Mixing
Quotient topology, finest topology
$begingroup$
let $X$ be a topological space and $sim$ equivalence relation on $X$.
$pi:Xto X/sim, xmapsto [x]$.
A subset $Usubseteq X/sim$ is open in the quotient topology, iff $pi^{-1}(U)$ is open in $X$.
This is the finest topology $tau$ on $X/sim$, where $pi$ is continuous.
Therefore the topology with the most open sets.
Is my proof correct?
Suppose $tau$ is not the finest topology on $X/sim$. Then there is a set $Unotintau$ with $pi^{-1}(U)$ is open in $X$. But then $U$ has to be in the quotient topology $tau$, which gives us the contradiction.
Thanks in advance.
general-topology
edited Apr 24 '17 at 23:03
Joshua Lin
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
$endgroup$
add a comment |
$begingroup$
let $X$ be a topological space and $sim$ equivalence relation on $X$.
$pi:Xto X/sim, xmapsto [x]$.
A subset $Usubseteq X/sim$ is open in the quotient topology, iff $pi^{-1}(U)$ is open in $X$.
This is the finest topology $tau$ on $X/sim$, where $pi$ is continuous.
Therefore the topology with the most open sets.
Is my proof correct?
Suppose $tau$ is not the finest topology on $X/sim$. Then there is a set $Unotintau$ with $pi^{-1}(U)$ is open in $X$. But then $U$ has to be in the quotient topology $tau$, which gives us the contradiction.
Thanks in advance.
general-topology
edited Apr 24 '17 at 23:03
Joshua Lin
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
$endgroup$
$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34
add a comment |
$begingroup$
let $X$ be a topological space and $sim$ equivalence relation on $X$.
$pi:Xto X/sim, xmapsto [x]$.
A subset $Usubseteq X/sim$ is open in the quotient topology, iff $pi^{-1}(U)$ is open in $X$.
This is the finest topology $tau$ on $X/sim$, where $pi$ is continuous.
Therefore the topology with the most open sets.
Is my proof correct?
Suppose $tau$ is not the finest topology on $X/sim$. Then there is a set $Unotintau$ with $pi^{-1}(U)$ is open in $X$. But then $U$ has to be in the quotient topology $tau$, which gives us the contradiction.
Thanks in advance.
general-topology
edited Apr 24 '17 at 23:03
Joshua Lin
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
$endgroup$
let $X$ be a topological space and $sim$ equivalence relation on $X$.
$pi:Xto X/sim, xmapsto [x]$.
A subset $Usubseteq X/sim$ is open in the quotient topology, iff $pi^{-1}(U)$ is open in $X$.
This is the finest topology $tau$ on $X/sim$, where $pi$ is continuous.
Therefore the topology with the most open sets.
Is my proof correct?
Suppose $tau$ is not the finest topology on $X/sim$. Then there is a set $Unotintau$ with $pi^{-1}(U)$ is open in $X$. But then $U$ has to be in the quotient topology $tau$, which gives us the contradiction.
Thanks in advance.
general-topology
general-topology
edited Apr 24 '17 at 23:03
Joshua Lin
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
edited Apr 24 '17 at 23:03
Joshua Lin
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
edited Apr 24 '17 at 23:03
Joshua Lin
802821
edited Apr 24 '17 at 23:03
Joshua Lin
802821
edited Apr 24 '17 at 23:03
Joshua Lin
802821
802821
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
asked Apr 24 '17 at 19:33
CornmanCornman
3,22121229
3,22121229
$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34
add a comment |
$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34
$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The easiest way to reason, I think is:
Suppose that $mathscr{T}$ is any topology on $X/ {sim} $ that makes $pi$ continuous. Let $U in mathscr{T}$, then by continuity of $pi$ gives us that $pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U in mathscr{T}_{text{quot}}$. So $mathscr{T} subseteq mathscr{T}_{text{quot}}$.
This shows that any topology that makes $pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $pi$ continuous. This shows that $mathscr{T}_{text{quot}}$ is the maximal (i.e. finest) topology that makes $pi$ continuous.
