Product of quotient map ando the identity is quotient map
$begingroup$
The following is an exercise in Tommo Tom Dieck's book "Algebriac Topology"
If $X$ is a topological space, $A$ is a compact subspace of $X$ ans $p:Xto X/A$ is the canonical quotient map then for any topological space $Y$ the product map $ptimes id_Y$ is a quotient map.
I need to prove It in order to demonstrate that $Sigma(Y)$, the reduced suspension, is the same than the smash product of $S^1$ and $Y$.
What I have done is try to prove that if $U$ is a subset such that $(ptimes id_Y)^{-1}$ is open, i.e., is in the form $bigcup_{iinalpha}O_itimes B_i$ then $Asubset bigcup_{iinalpha}O_i$, ando knowing that $A$ is compact this can be reduced yo a finite union.
I don't know how to proceed for now on.
It would be very useful if you could give me a hint.
general-topology algebraic-topology quotient-spaces
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
$endgroup$
add a comment |
$begingroup$
The following is an exercise in Tommo Tom Dieck's book "Algebriac Topology"
If $X$ is a topological space, $A$ is a compact subspace of $X$ ans $p:Xto X/A$ is the canonical quotient map then for any topological space $Y$ the product map $ptimes id_Y$ is a quotient map.
I need to prove It in order to demonstrate that $Sigma(Y)$, the reduced suspension, is the same than the smash product of $S^1$ and $Y$.
What I have done is try to prove that if $U$ is a subset such that $(ptimes id_Y)^{-1}$ is open, i.e., is in the form $bigcup_{iinalpha}O_itimes B_i$ then $Asubset bigcup_{iinalpha}O_i$, ando knowing that $A$ is compact this can be reduced yo a finite union.
I don't know how to proceed for now on.
It would be very useful if you could give me a hint.
general-topology algebraic-topology quotient-spaces
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
$endgroup$
1
$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23
add a comment |
$begingroup$
The following is an exercise in Tommo Tom Dieck's book "Algebriac Topology"
If $X$ is a topological space, $A$ is a compact subspace of $X$ ans $p:Xto X/A$ is the canonical quotient map then for any topological space $Y$ the product map $ptimes id_Y$ is a quotient map.
I need to prove It in order to demonstrate that $Sigma(Y)$, the reduced suspension, is the same than the smash product of $S^1$ and $Y$.
What I have done is try to prove that if $U$ is a subset such that $(ptimes id_Y)^{-1}$ is open, i.e., is in the form $bigcup_{iinalpha}O_itimes B_i$ then $Asubset bigcup_{iinalpha}O_i$, ando knowing that $A$ is compact this can be reduced yo a finite union.
I don't know how to proceed for now on.
It would be very useful if you could give me a hint.
general-topology algebraic-topology quotient-spaces
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
$endgroup$
The following is an exercise in Tommo Tom Dieck's book "Algebriac Topology"
If $X$ is a topological space, $A$ is a compact subspace of $X$ ans $p:Xto X/A$ is the canonical quotient map then for any topological space $Y$ the product map $ptimes id_Y$ is a quotient map.
I need to prove It in order to demonstrate that $Sigma(Y)$, the reduced suspension, is the same than the smash product of $S^1$ and $Y$.
What I have done is try to prove that if $U$ is a subset such that $(ptimes id_Y)^{-1}$ is open, i.e., is in the form $bigcup_{iinalpha}O_itimes B_i$ then $Asubset bigcup_{iinalpha}O_i$, ando knowing that $A$ is compact this can be reduced yo a finite union.
I don't know how to proceed for now on.
It would be very useful if you could give me a hint.
general-topology algebraic-topology quotient-spaces
general-topology algebraic-topology quotient-spaces
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
asked Sep 12 '17 at 18:04
Paula Cartagena AtaráPaula Cartagena Atará
1496
1496
1
$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23
add a comment |
1
$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23
1
1
$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23
add a comment |
1 Answer
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Let $U subset X/A times Y$ be such that $(p times id)^{-1} (U)$ is open. We want to show $U$ is also open. For any $(x,y) in (p times id)^{-1} (U)$, there is open $U_0 subset X$ and open $V_0 subset Y$ such that $(x,y) in U_0 times V_0 subset (p times id)^{-1} (U)$. If $U_0$ doesn't intersect $A$, then $p(U_0)$ is open in $X /A$. If $U_0$ has an nonempty intersection with $A$, then $A times leftlbrace yrightrbrace$ is contained in $ (p times id)^{-1} (U)$. By compactness of $A$, there exist open neighborhood $U_1$ containing $A$ and $x$, and open $V_1$ containing $y$ such that $U_1 times V_1 subset (p times id)^{-1}(U)$. $p$ maps $U_1$ to an open set in $X /A$. This shows that $p times id$ is a quotient map.
