Show that a proper locally invertible map is surjective [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This question already has an answer here:
Application of inverse function theorem for several variable functions
1 answer
Let $f:R^2to R^2$ be a continuously differentiable function such that $Df(x)$ is invertible for all $xin R^2$ and $f^-1(K)$ is compact for every compact set $K$. Show that $f$ is surjective.
The first condition gives by the inverse function theorem that $f$ is invertible in a neighborhood of any point. But how exactly to use properness?
calculus real-analysis general-topology multivariable-calculus derivatives
asked Aug 12 at 1:36
user531232
24413
marked as duplicate by Xander Henderson, Martin Argerami real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Aug 15 at 4:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Application of inverse function theorem for several variable functions
1 answer
Let $f:R^2to R^2$ be a continuously differentiable function such that $Df(x)$ is invertible for all $xin R^2$ and $f^-1(K)$ is compact for every compact set $K$. Show that $f$ is surjective.
The first condition gives by the inverse function theorem that $f$ is invertible in a neighborhood of any point. But how exactly to use properness?
calculus real-analysis general-topology multivariable-calculus derivatives
asked Aug 12 at 1:36
user531232
24413
marked as duplicate by Xander Henderson, Martin Argerami real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Aug 15 at 4:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Application of inverse function theorem for several variable functions
1 answer
Let $f:R^2to R^2$ be a continuously differentiable function such that $Df(x)$ is invertible for all $xin R^2$ and $f^-1(K)$ is compact for every compact set $K$. Show that $f$ is surjective.
The first condition gives by the inverse function theorem that $f$ is invertible in a neighborhood of any point. But how exactly to use properness?
calculus real-analysis general-topology multivariable-calculus derivatives
asked Aug 12 at 1:36
user531232
24413
This question already has an answer here:
Application of inverse function theorem for several variable functions
1 answer
Let $f:R^2to R^2$ be a continuously differentiable function such that $Df(x)$ is invertible for all $xin R^2$ and $f^-1(K)$ is compact for every compact set $K$. Show that $f$ is surjective.
The first condition gives by the inverse function theorem that $f$ is invertible in a neighborhood of any point. But how exactly to use properness?
This question already has an answer here:
Application of inverse function theorem for several variable functions
1 answer
calculus real-analysis general-topology multivariable-calculus derivatives
asked Aug 12 at 1:36
user531232
24413
asked Aug 12 at 1:36
user531232
24413
asked Aug 12 at 1:36
user531232
24413
asked Aug 12 at 1:36
user531232
24413
24413
marked as duplicate by Xander Henderson, Martin Argerami real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Aug 15 at 4:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, Martin Argerami real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
Aug 15 at 4:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
The fact that $Df$ is invertible implies that the image of $f$ is open. It is enough to show that the image of $f$ is closed.
Suppose that $y=lim_nf(x_n)$, let $B=B(y,1)$, the closed ball of radius $1$, there exists $N$ such that $n>N$ implies that $f(x_n)in B$, $f^-1(B)$ is compact and contains $x_n, n>N$, since $f^-1(B)$ compact, you can extract a converging subsequence $x_n_p$, write $x=lim_nx_n_p$ since $f$ is continuous, $f(x)=lim_nf(x_n_p)=y$.
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The fact that $Df$ is invertible implies that the image of $f$ is open. It is enough to show that the image of $f$ is closed.
Suppose that $y=lim_nf(x_n)$, let $B=B(y,1)$, the closed ball of radius $1$, there exists $N$ such that $n>N$ implies that $f(x_n)in B$, $f^-1(B)$ is compact and contains $x_n, n>N$, since $f^-1(B)$ compact, you can extract a converging subsequence $x_n_p$, write $x=lim_nx_n_p$ since $f$ is continuous, $f(x)=lim_nf(x_n_p)=y$.
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
add a comment |Â
up vote
1
down vote
The fact that $Df$ is invertible implies that the image of $f$ is open. It is enough to show that the image of $f$ is closed.
Suppose that $y=lim_nf(x_n)$, let $B=B(y,1)$, the closed ball of radius $1$, there exists $N$ such that $n>N$ implies that $f(x_n)in B$, $f^-1(B)$ is compact and contains $x_n, n>N$, since $f^-1(B)$ compact, you can extract a converging subsequence $x_n_p$, write $x=lim_nx_n_p$ since $f$ is continuous, $f(x)=lim_nf(x_n_p)=y$.
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The fact that $Df$ is invertible implies that the image of $f$ is open. It is enough to show that the image of $f$ is closed.
Suppose that $y=lim_nf(x_n)$, let $B=B(y,1)$, the closed ball of radius $1$, there exists $N$ such that $n>N$ implies that $f(x_n)in B$, $f^-1(B)$ is compact and contains $x_n, n>N$, since $f^-1(B)$ compact, you can extract a converging subsequence $x_n_p$, write $x=lim_nx_n_p$ since $f$ is continuous, $f(x)=lim_nf(x_n_p)=y$.
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
The fact that $Df$ is invertible implies that the image of $f$ is open. It is enough to show that the image of $f$ is closed.
Suppose that $y=lim_nf(x_n)$, let $B=B(y,1)$, the closed ball of radius $1$, there exists $N$ such that $n>N$ implies that $f(x_n)in B$, $f^-1(B)$ is compact and contains $x_n, n>N$, since $f^-1(B)$ compact, you can extract a converging subsequence $x_n_p$, write $x=lim_nx_n_p$ since $f$ is continuous, $f(x)=lim_nf(x_n_p)=y$.
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
answered Aug 12 at 1:49
Tsemo Aristide
51.6k11243
51.6k11243
add a comment |Â
add a comment |Â
