Czes Kosniowski: Homeomorphic topological spaces $S^1times I / sim$ and $D^2$ [duplicate]$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$Proof that $mathbb R/[0,1]$ and $mathbb R$ are homeomorphicShow $(S^1times [0,1])/$~ is homeomorphic to $D^2$Topological , Homeomorphic version of $|S times S|=|S| $Why is $D^n/sim$ homeomorphic to $mathbbRP^n$?Why $(mathbb Qtimesmathbb Q)/(mathbb Ztimes=)$ is not homeomorphic to $(mathbb Q/mathbb Z)times(mathbb Q/=)$?Quotient spaces homeomorphic to the realsProve that $(Xtimes [0,1])/sim'$ is homeomorphic to $(X/sim)times[0,1].$RP2 is homeomorphic to $(I times I)/ sim$Homeomorphism between a quotient topology on $mathbbR^2$ and $mathbbR$
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Czes Kosniowski: Homeomorphic topological spaces $S^1times I / sim$ and $D^2$ [duplicate]
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$Proof that $mathbb R/[0,1]$ and $mathbb R$ are homeomorphicShow $(S^1times [0,1])/$~ is homeomorphic to $D^2$Topological , Homeomorphic version of $|S times S|=|S| $Why is $D^n/sim$ homeomorphic to $mathbbRP^n$?Why $(mathbb Qtimesmathbb Q)/(mathbb Ztimes=)$ is not homeomorphic to $(mathbb Q/mathbb Z)times(mathbb Q/=)$?Quotient spaces homeomorphic to the realsProve that $(Xtimes [0,1])/sim'$ is homeomorphic to $(X/sim)times[0,1].$RP2 is homeomorphic to $(I times I)/ sim$Homeomorphism between a quotient topology on $mathbbR^2$ and $mathbbR$
$begingroup$
This question already has an answer here:
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$
1 answer
Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.
Prove that $X$/~ is homeomorphic to the unit disc $D^2=le1=xin mathbbC: $ with the induced topology.
Any hints would be appreciated.
general-topology algebraic-topology
edited yesterday
Paul Frost
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
$endgroup$
marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro yesterday
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$
1 answer
Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.
Prove that $X$/~ is homeomorphic to the unit disc $D^2=le1=xin mathbbC: $ with the induced topology.
Any hints would be appreciated.
general-topology algebraic-topology
edited yesterday
Paul Frost
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
$endgroup$
marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro yesterday
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
2
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago
add a comment |
$begingroup$
This question already has an answer here:
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$
1 answer
Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.
Prove that $X$/~ is homeomorphic to the unit disc $D^2=le1=xin mathbbC: $ with the induced topology.
Any hints would be appreciated.
general-topology algebraic-topology
edited yesterday
Paul Frost
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
$endgroup$
This question already has an answer here:
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$
1 answer
Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.
Prove that $X$/~ is homeomorphic to the unit disc $D^2=le1=xin mathbbC: $ with the induced topology.
Any hints would be appreciated.
This question already has an answer here:
$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$
1 answer
general-topology algebraic-topology
general-topology algebraic-topology
edited yesterday
Paul Frost
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
edited yesterday
Paul Frost
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
edited yesterday
Paul Frost
11.6k3934
edited yesterday
Paul Frost
11.6k3934
edited yesterday
Paul Frost
11.6k3934
11.6k3934
asked 2 days ago
AlfdavAlfdav
977
asked 2 days ago
AlfdavAlfdav
977
asked 2 days ago
AlfdavAlfdav
977
977
marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro yesterday
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro yesterday
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
2
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago
add a comment |
1
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
2
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago
1
1
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
2
2
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space
$$barfcolon (S^1 times I)/sim to D^2 $$
Now show that this function is a homeomorphism.
answered 2 days ago
WilliamWilliam
2,5851224
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space
$$barfcolon (S^1 times I)/sim to D^2 $$
Now show that this function is a homeomorphism.
answered 2 days ago
WilliamWilliam
2,5851224
$endgroup$
add a comment |
$begingroup$
Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space
$$barfcolon (S^1 times I)/sim to D^2 $$
Now show that this function is a homeomorphism.
answered 2 days ago
WilliamWilliam
2,5851224
$endgroup$
add a comment |
$begingroup$
Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space
$$barfcolon (S^1 times I)/sim to D^2 $$
Now show that this function is a homeomorphism.
answered 2 days ago
WilliamWilliam
2,5851224
$endgroup$
Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space
$$barfcolon (S^1 times I)/sim to D^2 $$
Now show that this function is a homeomorphism.
answered 2 days ago
WilliamWilliam
2,5851224
edited 2 days ago
answered 2 days ago
WilliamWilliam
2,5851224
answered 2 days ago
WilliamWilliam
2,5851224
answered 2 days ago
WilliamWilliam
2,5851224
2,5851224
add a comment |
add a comment |
1
$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
2 days ago
$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
2 days ago
2
$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
2 days ago