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $pi^{-1}(U)$ is open within the domain space of $pi$.
edited Jan 14 at 18:19
Henno Brandsma
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The easiest way to reason, I think is:
Suppose that $mathscr{T}$ is any topology on $X/ {sim} $ that makes $pi$ continuous. Let $U in mathscr{T}$, then by continuity of $pi$ gives us that $pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U in mathscr{T}_{text{quot}}$. So $mathscr{T} subseteq mathscr{T}_{text{quot}}$.
This shows that any topology that makes $pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $pi$ continuous. This shows that $mathscr{T}_{text{quot}}$ is the maximal (i.e. finest) topology that makes $pi$ continuous.
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
The easiest way to reason, I think is:
Suppose that $mathscr{T}$ is any topology on $X/ {sim} $ that makes $pi$ continuous. Let $U in mathscr{T}$, then by continuity of $pi$ gives us that $pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U in mathscr{T}_{text{quot}}$. So $mathscr{T} subseteq mathscr{T}_{text{quot}}$.
This shows that any topology that makes $pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $pi$ continuous. This shows that $mathscr{T}_{text{quot}}$ is the maximal (i.e. finest) topology that makes $pi$ continuous.
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
The easiest way to reason, I think is:
Suppose that $mathscr{T}$ is any topology on $X/ {sim} $ that makes $pi$ continuous. Let $U in mathscr{T}$, then by continuity of $pi$ gives us that $pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U in mathscr{T}_{text{quot}}$. So $mathscr{T} subseteq mathscr{T}_{text{quot}}$.
This shows that any topology that makes $pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $pi$ continuous. This shows that $mathscr{T}_{text{quot}}$ is the maximal (i.e. finest) topology that makes $pi$ continuous.
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
The easiest way to reason, I think is:
Suppose that $mathscr{T}$ is any topology on $X/ {sim} $ that makes $pi$ continuous. Let $U in mathscr{T}$, then by continuity of $pi$ gives us that $pi^{-1}[U]$ is open in $X$. But then the definition of the quotient topology says that $U in mathscr{T}_{text{quot}}$. So $mathscr{T} subseteq mathscr{T}_{text{quot}}$.
This shows that any topology that makes $pi$ continuous is a subset of the quotient topology. And clearly the quotient topology is one of the topologies that makes $pi$ continuous. This shows that $mathscr{T}_{text{quot}}$ is the maximal (i.e. finest) topology that makes $pi$ continuous.
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
answered Apr 25 '17 at 8:52
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
$begingroup$
Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $pi^{-1}(U)$ is open within the domain space of $pi$.
edited Jan 14 at 18:19
Henno Brandsma
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $pi^{-1}(U)$ is open within the domain space of $pi$.
edited Jan 14 at 18:19
Henno Brandsma
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $pi^{-1}(U)$ is open within the domain space of $pi$.
edited Jan 14 at 18:19
Henno Brandsma
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
$endgroup$
Your reasoning is correct. However the quotient topology is usually defined as the finest topology. Then it is to be proved that $U$ is open by quotient topology iff $pi^{-1}(U)$ is open within the domain space of $pi$.
edited Jan 14 at 18:19
Henno Brandsma
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
edited Jan 14 at 18:19
Henno Brandsma
110k347116
edited Jan 14 at 18:19
Henno Brandsma
110k347116
edited Jan 14 at 18:19
Henno Brandsma
110k347116
110k347116
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
answered Apr 24 '17 at 23:09
William ElliotWilliam Elliot
8,1622720
8,1622720
add a comment |
add a comment |
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$begingroup$
What is your definition of "accurate"?
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:22
$begingroup$
Excuse me, I do not know the correct translation. With "accurate" is meant, that $tau$ is the topology with the most open sets, such that $pi$ is continuous.
$endgroup$
– Cornman
Apr 24 '17 at 20:31
$begingroup$
The standard term in English is "fine". I'll go ahead and edit that.
$endgroup$
– Eric Wofsey
Apr 24 '17 at 20:33
$begingroup$
Thank you, so I should have trusted the dictionary...
$endgroup$
– Cornman
Apr 24 '17 at 20:34