Mixing
Show $S^2$ with 2 cells attached is equivalent to a wedge of spheres
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Show that a space obtained from $S^2$ by attaching n 2-cells along any collection of n circles in $S^2$ is homotopy equivalent to the wedge of n+1 spheres.
I'm a little confused here. I'm imagining a sphere, and putting 2 dimensional discs inside of it. Attaching one disc along one circle of $S^2$ and then collapsing that disc to a point would appear to me to create a wedge of two spheres.
However, if I attached 2 discs along 2 different circles of $S^2$ and collapsed each to a point, to me it seems I would have created a wedge of four spheres, one sphere for each division created by the two discs inside the sphere.
What is wrong with my thinking here? Thanks!!
algebraic-topology
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add a comment |
$begingroup$
Show that a space obtained from $S^2$ by attaching n 2-cells along any collection of n circles in $S^2$ is homotopy equivalent to the wedge of n+1 spheres.
I'm a little confused here. I'm imagining a sphere, and putting 2 dimensional discs inside of it. Attaching one disc along one circle of $S^2$ and then collapsing that disc to a point would appear to me to create a wedge of two spheres.
However, if I attached 2 discs along 2 different circles of $S^2$ and collapsed each to a point, to me it seems I would have created a wedge of four spheres, one sphere for each division created by the two discs inside the sphere.
What is wrong with my thinking here? Thanks!!
algebraic-topology
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1
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Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
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– user3482749
Jan 18 at 15:25
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You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
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– Alex Provost
Jan 18 at 16:07
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Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29
add a comment |
$begingroup$
Show that a space obtained from $S^2$ by attaching n 2-cells along any collection of n circles in $S^2$ is homotopy equivalent to the wedge of n+1 spheres.
I'm a little confused here. I'm imagining a sphere, and putting 2 dimensional discs inside of it. Attaching one disc along one circle of $S^2$ and then collapsing that disc to a point would appear to me to create a wedge of two spheres.
However, if I attached 2 discs along 2 different circles of $S^2$ and collapsed each to a point, to me it seems I would have created a wedge of four spheres, one sphere for each division created by the two discs inside the sphere.
What is wrong with my thinking here? Thanks!!
algebraic-topology
$endgroup$
Show that a space obtained from $S^2$ by attaching n 2-cells along any collection of n circles in $S^2$ is homotopy equivalent to the wedge of n+1 spheres.
I'm a little confused here. I'm imagining a sphere, and putting 2 dimensional discs inside of it. Attaching one disc along one circle of $S^2$ and then collapsing that disc to a point would appear to me to create a wedge of two spheres.
However, if I attached 2 discs along 2 different circles of $S^2$ and collapsed each to a point, to me it seems I would have created a wedge of four spheres, one sphere for each division created by the two discs inside the sphere.
What is wrong with my thinking here? Thanks!!
algebraic-topology
algebraic-topology
asked Jan 18 at 15:21
user624065
1
$begingroup$
Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
$endgroup$
– user3482749
Jan 18 at 15:25
$begingroup$
You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
$endgroup$
– Alex Provost
Jan 18 at 16:07
$begingroup$
Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29
add a comment |
1
$begingroup$
Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
$endgroup$
– user3482749
Jan 18 at 15:25
$begingroup$
You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
$endgroup$
– Alex Provost
Jan 18 at 16:07
$begingroup$
Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29
1
1
$begingroup$
Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
$endgroup$
– user3482749
Jan 18 at 15:25
$begingroup$
Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
$endgroup$
– user3482749
Jan 18 at 15:25
$begingroup$
You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
$endgroup$
– Alex Provost
Jan 18 at 16:07
$begingroup$
You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
$endgroup$
– Alex Provost
Jan 18 at 16:07
$begingroup$
Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29
$begingroup$
Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29
add a comment |
1 Answer
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If you form $Z= S^2 cup_f e^2$ then the attaching map- $f: S^1 to S^2$ is null homotopic, as $S^2$ is simply connected. So $Z simeq S^2 vee S^2 $. The general result you need is that if $Z= B cup_f X$ where $f: A to B$, and the inclusion $i : A to X$ is a closed cofibration, then the homotopy type of $Z$ depends only on the homotopy class of $f$.
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
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$begingroup$
If you form $Z= S^2 cup_f e^2$ then the attaching map- $f: S^1 to S^2$ is null homotopic, as $S^2$ is simply connected. So $Z simeq S^2 vee S^2 $. The general result you need is that if $Z= B cup_f X$ where $f: A to B$, and the inclusion $i : A to X$ is a closed cofibration, then the homotopy type of $Z$ depends only on the homotopy class of $f$.
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
If you form $Z= S^2 cup_f e^2$ then the attaching map- $f: S^1 to S^2$ is null homotopic, as $S^2$ is simply connected. So $Z simeq S^2 vee S^2 $. The general result you need is that if $Z= B cup_f X$ where $f: A to B$, and the inclusion $i : A to X$ is a closed cofibration, then the homotopy type of $Z$ depends only on the homotopy class of $f$.
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
If you form $Z= S^2 cup_f e^2$ then the attaching map- $f: S^1 to S^2$ is null homotopic, as $S^2$ is simply connected. So $Z simeq S^2 vee S^2 $. The general result you need is that if $Z= B cup_f X$ where $f: A to B$, and the inclusion $i : A to X$ is a closed cofibration, then the homotopy type of $Z$ depends only on the homotopy class of $f$.
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
If you form $Z= S^2 cup_f e^2$ then the attaching map- $f: S^1 to S^2$ is null homotopic, as $S^2$ is simply connected. So $Z simeq S^2 vee S^2 $. The general result you need is that if $Z= B cup_f X$ where $f: A to B$, and the inclusion $i : A to X$ is a closed cofibration, then the homotopy type of $Z$ depends only on the homotopy class of $f$.
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
edited Jan 19 at 16:11
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
answered Jan 19 at 15:40
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
add a comment |
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$begingroup$
Consider what happens if the circles are parallel. What, if anything, changes when we move the circles to not be parallel?
$endgroup$
– user3482749
Jan 18 at 15:25
$begingroup$
You can use an inductive argument: attaching $n$ 2-cells to $S^2$ is equivalent to attaching one 2-cell to the space obtained by attaching $(n-1)$ 2-cells to $S^2$. Then if you agree that attaching one 2-cell is equivalent to wedging with a new sphere, the claim follows.
$endgroup$
– Alex Provost
Jan 18 at 16:07
$begingroup$
Specifically, I don't understand why you are imagining four spheres instead of three. This is what I have in mind.
$endgroup$
– Alex Provost
Jan 18 at 16:29