A Lemma from Milnor's Topology from the Differentiable Viewpoint
$begingroup$
I have a question about an proof in Milnor's TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT (see pages 65-66): The aim is to show the classifying theorem that any smooth, connected $1$-dimensional manifold is difeomorphic
either to the circle $S^1$ or to some interval of real numbers.
In order to show it the author uses following lemma:
Here the proof with red tagged argument which isn't clear to me:
We take the graph $Gamma subset I times J$ consisting of all $(s,t)$ with $f(s)= g(t)$ ($f, g$ parametrisations; for the notation: see above)
My questions are following:
Why $Gamma$ is closed in $I times J$? (my considerations: I guess that because for small enough open $U subset M$ the diagonal of $U times U$ is closed (since $M$ Hausdorff) and $Gamma$ is just it's preimage. Is the argument ok?)
Why the lines of $Gamma$ cannot end in the interior $mathring{I} times mathring{J}$? Why does the fact that $g^{-1} circ f$ is a local isomorphism exclude it?
differential-topology
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
$endgroup$
add a comment |
$begingroup$
I have a question about an proof in Milnor's TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT (see pages 65-66): The aim is to show the classifying theorem that any smooth, connected $1$-dimensional manifold is difeomorphic
either to the circle $S^1$ or to some interval of real numbers.
In order to show it the author uses following lemma:
Here the proof with red tagged argument which isn't clear to me:
We take the graph $Gamma subset I times J$ consisting of all $(s,t)$ with $f(s)= g(t)$ ($f, g$ parametrisations; for the notation: see above)
My questions are following:
Why $Gamma$ is closed in $I times J$? (my considerations: I guess that because for small enough open $U subset M$ the diagonal of $U times U$ is closed (since $M$ Hausdorff) and $Gamma$ is just it's preimage. Is the argument ok?)
Why the lines of $Gamma$ cannot end in the interior $mathring{I} times mathring{J}$? Why does the fact that $g^{-1} circ f$ is a local isomorphism exclude it?
differential-topology
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
$endgroup$
add a comment |
$begingroup$
I have a question about an proof in Milnor's TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT (see pages 65-66): The aim is to show the classifying theorem that any smooth, connected $1$-dimensional manifold is difeomorphic
either to the circle $S^1$ or to some interval of real numbers.
In order to show it the author uses following lemma:
Here the proof with red tagged argument which isn't clear to me:
We take the graph $Gamma subset I times J$ consisting of all $(s,t)$ with $f(s)= g(t)$ ($f, g$ parametrisations; for the notation: see above)
My questions are following:
Why $Gamma$ is closed in $I times J$? (my considerations: I guess that because for small enough open $U subset M$ the diagonal of $U times U$ is closed (since $M$ Hausdorff) and $Gamma$ is just it's preimage. Is the argument ok?)
Why the lines of $Gamma$ cannot end in the interior $mathring{I} times mathring{J}$? Why does the fact that $g^{-1} circ f$ is a local isomorphism exclude it?
differential-topology
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
$endgroup$
I have a question about an proof in Milnor's TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT (see pages 65-66): The aim is to show the classifying theorem that any smooth, connected $1$-dimensional manifold is difeomorphic
either to the circle $S^1$ or to some interval of real numbers.
In order to show it the author uses following lemma:
Here the proof with red tagged argument which isn't clear to me:
We take the graph $Gamma subset I times J$ consisting of all $(s,t)$ with $f(s)= g(t)$ ($f, g$ parametrisations; for the notation: see above)
My questions are following:
Why $Gamma$ is closed in $I times J$? (my considerations: I guess that because for small enough open $U subset M$ the diagonal of $U times U$ is closed (since $M$ Hausdorff) and $Gamma$ is just it's preimage. Is the argument ok?)
Why the lines of $Gamma$ cannot end in the interior $mathring{I} times mathring{J}$? Why does the fact that $g^{-1} circ f$ is a local isomorphism exclude it?
differential-topology
differential-topology
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
asked Jan 18 at 18:12
KarlPeterKarlPeter
6101315
6101315
add a comment |
add a comment |
1 Answer
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$Gamma$ is closed since $ftimes g$ is continuous, $Gamma=(ftimes g)^{-1}(D)$ where $D={(x,x)}$ is the diagonal.
Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(ftimes g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
add a comment |
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1 Answer
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$begingroup$
$Gamma$ is closed since $ftimes g$ is continuous, $Gamma=(ftimes g)^{-1}(D)$ where $D={(x,x)}$ is the diagonal.
Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(ftimes g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
add a comment |
$begingroup$
$Gamma$ is closed since $ftimes g$ is continuous, $Gamma=(ftimes g)^{-1}(D)$ where $D={(x,x)}$ is the diagonal.
Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(ftimes g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
add a comment |
$begingroup$
$Gamma$ is closed since $ftimes g$ is continuous, $Gamma=(ftimes g)^{-1}(D)$ where $D={(x,x)}$ is the diagonal.
Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(ftimes g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$Gamma$ is closed since $ftimes g$ is continuous, $Gamma=(ftimes g)^{-1}(D)$ where $D={(x,x)}$ is the diagonal.
Suppose that a segment $c$ end to the interior, there exists a family $(u_n,v_n)$ with $f(u_n)=f(v_n)$ and $lim_n(u_n,v_n)=(u,v)$ is the end of $c$, since $f,g$ are continuous, $(ftimes g)(u,v)=lim_n(f(u_n),g(v_n))$ implies that $f(u)=f(v)$. Thus we can assume that the segment is closed. We have $g(^{-1}circ f)(u)=v$ and $u$ is in the interior of $I$, Write $u_t=u+t$ where $u+t$ is in $I$, write $v_t=(g^{-1}circ f)(u_t)$, we have $(u_t,v_t)$ extends $c$.
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
edited Jan 18 at 18:35
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
answered Jan 18 at 18:20
Tsemo AristideTsemo Aristide
58.1k11445
58.1k11445
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
add a comment |
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
$begingroup$
Thank you for the answer. Yes, it becomes clearer. One point seems unclear: Why is $(g^{-1}circ f)(u_t)$ is well defined when you shift $u$ to $u_t = u +t$? Or in other words why $f(u_t) in im(g)$? Obvioulsly it suffice to show that $f(u) in overset{circ}{im(g)}$. Does for a parametrisation always hold following statement: $t in mathring{J} Leftrightarrow g(t) in mathring{im(g)}$
$endgroup$
– KarlPeter
Jan 18 at 20:25
add a comment |
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