Homotopy type of $mathbb{R}^3 setminus { mathrm{wedge hspace{3pt} sum hspace{3pt} of hspace{3pt} 2...
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Is there something "nice" that the space $mathbb{R}^3 setminus { mathrm{wedge hspace{3pt} sum hspace{3pt} of hspace{3pt} 2 hspace{3pt} circles} }$ is homotopy equivalent to? I know that $mathbb{R}^3 setminus {mathrm{circle}} simeq S^{2} vee S^{1}$, and that $mathbb{R}^3 setminus { mathrm{2 hspace{3pt} linked hspace{3pt} circles}} simeq S^{2} vee T^{2}$. However, I can't seem to apply either of the "visual methods" used to prove these homotopy equivalences to my case.
My wild guess is that it might be homotopic to $M_{2}$, the orientable closed surface of genus $2$, but I'm not sure if this is true at all.
algebraic-topology homotopy-theory
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
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show 2 more comments
$begingroup$
Is there something "nice" that the space $mathbb{R}^3 setminus { mathrm{wedge hspace{3pt} sum hspace{3pt} of hspace{3pt} 2 hspace{3pt} circles} }$ is homotopy equivalent to? I know that $mathbb{R}^3 setminus {mathrm{circle}} simeq S^{2} vee S^{1}$, and that $mathbb{R}^3 setminus { mathrm{2 hspace{3pt} linked hspace{3pt} circles}} simeq S^{2} vee T^{2}$. However, I can't seem to apply either of the "visual methods" used to prove these homotopy equivalences to my case.
My wild guess is that it might be homotopic to $M_{2}$, the orientable closed surface of genus $2$, but I'm not sure if this is true at all.
algebraic-topology homotopy-theory
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
$endgroup$
$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
3
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
1
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18
|
show 2 more comments
$begingroup$
Is there something "nice" that the space $mathbb{R}^3 setminus { mathrm{wedge hspace{3pt} sum hspace{3pt} of hspace{3pt} 2 hspace{3pt} circles} }$ is homotopy equivalent to? I know that $mathbb{R}^3 setminus {mathrm{circle}} simeq S^{2} vee S^{1}$, and that $mathbb{R}^3 setminus { mathrm{2 hspace{3pt} linked hspace{3pt} circles}} simeq S^{2} vee T^{2}$. However, I can't seem to apply either of the "visual methods" used to prove these homotopy equivalences to my case.
My wild guess is that it might be homotopic to $M_{2}$, the orientable closed surface of genus $2$, but I'm not sure if this is true at all.
algebraic-topology homotopy-theory
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
$endgroup$
Is there something "nice" that the space $mathbb{R}^3 setminus { mathrm{wedge hspace{3pt} sum hspace{3pt} of hspace{3pt} 2 hspace{3pt} circles} }$ is homotopy equivalent to? I know that $mathbb{R}^3 setminus {mathrm{circle}} simeq S^{2} vee S^{1}$, and that $mathbb{R}^3 setminus { mathrm{2 hspace{3pt} linked hspace{3pt} circles}} simeq S^{2} vee T^{2}$. However, I can't seem to apply either of the "visual methods" used to prove these homotopy equivalences to my case.
My wild guess is that it might be homotopic to $M_{2}$, the orientable closed surface of genus $2$, but I'm not sure if this is true at all.
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
asked Dec 27 '18 at 21:32
Matija SreckovicMatija Sreckovic
928517
928517
$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
3
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
1
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18
|
show 2 more comments
$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
3
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
1
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18
$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
3
3
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
1
1
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18
|
show 2 more comments
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$begingroup$
Hmm, I'd imagine that the proof of $mathbb R^3setminus S^1simeq S^2wedge S^1$ would generalize ...
$endgroup$
– Henning Makholm
Dec 27 '18 at 21:42
$begingroup$
Get a big sphere around them and draw $2$ diameters?
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:45
$begingroup$
Obviously not diamaters, but... I'll check it out.
$endgroup$
– Matija Sreckovic
Dec 27 '18 at 21:46
3
$begingroup$
In general, if you remove a wedge of $n$ circles from $mathbb{R}^3$, then what is remaining is homotopy equivalent to a wedge of $S^2$ and $n$ circles.
$endgroup$
– Alvin Jin
Dec 27 '18 at 22:01
1
$begingroup$
Depends wildly on the way the circles are embedded.
$endgroup$
– user98602
Dec 27 '18 at 23:18