answered Jan 11 at 6:06
user559675user559675
627
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Let $U subset X/A times Y$ be such that $(p times id)^{-1} (U)$ is open. We want to show $U$ is also open. For any $(x,y) in (p times id)^{-1} (U)$, there is open $U_0 subset X$ and open $V_0 subset Y$ such that $(x,y) in U_0 times V_0 subset (p times id)^{-1} (U)$. If $U_0$ doesn't intersect $A$, then $p(U_0)$ is open in $X /A$. If $U_0$ has an nonempty intersection with $A$, then $A times leftlbrace yrightrbrace$ is contained in $ (p times id)^{-1} (U)$. By compactness of $A$, there exist open neighborhood $U_1$ containing $A$ and $x$, and open $V_1$ containing $y$ such that $U_1 times V_1 subset (p times id)^{-1}(U)$. $p$ maps $U_1$ to an open set in $X /A$. This shows that $p times id$ is a quotient map.
answered Jan 11 at 6:06
user559675user559675
627
$endgroup$
add a comment |
$begingroup$
Let $U subset X/A times Y$ be such that $(p times id)^{-1} (U)$ is open. We want to show $U$ is also open. For any $(x,y) in (p times id)^{-1} (U)$, there is open $U_0 subset X$ and open $V_0 subset Y$ such that $(x,y) in U_0 times V_0 subset (p times id)^{-1} (U)$. If $U_0$ doesn't intersect $A$, then $p(U_0)$ is open in $X /A$. If $U_0$ has an nonempty intersection with $A$, then $A times leftlbrace yrightrbrace$ is contained in $ (p times id)^{-1} (U)$. By compactness of $A$, there exist open neighborhood $U_1$ containing $A$ and $x$, and open $V_1$ containing $y$ such that $U_1 times V_1 subset (p times id)^{-1}(U)$. $p$ maps $U_1$ to an open set in $X /A$. This shows that $p times id$ is a quotient map.
answered Jan 11 at 6:06
user559675user559675
627
$endgroup$
add a comment |
$begingroup$
Let $U subset X/A times Y$ be such that $(p times id)^{-1} (U)$ is open. We want to show $U$ is also open. For any $(x,y) in (p times id)^{-1} (U)$, there is open $U_0 subset X$ and open $V_0 subset Y$ such that $(x,y) in U_0 times V_0 subset (p times id)^{-1} (U)$. If $U_0$ doesn't intersect $A$, then $p(U_0)$ is open in $X /A$. If $U_0$ has an nonempty intersection with $A$, then $A times leftlbrace yrightrbrace$ is contained in $ (p times id)^{-1} (U)$. By compactness of $A$, there exist open neighborhood $U_1$ containing $A$ and $x$, and open $V_1$ containing $y$ such that $U_1 times V_1 subset (p times id)^{-1}(U)$. $p$ maps $U_1$ to an open set in $X /A$. This shows that $p times id$ is a quotient map.
answered Jan 11 at 6:06
user559675user559675
627
$endgroup$
Let $U subset X/A times Y$ be such that $(p times id)^{-1} (U)$ is open. We want to show $U$ is also open. For any $(x,y) in (p times id)^{-1} (U)$, there is open $U_0 subset X$ and open $V_0 subset Y$ such that $(x,y) in U_0 times V_0 subset (p times id)^{-1} (U)$. If $U_0$ doesn't intersect $A$, then $p(U_0)$ is open in $X /A$. If $U_0$ has an nonempty intersection with $A$, then $A times leftlbrace yrightrbrace$ is contained in $ (p times id)^{-1} (U)$. By compactness of $A$, there exist open neighborhood $U_1$ containing $A$ and $x$, and open $V_1$ containing $y$ such that $U_1 times V_1 subset (p times id)^{-1}(U)$. $p$ maps $U_1$ to an open set in $X /A$. This shows that $p times id$ is a quotient map.
answered Jan 11 at 6:06
user559675user559675
627
answered Jan 11 at 6:06
user559675user559675
627
answered Jan 11 at 6:06
user559675user559675
627
answered Jan 11 at 6:06
user559675user559675
627
627
add a comment |
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$begingroup$
Use the fact f is a quotient iff for all V, (V is open iff $f^{-1}$(V) is open).
$endgroup$
– William Elliot
Sep 12 '17 at 19:49
$begingroup$
Thanks William, that is what I am trying yo prove but It not clear for me, because for continuity one of the implication is clear, but the other one not.
$endgroup$
– Paula Cartagena Atará
Sep 12 '17 at 20:43
$begingroup$
Show each for each U_i x B_i that the inverse image is open and use the inverse image of a union of sets = the union of the inverse images of each set.
$endgroup$
– William Elliot
Sep 13 '17 at 4:25
$begingroup$
I first thought of the Whitehead theorem : The product of a quotient map and the identity on a locally compact space is a quotient map, but that's too restrictive, as we have general spaces here. So if its true it must lie in the simple nature of the quotient map.
$endgroup$
– Henno Brandsma
Sep 13 '17 at 21:23