Mixing
Quotients and subspaces that intersect each equivalence class
$begingroup$
My question is not really a question, but a proof checking and a clarification. Suppose that $X$ is a topological space, and let $sim$ an equivalence relation on this space. Suppose that there exist a subspace $Y$ of $X$ that intersect each equivalence class. I want to prove that the two spaces $X/sim$ And $Y/sim$ are homeomorphic. I think this is quite useful in a lot of situations, but I have not found this result in my books, so I have thought to ask, just to be sure.
I have thought about a proof: we denote with $pi_X:Xto Xsim$ and with $pi_Y:Yto Y/sim$ the two projection maps, and with $i:Yto X$ the canonical topological embedding. We consider now the following diagram:
$require{AMScd}$
begin{CD}
Y @>{pi_Y}>> Y/sim\
@V{pi_Xcirc i}VV\
X/sim
end{CD}
Now, if the maps $pi_Y$ and $pi_Xcirc i$ are two identifications, since they are constant on the fibres of the other, the universal property of identifications gives us the existence of two maps $f:X/sim to Y/sim$ and $g:Y/simto X/sim$ continuous that make the diagram commute, and we have that $g=f^{-1}:$
$$pi_Y=fcirc(pi_Xcirc i)=fcirc gcirc pi_Y Rightarrow fcirc g=id_{Y/sim},$$
and similarly $fcirc g=id_{X/sim}.$
If all the reasoning is right, what are the most simple conditions on $Y$ to have that $pi_Xcirc i$ is an identification?
general-topology proof-verification
asked Jan 29 at 18:25
MatPMatP
1367
$endgroup$
add a comment |
$begingroup$
My question is not really a question, but a proof checking and a clarification. Suppose that $X$ is a topological space, and let $sim$ an equivalence relation on this space. Suppose that there exist a subspace $Y$ of $X$ that intersect each equivalence class. I want to prove that the two spaces $X/sim$ And $Y/sim$ are homeomorphic. I think this is quite useful in a lot of situations, but I have not found this result in my books, so I have thought to ask, just to be sure.
I have thought about a proof: we denote with $pi_X:Xto Xsim$ and with $pi_Y:Yto Y/sim$ the two projection maps, and with $i:Yto X$ the canonical topological embedding. We consider now the following diagram:
$require{AMScd}$
begin{CD}
Y @>{pi_Y}>> Y/sim\
@V{pi_Xcirc i}VV\
X/sim
end{CD}
Now, if the maps $pi_Y$ and $pi_Xcirc i$ are two identifications, since they are constant on the fibres of the other, the universal property of identifications gives us the existence of two maps $f:X/sim to Y/sim$ and $g:Y/simto X/sim$ continuous that make the diagram commute, and we have that $g=f^{-1}:$
$$pi_Y=fcirc(pi_Xcirc i)=fcirc gcirc pi_Y Rightarrow fcirc g=id_{Y/sim},$$
and similarly $fcirc g=id_{X/sim}.$
If all the reasoning is right, what are the most simple conditions on $Y$ to have that $pi_Xcirc i$ is an identification?
general-topology proof-verification
asked Jan 29 at 18:25
MatPMatP
1367
$endgroup$
$begingroup$
Would you mind clarifying what is the definition of an identification?
$endgroup$
– Guido A.
Jan 29 at 19:00
$begingroup$
I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
$endgroup$
– MatP
Jan 29 at 21:03
$begingroup$
Where do you use $Y$ is open instead of any selecting subset?
$endgroup$
– Henno Brandsma
Jan 30 at 5:24
$begingroup$
No I don't need open, I think, it is only a misprint
$endgroup$
– MatP
Jan 30 at 11:29
add a comment |
$begingroup$
My question is not really a question, but a proof checking and a clarification. Suppose that $X$ is a topological space, and let $sim$ an equivalence relation on this space. Suppose that there exist a subspace $Y$ of $X$ that intersect each equivalence class. I want to prove that the two spaces $X/sim$ And $Y/sim$ are homeomorphic. I think this is quite useful in a lot of situations, but I have not found this result in my books, so I have thought to ask, just to be sure.
I have thought about a proof: we denote with $pi_X:Xto Xsim$ and with $pi_Y:Yto Y/sim$ the two projection maps, and with $i:Yto X$ the canonical topological embedding. We consider now the following diagram:
$require{AMScd}$
begin{CD}
Y @>{pi_Y}>> Y/sim\
@V{pi_Xcirc i}VV\
X/sim
end{CD}
Now, if the maps $pi_Y$ and $pi_Xcirc i$ are two identifications, since they are constant on the fibres of the other, the universal property of identifications gives us the existence of two maps $f:X/sim to Y/sim$ and $g:Y/simto X/sim$ continuous that make the diagram commute, and we have that $g=f^{-1}:$
$$pi_Y=fcirc(pi_Xcirc i)=fcirc gcirc pi_Y Rightarrow fcirc g=id_{Y/sim},$$
and similarly $fcirc g=id_{X/sim}.$
If all the reasoning is right, what are the most simple conditions on $Y$ to have that $pi_Xcirc i$ is an identification?
