An explicit homotopy equivalent from 3 circles that are tangent to neighbor to bouquet of 3 circles
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This problems is from Jack Lee's Introduction to topological manifold, problem 7-15.
Suppose you have 3 circles sitting in $mathbb{R}^2$ along the x axis and are tangent to neighboring circles (so the middle circle is tangent to the circle on the left and right), prove that it is homotopic equivalent to the bouquet of 3 circles.
I am aware of the "standard answer" using the hint given by the problem, by showing that both spaces are strong deformation retract of $mathbb{R}^2$ with 3 holes. However, is there a more explicity homotopy equivalent one can construct between the two spaces to show that they are homotopic equivalent? (Which is clearly, not a strong deformation retract of any kind).
general-topology manifolds
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
$endgroup$
add a comment |
$begingroup$
This problems is from Jack Lee's Introduction to topological manifold, problem 7-15.
Suppose you have 3 circles sitting in $mathbb{R}^2$ along the x axis and are tangent to neighboring circles (so the middle circle is tangent to the circle on the left and right), prove that it is homotopic equivalent to the bouquet of 3 circles.
I am aware of the "standard answer" using the hint given by the problem, by showing that both spaces are strong deformation retract of $mathbb{R}^2$ with 3 holes. However, is there a more explicity homotopy equivalent one can construct between the two spaces to show that they are homotopic equivalent? (Which is clearly, not a strong deformation retract of any kind).
general-topology manifolds
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
$endgroup$
1
$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58
add a comment |
$begingroup$
This problems is from Jack Lee's Introduction to topological manifold, problem 7-15.
Suppose you have 3 circles sitting in $mathbb{R}^2$ along the x axis and are tangent to neighboring circles (so the middle circle is tangent to the circle on the left and right), prove that it is homotopic equivalent to the bouquet of 3 circles.
I am aware of the "standard answer" using the hint given by the problem, by showing that both spaces are strong deformation retract of $mathbb{R}^2$ with 3 holes. However, is there a more explicity homotopy equivalent one can construct between the two spaces to show that they are homotopic equivalent? (Which is clearly, not a strong deformation retract of any kind).
general-topology manifolds
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
$endgroup$
This problems is from Jack Lee's Introduction to topological manifold, problem 7-15.
Suppose you have 3 circles sitting in $mathbb{R}^2$ along the x axis and are tangent to neighboring circles (so the middle circle is tangent to the circle on the left and right), prove that it is homotopic equivalent to the bouquet of 3 circles.
I am aware of the "standard answer" using the hint given by the problem, by showing that both spaces are strong deformation retract of $mathbb{R}^2$ with 3 holes. However, is there a more explicity homotopy equivalent one can construct between the two spaces to show that they are homotopic equivalent? (Which is clearly, not a strong deformation retract of any kind).
general-topology manifolds
general-topology manifolds
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
edited Dec 19 '18 at 12:51
Paul Frost
10.8k3934
10.8k3934
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
asked Dec 18 '18 at 23:03
EcotisticianEcotistician
32018
32018
1
$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58
add a comment |
1
$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58
1
1
$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58
add a comment |
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$begingroup$
You can use the "standard answer". If $i:Ytomathbb{R}^2backslash{x,y,z}$ is the inclusion and $r:mathbb{R}^2backslash{x,y,z}to X$ is a deformation retraction then your homotopy equivalence is simply $rcirc i$, right? So all you have to do is construct $r$.
$endgroup$
– freakish
Dec 19 '18 at 10:56
$begingroup$
I see, is it possible to consider the 3 circle space in the problem as a strong deformation retract of 3 non-linear points (so each circle is a strong deformation retract of a hole that's not sitting at its center? (Namely, the space we know to have an obvious strong deformation retract to the bonquet of 3 circles.
$endgroup$
– Ecotistician
Dec 19 '18 at 17:58