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Show that $mathbbR^2/∼$ is homeomorphic to $mathbbR$
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Question: Let $∼$ be the equivalence relation on $mathbbR^2$ given by ($x_1,x_2)∼ (y_1,y_2)$ if and only if $x_2 = y_2$. Show that the identiï¬Âcation space $mathbbR^2/∼$ is homeomorphic to $mathbbR$.
I have tried to define a function $f:mathbbR^2/∼ to mathbbR$ by $f(x,y)=y$ for all $(x,y)inmathbbR^2$. Then $f$ is clearly a bijection.
My trouble is to show $f$ and $f^-1$ are continuous. I think $f$ is a projection map, so it is continuous, am I right?
Any help is appreciated.
general-topology equivalence-relations quotient-spaces
asked Jul 27 at 3:01
user138017
446212
add a comment |Â
up vote
0
down vote
favorite
Question: Let $∼$ be the equivalence relation on $mathbbR^2$ given by ($x_1,x_2)∼ (y_1,y_2)$ if and only if $x_2 = y_2$. Show that the identiï¬Âcation space $mathbbR^2/∼$ is homeomorphic to $mathbbR$.
I have tried to define a function $f:mathbbR^2/∼ to mathbbR$ by $f(x,y)=y$ for all $(x,y)inmathbbR^2$. Then $f$ is clearly a bijection.
My trouble is to show $f$ and $f^-1$ are continuous. I think $f$ is a projection map, so it is continuous, am I right?
Any help is appreciated.
general-topology equivalence-relations quotient-spaces
asked Jul 27 at 3:01
user138017
446212
1
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
1
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
1
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
1
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question: Let $∼$ be the equivalence relation on $mathbbR^2$ given by ($x_1,x_2)∼ (y_1,y_2)$ if and only if $x_2 = y_2$. Show that the identiï¬Âcation space $mathbbR^2/∼$ is homeomorphic to $mathbbR$.
I have tried to define a function $f:mathbbR^2/∼ to mathbbR$ by $f(x,y)=y$ for all $(x,y)inmathbbR^2$. Then $f$ is clearly a bijection.
My trouble is to show $f$ and $f^-1$ are continuous. I think $f$ is a projection map, so it is continuous, am I right?
Any help is appreciated.
general-topology equivalence-relations quotient-spaces
asked Jul 27 at 3:01
user138017
446212
Question: Let $∼$ be the equivalence relation on $mathbbR^2$ given by ($x_1,x_2)∼ (y_1,y_2)$ if and only if $x_2 = y_2$. Show that the identiï¬Âcation space $mathbbR^2/∼$ is homeomorphic to $mathbbR$.
I have tried to define a function $f:mathbbR^2/∼ to mathbbR$ by $f(x,y)=y$ for all $(x,y)inmathbbR^2$. Then $f$ is clearly a bijection.
My trouble is to show $f$ and $f^-1$ are continuous. I think $f$ is a projection map, so it is continuous, am I right?
Any help is appreciated.
general-topology equivalence-relations quotient-spaces
asked Jul 27 at 3:01
user138017
446212
asked Jul 27 at 3:01
user138017
446212
asked Jul 27 at 3:01
user138017
446212
asked Jul 27 at 3:01
user138017
446212
446212
1
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
1
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
1
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
1
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31
add a comment |Â
1
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
1
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
1
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
1
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31
1
1
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
1
1
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
1
1
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
1
1
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31
add a comment |Â
1 Answer
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Let $[y]$ denote the equivalence class of any point $(x,y) in mathbbR^2$, according to $sim$. That is, $[y] = (x,y) : x in mathbbR$. Now let $U$ be an open subset of $mathbbR$. Then $f^-1(U) = [y] : y in U$ and this set is open in $mathbbR^2/ sim$ by the definition of quotient toplogy, since if $pi : mathbbR^2 to mathbbR^2 / sim$ given by $pi(x,y) = [y]$ is the quotient map, then $pi^-1(f^-1(U)) = mathbbRtimes U$, which is open in $mathbbR^2$.
That $f^-1$ is continuous means that if $W$ is open in $mathbbR^2/ sim$, then $f(W)$ is open in $mathbbR$. So fix such $W$ and note that $pi^-1(W) = mathbbRtimes W$ must be open in $mathbbR^2$, so $W$ is open in $mathbbR$. But $W = f(W)$, so we're done.
answered Jul 27 at 3:26
mathworker21
6,4301727
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $[y]$ denote the equivalence class of any point $(x,y) in mathbbR^2$, according to $sim$. That is, $[y] = (x,y) : x in mathbbR$. Now let $U$ be an open subset of $mathbbR$. Then $f^-1(U) = [y] : y in U$ and this set is open in $mathbbR^2/ sim$ by the definition of quotient toplogy, since if $pi : mathbbR^2 to mathbbR^2 / sim$ given by $pi(x,y) = [y]$ is the quotient map, then $pi^-1(f^-1(U)) = mathbbRtimes U$, which is open in $mathbbR^2$.
