1,2}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 pi)rightarrow S_1$ (the unit circle in $mathbb{R}^2$ defined by
$f(phi )=(cos phi ,sin phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2pi )$ is not).
The function
$f^{-1}$ is not continuous at the point
$(1,0)$, because although
$f^{-1}$ maps
$(1,0)$ to
$0$, any neighbourhood of this point also includes points that the function maps close to
$2pi $, but the points it maps to numbers in between lie outside the neighbourhood.
answered Jan 20 at 10:08
user289143user289143
903313
$endgroup$
add a comment |
$begingroup$
The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 pi)rightarrow S_1$ (the unit circle in $mathbb{R}^2$ defined by
$f(phi )=(cos phi ,sin phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2pi )$ is not).
The function
$f^{-1}$ is not continuous at the point
$(1,0)$, because although
$f^{-1}$ maps
$(1,0)$ to
$0$, any neighbourhood of this point also includes points that the function maps close to
$2pi $, but the points it maps to numbers in between lie outside the neighbourhood.
answered Jan 20 at 10:08
user289143user289143
903313
$endgroup$
add a comment |
$begingroup$
The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 pi)rightarrow S_1$ (the unit circle in $mathbb{R}^2$ defined by
$f(phi )=(cos phi ,sin phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2pi )$ is not).
The function
$f^{-1}$ is not continuous at the point
$(1,0)$, because although
$f^{-1}$ maps
$(1,0)$ to
$0$, any neighbourhood of this point also includes points that the function maps close to
$2pi $, but the points it maps to numbers in between lie outside the neighbourhood.
answered Jan 20 at 10:08
user289143user289143
903313
$endgroup$
The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 pi)rightarrow S_1$ (the unit circle in $mathbb{R}^2$ defined by
$f(phi )=(cos phi ,sin phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2pi )$ is not).
The function
$f^{-1}$ is not continuous at the point
$(1,0)$, because although
$f^{-1}$ maps
$(1,0)$ to
$0$, any neighbourhood of this point also includes points that the function maps close to
$2pi $, but the points it maps to numbers in between lie outside the neighbourhood.
answered Jan 20 at 10:08
user289143user289143
903313
edited Jan 20 at 10:15
answered Jan 20 at 10:08
user289143user289143
903313
answered Jan 20 at 10:08
user289143user289143
903313
answered Jan 20 at 10:08
user289143user289143
903313
903313
add a comment |
add a comment |
$begingroup$
You have to assume this, it is not automatic:
If $X={1,2}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
add a comment |
$begingroup$
You have to assume this, it is not automatic:
If $X={1,2}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
add a comment |
$begingroup$
You have to assume this, it is not automatic:
If $X={1,2}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
You have to assume this, it is not automatic:
If $X={1,2}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
answered Jan 20 at 10:06
Henno BrandsmaHenno Brandsma
110k347117
110k347117
add a comment |
add a comment |
1
$begingroup$
Sure. Take the identity on $Bbb R$ with the discrete topology onto $Bbb R$ with the usual topology.
$endgroup$
– David Mitra
Jan 20 at 10:04