Mixing
$f_n to f$ then $tilde{f_n}to tilde{f}$ where $tilde{f_n}$ and $tilde{f}$ are the lifting of $f_n$ and $f$.
Let $p:(mathbb{R}^2,tilde{u})to (mathbb{R}^2setminus{0},u_0) $ be a covering map and let $f_n:(I^2,u)to (mathbb{R}^2setminus{0},u_0)$ be a sequence of function which converges uniformly to $f$ on every compact subset of $I^2=[0,1]times [0,1]$. Let $tilde{f_n}$ and $tilde{f}$ be unique lifting of $f_n$ and $f$. Then is it true that $tilde{f_n}to tilde{f}$ uniformly on each compact subset of $I^2$?
analysis algebraic-topology
asked Nov 8 '18 at 14:14
XYZABC
304110
add a comment |
Let $p:(mathbb{R}^2,tilde{u})to (mathbb{R}^2setminus{0},u_0) $ be a covering map and let $f_n:(I^2,u)to (mathbb{R}^2setminus{0},u_0)$ be a sequence of function which converges uniformly to $f$ on every compact subset of $I^2=[0,1]times [0,1]$. Let $tilde{f_n}$ and $tilde{f}$ be unique lifting of $f_n$ and $f$. Then is it true that $tilde{f_n}to tilde{f}$ uniformly on each compact subset of $I^2$?
analysis algebraic-topology
asked Nov 8 '18 at 14:14
XYZABC
304110
add a comment |
Let $p:(mathbb{R}^2,tilde{u})to (mathbb{R}^2setminus{0},u_0) $ be a covering map and let $f_n:(I^2,u)to (mathbb{R}^2setminus{0},u_0)$ be a sequence of function which converges uniformly to $f$ on every compact subset of $I^2=[0,1]times [0,1]$. Let $tilde{f_n}$ and $tilde{f}$ be unique lifting of $f_n$ and $f$. Then is it true that $tilde{f_n}to tilde{f}$ uniformly on each compact subset of $I^2$?
analysis algebraic-topology
asked Nov 8 '18 at 14:14
XYZABC
304110
Let $p:(mathbb{R}^2,tilde{u})to (mathbb{R}^2setminus{0},u_0) $ be a covering map and let $f_n:(I^2,u)to (mathbb{R}^2setminus{0},u_0)$ be a sequence of function which converges uniformly to $f$ on every compact subset of $I^2=[0,1]times [0,1]$. Let $tilde{f_n}$ and $tilde{f}$ be unique lifting of $f_n$ and $f$. Then is it true that $tilde{f_n}to tilde{f}$ uniformly on each compact subset of $I^2$?
analysis algebraic-topology
analysis algebraic-topology
asked Nov 8 '18 at 14:14
XYZABC
304110
asked Nov 8 '18 at 14:14
XYZABC
304110
asked Nov 8 '18 at 14:14
XYZABC
304110
asked Nov 8 '18 at 14:14
XYZABC
304110
asked Nov 8 '18 at 14:14
XYZABC
304110
304110
add a comment |
add a comment |
1 Answer
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First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y in f(I^2)$. Label these local inverses by points $x in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $tilde{f}(x) = p_x^{-1} (f(x))$ for all $x in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough and get uniform convergence.
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
add a comment |
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1 Answer
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1 Answer
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votes
First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y in f(I^2)$. Label these local inverses by points $x in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $tilde{f}(x) = p_x^{-1} (f(x))$ for all $x in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough and get uniform convergence.
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
add a comment |
First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y in f(I^2)$. Label these local inverses by points $x in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $tilde{f}(x) = p_x^{-1} (f(x))$ for all $x in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough and get uniform convergence.
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
add a comment |
First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y in f(I^2)$. Label these local inverses by points $x in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $tilde{f}(x) = p_x^{-1} (f(x))$ for all $x in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough and get uniform convergence.
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
First of all, $I^2$ is itself compact, so uniform convergence on compact subsets is just uniform convergence in this case. Otherwise this is true by standard lifting and compactness arguments. By compactness, the image $f(I^2)$ has some positive distance $r>0$ to $0$, so local inverses of $p$ exist in $r$-neighborhoods of every $y in f(I^2)$. Label these local inverses by points $x in I$ such that $p_x^{-1}$ is the local inverse of $p$ in the $r$-neighborhood of $f(x)$ which is encountered by path lifting $f$ along the straight line from $u$ to $x$. (By simple connectedness this is really independent of the path.) This means that $tilde{f}(x) = p_x^{-1} (f(x))$ for all $x in I^2$. We also have that $p_x^{-1} = p_y^{-1}$ if $|x-y|$ is small enough, in the intersection of the $r$-neighborhoods abour $f(x)$ and $f(y)$. And by compactness we can use finitely many of these inverse branches to construct the lifting, which shows that these are uniformly continuous in some neighborhood of $f(I^2)$. Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough and get uniform convergence.
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
answered Nov 20 '18 at 21:33
Lukas Geyer
13.3k1454
13.3k1454
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
add a comment |
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
There are certain things that I have not understood here: (1). Why does the local inverse of $p$ exist in $r$ neighbourhood of $y$. (2). "Then you can use these same inverse branches of $p$ to lift $tilde{f_n}$ for $n$ large enough." This is also unclear to me why do all will go in the same branch after a certain $n$.
– XYZABC
Nov 24 '18 at 6:31
add a comment |
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