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How to prove $pin S^1$ s.t., $d_p(f|_S^1)=0$
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Proposition: Let $D:=(x,y)$ and $S^1$ is boundary of $D$. Then, for all differentiable function $f:Dto S^1$, there exist $pin S^1$ s.t., $d_p(f|_S^1)=0$
$f|_S^1(S^1)$ is compact, so I want to prove above proposition by similar way in case of $f$ is $mathbbR$-valued. But I don't know how to prove it.
real-analysis manifolds differential-topology
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Proposition: Let $D:=(x,y)$ and $S^1$ is boundary of $D$. Then, for all differentiable function $f:Dto S^1$, there exist $pin S^1$ s.t., $d_p(f|_S^1)=0$
$f|_S^1(S^1)$ is compact, so I want to prove above proposition by similar way in case of $f$ is $mathbbR$-valued. But I don't know how to prove it.
real-analysis manifolds differential-topology
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Proposition: Let $D:=(x,y)$ and $S^1$ is boundary of $D$. Then, for all differentiable function $f:Dto S^1$, there exist $pin S^1$ s.t., $d_p(f|_S^1)=0$
$f|_S^1(S^1)$ is compact, so I want to prove above proposition by similar way in case of $f$ is $mathbbR$-valued. But I don't know how to prove it.
real-analysis manifolds differential-topology
Proposition: Let $D:=(x,y)$ and $S^1$ is boundary of $D$. Then, for all differentiable function $f:Dto S^1$, there exist $pin S^1$ s.t., $d_p(f|_S^1)=0$
$f|_S^1(S^1)$ is compact, so I want to prove above proposition by similar way in case of $f$ is $mathbbR$-valued. But I don't know how to prove it.
real-analysis manifolds differential-topology
asked Jul 15 at 8:17
user577275
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1 Answer
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This follows immediately from the case of $mathbbR$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:Dto mathbbR$ such that $f=pcirc g$ where $p:mathbbRto S^1$ is the universal covering map $p(t)=(cos t,sin t)$.
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This follows immediately from the case of $mathbbR$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:Dto mathbbR$ such that $f=pcirc g$ where $p:mathbbRto S^1$ is the universal covering map $p(t)=(cos t,sin t)$.
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
add a comment |Â
up vote
1
down vote
accepted
This follows immediately from the case of $mathbbR$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:Dto mathbbR$ such that $f=pcirc g$ where $p:mathbbRto S^1$ is the universal covering map $p(t)=(cos t,sin t)$.
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This follows immediately from the case of $mathbbR$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:Dto mathbbR$ such that $f=pcirc g$ where $p:mathbbRto S^1$ is the universal covering map $p(t)=(cos t,sin t)$.
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
This follows immediately from the case of $mathbbR$-valued functions, since $D$ is simply connected so $f$ lifts to the universal cover. That is, there is a differentiable map $g:Dto mathbbR$ such that $f=pcirc g$ where $p:mathbbRto S^1$ is the universal covering map $p(t)=(cos t,sin t)$.
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
answered Jul 15 at 8:24
Eric Wofsey
163k12189300
163k12189300
add a comment |Â
add a comment |Â
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