Quotient maps and open maps












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$begingroup$


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.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












1 Answer
1






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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.






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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.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 11:08

























        answered Jan 3 at 22:50









        Paul FrostPaul Frost

        11.6k3934




        11.6k3934






























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