Let $A: X rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^-1$ is also continous? The 2019 Stack Overflow Developer Survey Results Are InIs it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?Existence of a continuous function which does not achieve a maximum.Unbounded continuous function on non-compact metric spaceTopology of bijective functions between Banach spacesContinuity criterions for linear maps between Banach spaces?When linear map is bounded and bijective then $X$ is reflexive.proving that the quotient linear map of a continuous linear map is also continuous (normed spaces)Can natural quotient map between Banach spaces be closed?The image of an bounded and linear operator between two banach spaces is closed if the image is finite codimension?Does any continuous map between two Banach spaces over a nonarchimedian field is a closed map?bijective continuous map from X to Y and Y to XFind an homeomorphism between two topological spacesMap which is bijective and continuous, but not open
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Let $A: X rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^-1$ is also continous?
The 2019 Stack Overflow Developer Survey Results Are InIs it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?Existence of a continuous function which does not achieve a maximum.Unbounded continuous function on non-compact metric spaceTopology of bijective functions between Banach spacesContinuity criterions for linear maps between Banach spaces?When linear map is bounded and bijective then $X$ is reflexive.proving that the quotient linear map of a continuous linear map is also continuous (normed spaces)Can natural quotient map between Banach spaces be closed?The image of an bounded and linear operator between two banach spaces is closed if the image is finite codimension?Does any continuous map between two Banach spaces over a nonarchimedian field is a closed map?bijective continuous map from X to Y and Y to XFind an homeomorphism between two topological spacesMap which is bijective and continuous, but not open
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
add a comment |
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
$begingroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
$endgroup$
We know if A is a map continuous, bijective and linear then the answer is yes, $A^-1$ is continuous. But, if $A$ is no linear then $A^-1$ is also continuous ?
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
general-topology functional-analysis operator-theory banach-spaces nonlinear-analysis
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
edited Mar 30 at 22:12
Paul Frost
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
edited Mar 30 at 22:12
Paul Frost
12.6k31035
edited Mar 30 at 22:12
Paul Frost
12.6k31035
edited Mar 30 at 22:12
Paul Frost
12.6k31035
12.6k31035
asked Mar 30 at 21:59
jugartecjugartec
111
asked Mar 30 at 21:59
jugartecjugartec
111
asked Mar 30 at 21:59
jugartecjugartec
111
111
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
2
2
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25
add a comment |
1 Answer
1
active
oldest
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Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $Ssubset X$ is noncompact. Hence, there is a continuous surjective function $h: Sto (0,1]$. For $xin X-0$ set $barx:= x/||x||$.
Define the self-map
$$
f: Xto X, ~~f(x)=h(barx) x, ~~hboxif~ xne 0; ~~f(0)=0.
$$
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_nin S$ such that $lim_ntoinfty h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $Xto Y$ is a homeomorphism by Brouwer's invariance of domain theorem.
Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here.
Every noncompact metric space admits a continuous surjective function to $mathbb R$; this was again discussed many times, see for instance here and here. Composing with the function $tmapsto e^-t^2$ gives a surjective continuous function to $(0,1]$.
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
edited Apr 1 at 17:48
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
answered Mar 31 at 23:40
Moishe KohanMoishe Kohan
48.5k344110
48.5k344110
add a comment |
add a comment |
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$begingroup$
Such examples must be infinite-dimensional (due to invariance of domain).
$endgroup$
– Henno Brandsma
Mar 30 at 23:25