Quotient maps and open maps
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I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:
Let $q: X rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = {(x_1, x_2) mid q(x_1) = q(x_2) }$ is closed in $X times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q times q$ is open), continuous and surjective. Can someone please clarify?
On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.
general-topology quotient-spaces open-map
$endgroup$
|
show 2 more comments
$begingroup$
I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:
Let $q: X rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = {(x_1, x_2) mid q(x_1) = q(x_2) }$ is closed in $X times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q times q$ is open), continuous and surjective. Can someone please clarify?
On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.
general-topology quotient-spaces open-map
$endgroup$
$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
1
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23
|
show 2 more comments
$begingroup$
I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:
Let $q: X rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = {(x_1, x_2) mid q(x_1) = q(x_2) }$ is closed in $X times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q times q$ is open), continuous and surjective. Can someone please clarify?
On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.
general-topology quotient-spaces open-map
$endgroup$
I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:
Let $q: X rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = {(x_1, x_2) mid q(x_1) = q(x_2) }$ is closed in $X times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q times q$ is open), continuous and surjective. Can someone please clarify?
On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.
general-topology quotient-spaces open-map
general-topology quotient-spaces open-map
asked Jan 3 at 20:32
Thomas BakxThomas Bakx
3359
3359
$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
1
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23
|
show 2 more comments
$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
1
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23
$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
1
1
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23
|
show 2 more comments
1 Answer
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oldest
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$begingroup$
Let us agree that a map is a continuous function.
A quotient map $q : X to Y$ is a surjective map such that $V subset Y$ is open if and only $q^{-1}(V) subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X to Y$ be an open quotient map" is therefore the same as "Let $q : X to Y$ be an open map".
What is the relation to the post $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$?
In this post we do not start with a map $q : X to Y$ between topological spaces, but with a space $X$ and an equivalence relation $sim$ on $X$ and then define $Y = X / sim$. The function
$$q : X to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X to Y$ is an open map, and that is the reason why we need the information that $q$ is the quotient map.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
$endgroup$
add a comment |
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$begingroup$
Let us agree that a map is a continuous function.
A quotient map $q : X to Y$ is a surjective map such that $V subset Y$ is open if and only $q^{-1}(V) subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X to Y$ be an open quotient map" is therefore the same as "Let $q : X to Y$ be an open map".
What is the relation to the post $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$?
In this post we do not start with a map $q : X to Y$ between topological spaces, but with a space $X$ and an equivalence relation $sim$ on $X$ and then define $Y = X / sim$. The function
$$q : X to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X to Y$ is an open map, and that is the reason why we need the information that $q$ is the quotient map.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
$endgroup$
add a comment |
$begingroup$
Let us agree that a map is a continuous function.
A quotient map $q : X to Y$ is a surjective map such that $V subset Y$ is open if and only $q^{-1}(V) subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X to Y$ be an open quotient map" is therefore the same as "Let $q : X to Y$ be an open map".
What is the relation to the post $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$?
In this post we do not start with a map $q : X to Y$ between topological spaces, but with a space $X$ and an equivalence relation $sim$ on $X$ and then define $Y = X / sim$. The function
$$q : X to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X to Y$ is an open map, and that is the reason why we need the information that $q$ is the quotient map.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
$endgroup$
add a comment |
$begingroup$
Let us agree that a map is a continuous function.
A quotient map $q : X to Y$ is a surjective map such that $V subset Y$ is open if and only $q^{-1}(V) subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X to Y$ be an open quotient map" is therefore the same as "Let $q : X to Y$ be an open map".
What is the relation to the post $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$?
In this post we do not start with a map $q : X to Y$ between topological spaces, but with a space $X$ and an equivalence relation $sim$ on $X$ and then define $Y = X / sim$. The function
$$q : X to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X to Y$ is an open map, and that is the reason why we need the information that $q$ is the quotient map.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
$endgroup$
Let us agree that a map is a continuous function.
A quotient map $q : X to Y$ is a surjective map such that $V subset Y$ is open if and only $q^{-1}(V) subset X$ is open.
A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).
The assumption "Let $q : X to Y$ be an open quotient map" is therefore the same as "Let $q : X to Y$ be an open map".
What is the relation to the post $X/{sim}$ is Hausdorff if and only if $sim$ is closed in $X times X$?
In this post we do not start with a map $q : X to Y$ between topological spaces, but with a space $X$ and an equivalence relation $sim$ on $X$ and then define $Y = X / sim$. The function
$$q : X to Y, q(x) = [x] ,$$
where $[x]$ denotes the equivalence class of $x$ with respect to $sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).
Thus, the phrase "If the quotient map is open" tells us two facts:
(1) The set $Y$ is endowed with the quotient topology.
(2) $q$ is an open map.
For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X to Y$ is an open map, and that is the reason why we need the information that $q$ is the quotient map.
In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.
Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.
edited Jan 4 at 11:08
answered Jan 3 at 22:50
Paul FrostPaul Frost
11.6k3934
11.6k3934
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$begingroup$
Okay, so I guess a quotient map $f:Xto Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map.
$endgroup$
– Ben W
Jan 3 at 20:46
$begingroup$
Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness.
$endgroup$
– Thomas Bakx
Jan 3 at 20:49
$begingroup$
Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis.,
$endgroup$
– Ben W
Jan 3 at 20:50
$begingroup$
All that is clear, this is not answering my question.
$endgroup$
– Thomas Bakx
Jan 3 at 21:25
1
$begingroup$
@ThomasBakx What makes you think that a projection $mathbb{R}^2tomathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $Vsubseteq Y$ if $f:Xto Y$ is surjective.
$endgroup$
– freakish
Jan 3 at 22:23