general-topology proof-verification
asked Jan 29 at 18:25
MatPMatP
1367
$endgroup$
My question is not really a question, but a proof checking and a clarification. Suppose that $X$ is a topological space, and let $sim$ an equivalence relation on this space. Suppose that there exist a subspace $Y$ of $X$ that intersect each equivalence class. I want to prove that the two spaces $X/sim$ And $Y/sim$ are homeomorphic. I think this is quite useful in a lot of situations, but I have not found this result in my books, so I have thought to ask, just to be sure.
I have thought about a proof: we denote with $pi_X:Xto Xsim$ and with $pi_Y:Yto Y/sim$ the two projection maps, and with $i:Yto X$ the canonical topological embedding. We consider now the following diagram:
$require{AMScd}$
begin{CD}
Y @>{pi_Y}>> Y/sim\
@V{pi_Xcirc i}VV\
X/sim
end{CD}
Now, if the maps $pi_Y$ and $pi_Xcirc i$ are two identifications, since they are constant on the fibres of the other, the universal property of identifications gives us the existence of two maps $f:X/sim to Y/sim$ and $g:Y/simto X/sim$ continuous that make the diagram commute, and we have that $g=f^{-1}:$
$$pi_Y=fcirc(pi_Xcirc i)=fcirc gcirc pi_Y Rightarrow fcirc g=id_{Y/sim},$$
and similarly $fcirc g=id_{X/sim}.$
If all the reasoning is right, what are the most simple conditions on $Y$ to have that $pi_Xcirc i$ is an identification?
general-topology proof-verification
general-topology proof-verification
asked Jan 29 at 18:25
MatPMatP
1367
asked Jan 29 at 18:25
MatPMatP
1367
edited Feb 1 at 15:25
MatP
asked Jan 29 at 18:25
MatPMatP
1367
asked Jan 29 at 18:25
MatPMatP
1367
asked Jan 29 at 18:25
MatPMatP
1367
1367
$begingroup$
Would you mind clarifying what is the definition of an identification?
$endgroup$
– Guido A.
Jan 29 at 19:00
$begingroup$
I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
$endgroup$
– MatP
Jan 29 at 21:03
$begingroup$
Where do you use $Y$ is open instead of any selecting subset?
$endgroup$
– Henno Brandsma
Jan 30 at 5:24
$begingroup$
No I don't need open, I think, it is only a misprint
$endgroup$
– MatP
Jan 30 at 11:29
add a comment |
$begingroup$
Would you mind clarifying what is the definition of an identification?
$endgroup$
– Guido A.
Jan 29 at 19:00
$begingroup$
I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
$endgroup$
– MatP
Jan 29 at 21:03
$begingroup$
Where do you use $Y$ is open instead of any selecting subset?
$endgroup$
– Henno Brandsma
Jan 30 at 5:24
$begingroup$
No I don't need open, I think, it is only a misprint
$endgroup$
– MatP
Jan 30 at 11:29
$begingroup$
Would you mind clarifying what is the definition of an identification?
$endgroup$
– Guido A.
Jan 29 at 19:00
$begingroup$
Would you mind clarifying what is the definition of an identification?
$endgroup$
– Guido A.
Jan 29 at 19:00
$begingroup$
I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
$endgroup$
– MatP
Jan 29 at 21:03
$begingroup$
I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
$endgroup$
– MatP
Jan 29 at 21:03
$begingroup$
Where do you use $Y$ is open instead of any selecting subset?
$endgroup$
– Henno Brandsma
Jan 30 at 5:24
$begingroup$
Where do you use $Y$ is open instead of any selecting subset?
$endgroup$
– Henno Brandsma
Jan 30 at 5:24
$begingroup$
No I don't need open, I think, it is only a misprint
$endgroup$
– MatP
Jan 30 at 11:29
$begingroup$
No I don't need open, I think, it is only a misprint
$endgroup$
– MatP
Jan 30 at 11:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$pi_{X}circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]cup[2,3], Y={q| qtext{ is a rational in } [0,1]}cup {r| r text{ is an irrational in }[2,3]},$$
and let $xsim y$ if $|x-y|=0text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A={[x]|xin
mathbb{Q}cap X}$ is not open in the space $X/sim$, for $pi_{X}^{-1}(A)=Xcapmathbb{Q}$ is not open in $A$. But $(pi_{X}circ i)^{-1}(A)=[0,1]capmathbb{Q}$ is open in $Y$. Thus $pi_{X}circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$pi_{X}circ i$ is the composition of an identification, $pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Yneq X$, $i$ is not surjective.)
