Equal winding number implies two paths are path homotopic? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If winding number of two curves are same then are they homotopic?Why is the winding number homotopy invariant?Are the maps 'in between' two homotopic paths paths themselves?Homotopic paths implies equal winding numbersShowing that identity and g are not homotopic (without Homology)Are all paths with the same endpoints homotopic in a simply connected region?Is the projection of two homotopic maps path homotopic?Using gluing/pasting lemma to show two paths are homotopic?Showing two paths in $mathbbR^2 setminus (0,0)$ are not homotopic.Lifts of relative homotopic paths to the same base point $Rightarrow$ The lifts homotopic and agree on the end point.Lifting homotopic paths
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Equal winding number implies two paths are path homotopic?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)If winding number of two curves are same then are they homotopic?Why is the winding number homotopy invariant?Are the maps 'in between' two homotopic paths paths themselves?Homotopic paths implies equal winding numbersShowing that identity and g are not homotopic (without Homology)Are all paths with the same endpoints homotopic in a simply connected region?Is the projection of two homotopic maps path homotopic?Using gluing/pasting lemma to show two paths are homotopic?Showing two paths in $mathbbR^2 setminus (0,0)$ are not homotopic.Lifts of relative homotopic paths to the same base point $Rightarrow$ The lifts homotopic and agree on the end point.Lifting homotopic paths
$begingroup$
Let $alpha,beta:[0,1]rightarrowmathbbCsetminusp$ be two (continuous) paths (not necessarily closed) with same endpoints ($alpha(0)=beta(0)$, $alpha(1)=beta(1)$), we know that if $alphasimeq_mathrmpbeta$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ by using lifting lemma. However, is the converse also true? If so, how to construct a homotopy?
algebraic-topology
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
$endgroup$
add a comment |
$begingroup$
Let $alpha,beta:[0,1]rightarrowmathbbCsetminusp$ be two (continuous) paths (not necessarily closed) with same endpoints ($alpha(0)=beta(0)$, $alpha(1)=beta(1)$), we know that if $alphasimeq_mathrmpbeta$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ by using lifting lemma. However, is the converse also true? If so, how to construct a homotopy?
algebraic-topology
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
$endgroup$
$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
1
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31
add a comment |
$begingroup$
Let $alpha,beta:[0,1]rightarrowmathbbCsetminusp$ be two (continuous) paths (not necessarily closed) with same endpoints ($alpha(0)=beta(0)$, $alpha(1)=beta(1)$), we know that if $alphasimeq_mathrmpbeta$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ by using lifting lemma. However, is the converse also true? If so, how to construct a homotopy?
algebraic-topology
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
$endgroup$
Let $alpha,beta:[0,1]rightarrowmathbbCsetminusp$ be two (continuous) paths (not necessarily closed) with same endpoints ($alpha(0)=beta(0)$, $alpha(1)=beta(1)$), we know that if $alphasimeq_mathrmpbeta$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ by using lifting lemma. However, is the converse also true? If so, how to construct a homotopy?
algebraic-topology
algebraic-topology
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
edited Jan 30 '14 at 23:30
Kaa1el
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
asked Jan 30 '14 at 23:17
Kaa1elKaa1el
1,345718
1,345718
$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
1
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31
add a comment |
$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
1
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31
$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
1
1
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31
add a comment |
1 Answer
1
active
oldest
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$begingroup$
OK, I found a possible proof:
Let $overlinealpha$, $overlinebeta$ be the lifting of $alpha$, $beta$ such that $overlinealpha(0)=overlinebeta(0)$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ implies there is a path homotopy between $overlinealpha$, $overlinebeta$, (both coordinate use path homotopy since they are paths in $mathbbR_>0timesmathbbR$). Then composite with function $(r,theta)mapsto rexp(itheta)$ would be a path homotopy between $alpha$ and $beta$.
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
$endgroup$
add a comment |
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$begingroup$
OK, I found a possible proof:
Let $overlinealpha$, $overlinebeta$ be the lifting of $alpha$, $beta$ such that $overlinealpha(0)=overlinebeta(0)$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ implies there is a path homotopy between $overlinealpha$, $overlinebeta$, (both coordinate use path homotopy since they are paths in $mathbbR_>0timesmathbbR$). Then composite with function $(r,theta)mapsto rexp(itheta)$ would be a path homotopy between $alpha$ and $beta$.
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
$endgroup$
add a comment |
$begingroup$
OK, I found a possible proof:
Let $overlinealpha$, $overlinebeta$ be the lifting of $alpha$, $beta$ such that $overlinealpha(0)=overlinebeta(0)$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ implies there is a path homotopy between $overlinealpha$, $overlinebeta$, (both coordinate use path homotopy since they are paths in $mathbbR_>0timesmathbbR$). Then composite with function $(r,theta)mapsto rexp(itheta)$ would be a path homotopy between $alpha$ and $beta$.
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
$endgroup$
add a comment |
$begingroup$
OK, I found a possible proof:
Let $overlinealpha$, $overlinebeta$ be the lifting of $alpha$, $beta$ such that $overlinealpha(0)=overlinebeta(0)$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ implies there is a path homotopy between $overlinealpha$, $overlinebeta$, (both coordinate use path homotopy since they are paths in $mathbbR_>0timesmathbbR$). Then composite with function $(r,theta)mapsto rexp(itheta)$ would be a path homotopy between $alpha$ and $beta$.
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
$endgroup$
OK, I found a possible proof:
Let $overlinealpha$, $overlinebeta$ be the lifting of $alpha$, $beta$ such that $overlinealpha(0)=overlinebeta(0)$, then $mathrmW(alpha,p)=mathrmW(beta,p)$ implies there is a path homotopy between $overlinealpha$, $overlinebeta$, (both coordinate use path homotopy since they are paths in $mathbbR_>0timesmathbbR$). Then composite with function $(r,theta)mapsto rexp(itheta)$ would be a path homotopy between $alpha$ and $beta$.
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
answered Jan 31 '14 at 4:46
Kaa1elKaa1el
1,345718
1,345718
add a comment |
add a comment |
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$begingroup$
First, the winding number is not defined for non-closed paths, you need a loop for this. Second, you have to learn about the fundamental group of the circle and how to prove that it is isomorphic to $mathbb Z$.
$endgroup$
– Moishe Kohan
Jan 30 '14 at 23:21
$begingroup$
No, the winding number can be defined for any path by lifting with respect to polar coordinate function; and for closed loop, it is indeed an integer.
$endgroup$
– Kaa1el
Jan 30 '14 at 23:27
1
$begingroup$
Then you would surely need the added condition that $alpha(0)=beta(0)$ and $alpha(1)=beta(1)$ and then talk about homotopies relative to end points.
$endgroup$
– Dan Rust
Jan 30 '14 at 23:29
$begingroup$
yes, you're correct, my mistake:)
$endgroup$
– Kaa1el
Jan 30 '14 at 23:31