How is going around the circle once in each direction homotopic to a point?
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Two paths are homotopic if one can be continuously deformed to the other, right? So I've been told that the fundamental group of the circle is isomorphic to the integers, since you can't deform e.g., the path that goes around once clockwise to that which goes around twice clockwise, since the disc gets in the way, so it's generated by the the clockwise and counterclockwise paths. But that seems to imply that if you go around once clockwise and once counterclockwise, you get a path that can be continuously deformed to the base point? How?
algebraic-topology fundamental-groups
asked 4 hours ago
user361424user361424
1,293414
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add a comment |
$begingroup$
Two paths are homotopic if one can be continuously deformed to the other, right? So I've been told that the fundamental group of the circle is isomorphic to the integers, since you can't deform e.g., the path that goes around once clockwise to that which goes around twice clockwise, since the disc gets in the way, so it's generated by the the clockwise and counterclockwise paths. But that seems to imply that if you go around once clockwise and once counterclockwise, you get a path that can be continuously deformed to the base point? How?
algebraic-topology fundamental-groups
asked 4 hours ago
user361424user361424
1,293414
$endgroup$
add a comment |
$begingroup$
Two paths are homotopic if one can be continuously deformed to the other, right? So I've been told that the fundamental group of the circle is isomorphic to the integers, since you can't deform e.g., the path that goes around once clockwise to that which goes around twice clockwise, since the disc gets in the way, so it's generated by the the clockwise and counterclockwise paths. But that seems to imply that if you go around once clockwise and once counterclockwise, you get a path that can be continuously deformed to the base point? How?
algebraic-topology fundamental-groups
asked 4 hours ago
user361424user361424
1,293414
$endgroup$
Two paths are homotopic if one can be continuously deformed to the other, right? So I've been told that the fundamental group of the circle is isomorphic to the integers, since you can't deform e.g., the path that goes around once clockwise to that which goes around twice clockwise, since the disc gets in the way, so it's generated by the the clockwise and counterclockwise paths. But that seems to imply that if you go around once clockwise and once counterclockwise, you get a path that can be continuously deformed to the base point? How?
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked 4 hours ago
user361424user361424
1,293414
asked 4 hours ago
user361424user361424
1,293414
asked 4 hours ago
user361424user361424
1,293414
asked 4 hours ago
user361424user361424
1,293414
asked 4 hours ago
user361424user361424
1,293414
1,293414
add a comment |
add a comment |
2 Answers
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The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.
Concretely, if
$$f(t) = begin{cases} e^{4pi i t}&tleq 1/2\ e^{-4pi i t}&1/2leq t leq 1 end{cases}$$
then there is a nulhomotopy
$$h(t,s) = begin{cases} f(st) & t leq 1/2\ f(t(1-s)) & 1/2leq t leq 1end{cases}$$
answered 3 hours ago
BenBen
3,523616
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1
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Presumably start and beginning should be beginning and end?
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– timtfj
2 hours ago
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@timtfj Fair enough :)
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– Ben
1 hour ago
add a comment |
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As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.
If you wrap a rubber band once around a post and let go, it will stay around the post:
On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):
answered 57 mins ago
cspruncsprun
56616
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.
Concretely, if
$$f(t) = begin{cases} e^{4pi i t}&tleq 1/2\ e^{-4pi i t}&1/2leq t leq 1 end{cases}$$
then there is a nulhomotopy
$$h(t,s) = begin{cases} f(st) & t leq 1/2\ f(t(1-s)) & 1/2leq t leq 1end{cases}$$
answered 3 hours ago
BenBen
3,523616
$endgroup$
1
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
add a comment |
$begingroup$
The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.
Concretely, if
$$f(t) = begin{cases} e^{4pi i t}&tleq 1/2\ e^{-4pi i t}&1/2leq t leq 1 end{cases}$$
then there is a nulhomotopy
$$h(t,s) = begin{cases} f(st) & t leq 1/2\ f(t(1-s)) & 1/2leq t leq 1end{cases}$$
answered 3 hours ago
BenBen
3,523616
$endgroup$
1
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
add a comment |
$begingroup$
The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.
Concretely, if
$$f(t) = begin{cases} e^{4pi i t}&tleq 1/2\ e^{-4pi i t}&1/2leq t leq 1 end{cases}$$
then there is a nulhomotopy
$$h(t,s) = begin{cases} f(st) & t leq 1/2\ f(t(1-s)) & 1/2leq t leq 1end{cases}$$
answered 3 hours ago
BenBen
3,523616
$endgroup$
The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.
Concretely, if
$$f(t) = begin{cases} e^{4pi i t}&tleq 1/2\ e^{-4pi i t}&1/2leq t leq 1 end{cases}$$
then there is a nulhomotopy
$$h(t,s) = begin{cases} f(st) & t leq 1/2\ f(t(1-s)) & 1/2leq t leq 1end{cases}$$
answered 3 hours ago
BenBen
3,523616
edited 1 hour ago
answered 3 hours ago
BenBen
3,523616
answered 3 hours ago
BenBen
3,523616
answered 3 hours ago
BenBen
3,523616
3,523616
1
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
add a comment |
1
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
1
1
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
Presumably start and beginning should be beginning and end?
$endgroup$
– timtfj
2 hours ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
$begingroup$
@timtfj Fair enough :)
$endgroup$
– Ben
1 hour ago
add a comment |
$begingroup$
As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.
If you wrap a rubber band once around a post and let go, it will stay around the post:
On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):
answered 57 mins ago
cspruncsprun
56616
$endgroup$
add a comment |
$begingroup$
As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.
If you wrap a rubber band once around a post and let go, it will stay around the post:
On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):
answered 57 mins ago
cspruncsprun
56616
$endgroup$
add a comment |
$begingroup$
As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.
If you wrap a rubber band once around a post and let go, it will stay around the post:
On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):
answered 57 mins ago
cspruncsprun
56616
$endgroup$
As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.
If you wrap a rubber band once around a post and let go, it will stay around the post:
On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):
answered 57 mins ago
cspruncsprun
56616
answered 57 mins ago
cspruncsprun
56616
answered 57 mins ago
cspruncsprun
56616
answered 57 mins ago
cspruncsprun
56616
56616
add a comment |
add a comment |
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