That $f^-1$ is continuous means that if $W$ is open in $mathbbR^2/ sim$, then $f(W)$ is open in $mathbbR$. So fix such $W$ and note that $pi^-1(W) = mathbbRtimes W$ must be open in $mathbbR^2$, so $W$ is open in $mathbbR$. But $W = f(W)$, so we're done.
answered Jul 27 at 3:26
mathworker21
6,4301727
add a comment |Â
up vote
2
down vote
Let $[y]$ denote the equivalence class of any point $(x,y) in mathbbR^2$, according to $sim$. That is, $[y] = (x,y) : x in mathbbR$. Now let $U$ be an open subset of $mathbbR$. Then $f^-1(U) = [y] : y in U$ and this set is open in $mathbbR^2/ sim$ by the definition of quotient toplogy, since if $pi : mathbbR^2 to mathbbR^2 / sim$ given by $pi(x,y) = [y]$ is the quotient map, then $pi^-1(f^-1(U)) = mathbbRtimes U$, which is open in $mathbbR^2$.
That $f^-1$ is continuous means that if $W$ is open in $mathbbR^2/ sim$, then $f(W)$ is open in $mathbbR$. So fix such $W$ and note that $pi^-1(W) = mathbbRtimes W$ must be open in $mathbbR^2$, so $W$ is open in $mathbbR$. But $W = f(W)$, so we're done.
answered Jul 27 at 3:26
mathworker21
6,4301727
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $[y]$ denote the equivalence class of any point $(x,y) in mathbbR^2$, according to $sim$. That is, $[y] = (x,y) : x in mathbbR$. Now let $U$ be an open subset of $mathbbR$. Then $f^-1(U) = [y] : y in U$ and this set is open in $mathbbR^2/ sim$ by the definition of quotient toplogy, since if $pi : mathbbR^2 to mathbbR^2 / sim$ given by $pi(x,y) = [y]$ is the quotient map, then $pi^-1(f^-1(U)) = mathbbRtimes U$, which is open in $mathbbR^2$.
That $f^-1$ is continuous means that if $W$ is open in $mathbbR^2/ sim$, then $f(W)$ is open in $mathbbR$. So fix such $W$ and note that $pi^-1(W) = mathbbRtimes W$ must be open in $mathbbR^2$, so $W$ is open in $mathbbR$. But $W = f(W)$, so we're done.
answered Jul 27 at 3:26
mathworker21
6,4301727
Let $[y]$ denote the equivalence class of any point $(x,y) in mathbbR^2$, according to $sim$. That is, $[y] = (x,y) : x in mathbbR$. Now let $U$ be an open subset of $mathbbR$. Then $f^-1(U) = [y] : y in U$ and this set is open in $mathbbR^2/ sim$ by the definition of quotient toplogy, since if $pi : mathbbR^2 to mathbbR^2 / sim$ given by $pi(x,y) = [y]$ is the quotient map, then $pi^-1(f^-1(U)) = mathbbRtimes U$, which is open in $mathbbR^2$.
That $f^-1$ is continuous means that if $W$ is open in $mathbbR^2/ sim$, then $f(W)$ is open in $mathbbR$. So fix such $W$ and note that $pi^-1(W) = mathbbRtimes W$ must be open in $mathbbR^2$, so $W$ is open in $mathbbR$. But $W = f(W)$, so we're done.
answered Jul 27 at 3:26
mathworker21
6,4301727
answered Jul 27 at 3:26
mathworker21
6,4301727
answered Jul 27 at 3:26
mathworker21
6,4301727
answered Jul 27 at 3:26
mathworker21
6,4301727
6,4301727
add a comment |Â
add a comment |Â
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1
Do you know the topology on $mathbbR^2 / sim$?
– mathworker21
Jul 27 at 3:18
1
@mathworker21 Not really, the question has not mentioned the topology, I guess just the quotient topology
– user138017
Jul 27 at 3:23
1
That $f$ is continuous (with quotient topology on $mathbbR^2 / sim$) comes from the continuity of $mathbbR^2 to mathbbR$ given by $(x,y) mapsto y$, which induces a continuous map on the quotient. That $f$ is a bijection is clear. That $f^-1$ is continuous can be shown by explicitly describing it as a composition of maps $mathbbR to mathbbR^2 to mathbbR^2 / sim$, where the first is $t mapsto (0 , t)$ and the second is the quotient map, both of which maps are continuous.
– Point
Jul 27 at 3:28
1
@Point nice! you're missing a parenthesis somewhere though
– mathworker21
Jul 27 at 3:30
@mathworker21 Thanks! I have tried to fix the paranthesis problem
– Point
Jul 27 at 3:31