answered Jan 31 at 5:22
KenKen
186110
$endgroup$
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
|
show 2 more comments
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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
$begingroup$
$pi_{X}circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]cup[2,3], Y={q| qtext{ is a rational in } [0,1]}cup {r| r text{ is an irrational in }[2,3]},$$
and let $xsim y$ if $|x-y|=0text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A={[x]|xin
mathbb{Q}cap X}$ is not open in the space $X/sim$, for $pi_{X}^{-1}(A)=Xcapmathbb{Q}$ is not open in $A$. But $(pi_{X}circ i)^{-1}(A)=[0,1]capmathbb{Q}$ is open in $Y$. Thus $pi_{X}circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$pi_{X}circ i$ is the composition of an identification, $pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Yneq X$, $i$ is not surjective.)
answered Jan 31 at 5:22
KenKen
186110
$endgroup$
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
|
show 2 more comments
$begingroup$
$pi_{X}circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]cup[2,3], Y={q| qtext{ is a rational in } [0,1]}cup {r| r text{ is an irrational in }[2,3]},$$
and let $xsim y$ if $|x-y|=0text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A={[x]|xin
mathbb{Q}cap X}$ is not open in the space $X/sim$, for $pi_{X}^{-1}(A)=Xcapmathbb{Q}$ is not open in $A$. But $(pi_{X}circ i)^{-1}(A)=[0,1]capmathbb{Q}$ is open in $Y$. Thus $pi_{X}circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$pi_{X}circ i$ is the composition of an identification, $pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Yneq X$, $i$ is not surjective.)
answered Jan 31 at 5:22
KenKen
186110
$endgroup$
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
|
show 2 more comments
$begingroup$
$pi_{X}circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]cup[2,3], Y={q| qtext{ is a rational in } [0,1]}cup {r| r text{ is an irrational in }[2,3]},$$
and let $xsim y$ if $|x-y|=0text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A={[x]|xin
mathbb{Q}cap X}$ is not open in the space $X/sim$, for $pi_{X}^{-1}(A)=Xcapmathbb{Q}$ is not open in $A$. But $(pi_{X}circ i)^{-1}(A)=[0,1]capmathbb{Q}$ is open in $Y$. Thus $pi_{X}circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$pi_{X}circ i$ is the composition of an identification, $pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Yneq X$, $i$ is not surjective.)
answered Jan 31 at 5:22
KenKen
186110
$endgroup$
$pi_{X}circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]cup[2,3], Y={q| qtext{ is a rational in } [0,1]}cup {r| r text{ is an irrational in }[2,3]},$$
and let $xsim y$ if $|x-y|=0text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A={[x]|xin
mathbb{Q}cap X}$ is not open in the space $X/sim$, for $pi_{X}^{-1}(A)=Xcapmathbb{Q}$ is not open in $A$. But $(pi_{X}circ i)^{-1}(A)=[0,1]capmathbb{Q}$ is open in $Y$. Thus $pi_{X}circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$pi_{X}circ i$ is the composition of an identification, $pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Yneq X$, $i$ is not surjective.)
answered Jan 31 at 5:22
KenKen
186110
edited Jan 31 at 23:26
answered Jan 31 at 5:22
KenKen
186110
answered Jan 31 at 5:22
KenKen
186110
answered Jan 31 at 5:22
KenKen
186110
186110
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
|
show 2 more comments
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
I am sorry I cannot be of help with your question on the conditions that makes $pi_{X}circ i$ into a quotient map.
$endgroup$
– Ken
Jan 31 at 5:30
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
Why do you say that $sim$ is an equivalence relation? $xnotsim x,$ since $|x-x|=0neq2dots$
$endgroup$
– MatP
Jan 31 at 16:09
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP I am sorry. Of course you are right. I should have written $xsim y iff |x-y|=0text{ or }2$. I edited.
$endgroup$
– Ken
Jan 31 at 23:25
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
@MatP $pi_{X}|_Y$ is not equal to $pi_{Y}$, for the codomain of them are $X/sim$ and $Y/sim$, respectively, which are different sets. (Even though there is a bijective correspondence between these two sets.)
$endgroup$
– Ken
Jan 31 at 23:30
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
$begingroup$
Sorry, but I have another question: why $Xcapmathbb{Q}$ is not open in $X,$ while $[0,1]capmathbb{Q}$ is open in $Y$?
$endgroup$
– MatP
Feb 1 at 15:40
|
show 2 more comments
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Would you mind clarifying what is the definition of an identification?
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– Guido A.
Jan 29 at 19:00
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I mean a continuous map $f:Xto Y$ such that for all $Asubset Y,$ $A$ is open if and only if $f^{-1}(A)$ is open.
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– MatP
Jan 29 at 21:03
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Where do you use $Y$ is open instead of any selecting subset?
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– Henno Brandsma
Jan 30 at 5:24
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No I don't need open, I think, it is only a misprint
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– MatP
Jan 30 at 11:29