Construction of a function from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ where $Gamma =mathbb R^2/_sim$.
$begingroup$
Let $sim$ an equivalence relation, and denote $Gamma=mathbb R^2/_sim$ the quotient space. We denote $pi:mathbb R^2to Gamma $ the natural projection. And now I want to identify $mathbb R^Gamma $ with $mathbb R^{mathbb R^2}$ in a natural way.
Is there a natural way map from a set $A$ and $mathbb R^A$ so that I can define easily the map $mathbb R^{Gamma }to mathbb R^{mathbb R^2}$ or the map $mathbb R^{mathbb R^2}to mathbb R^Gamma $ ?
general-topology functional-analysis
asked Jan 18 at 9:00
idmidm
8,61521446
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add a comment |
$begingroup$
Let $sim$ an equivalence relation, and denote $Gamma=mathbb R^2/_sim$ the quotient space. We denote $pi:mathbb R^2to Gamma $ the natural projection. And now I want to identify $mathbb R^Gamma $ with $mathbb R^{mathbb R^2}$ in a natural way.
Is there a natural way map from a set $A$ and $mathbb R^A$ so that I can define easily the map $mathbb R^{Gamma }to mathbb R^{mathbb R^2}$ or the map $mathbb R^{mathbb R^2}to mathbb R^Gamma $ ?
general-topology functional-analysis
asked Jan 18 at 9:00
idmidm
8,61521446
$endgroup$
add a comment |
$begingroup$
Let $sim$ an equivalence relation, and denote $Gamma=mathbb R^2/_sim$ the quotient space. We denote $pi:mathbb R^2to Gamma $ the natural projection. And now I want to identify $mathbb R^Gamma $ with $mathbb R^{mathbb R^2}$ in a natural way.
Is there a natural way map from a set $A$ and $mathbb R^A$ so that I can define easily the map $mathbb R^{Gamma }to mathbb R^{mathbb R^2}$ or the map $mathbb R^{mathbb R^2}to mathbb R^Gamma $ ?
general-topology functional-analysis
asked Jan 18 at 9:00
idmidm
8,61521446
$endgroup$
Let $sim$ an equivalence relation, and denote $Gamma=mathbb R^2/_sim$ the quotient space. We denote $pi:mathbb R^2to Gamma $ the natural projection. And now I want to identify $mathbb R^Gamma $ with $mathbb R^{mathbb R^2}$ in a natural way.
Is there a natural way map from a set $A$ and $mathbb R^A$ so that I can define easily the map $mathbb R^{Gamma }to mathbb R^{mathbb R^2}$ or the map $mathbb R^{mathbb R^2}to mathbb R^Gamma $ ?
general-topology functional-analysis
general-topology functional-analysis
asked Jan 18 at 9:00
idmidm
8,61521446
asked Jan 18 at 9:00
idmidm
8,61521446
asked Jan 18 at 9:00
idmidm
8,61521446
asked Jan 18 at 9:00
idmidm
8,61521446
asked Jan 18 at 9:00
idmidm
8,61521446
8,61521446
add a comment |
add a comment |
1 Answer
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$$gmapsto gcircpi$$ serves as prescription of a map: $$mathbb R^{Gamma}tomathbb R^{mathbb R^2}$$
answered Jan 18 at 9:13
drhabdrhab
104k545136
$endgroup$
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
$$gmapsto gcircpi$$ serves as prescription of a map: $$mathbb R^{Gamma}tomathbb R^{mathbb R^2}$$
answered Jan 18 at 9:13
drhabdrhab
104k545136
$endgroup$
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
add a comment |
$begingroup$
$$gmapsto gcircpi$$ serves as prescription of a map: $$mathbb R^{Gamma}tomathbb R^{mathbb R^2}$$
answered Jan 18 at 9:13
drhabdrhab
104k545136
$endgroup$
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
add a comment |
$begingroup$
$$gmapsto gcircpi$$ serves as prescription of a map: $$mathbb R^{Gamma}tomathbb R^{mathbb R^2}$$
answered Jan 18 at 9:13
drhabdrhab
104k545136
$endgroup$
$$gmapsto gcircpi$$ serves as prescription of a map: $$mathbb R^{Gamma}tomathbb R^{mathbb R^2}$$
answered Jan 18 at 9:13
drhabdrhab
104k545136
answered Jan 18 at 9:13
drhabdrhab
104k545136
answered Jan 18 at 9:13
drhabdrhab
104k545136
answered Jan 18 at 9:13
drhabdrhab
104k545136
104k545136
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
add a comment |
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
What do you mean by prescription map ? Surprinsingly, this way doesn't really look natural for me... Normally, I would go from $mathbb R^{mathbb R^2}to mathbb R^Gamma $ (like for the projection...)
$endgroup$
– idm
Jan 18 at 9:15
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
$begingroup$
I do not call it a "prescription map" but just a map. Every map can be described/defined by means of a prescription. The map in my answer sends element $ginmathbb R^{Gamma}$ to element $gcircpiinmathbb R^{mathbb R^2}$. A natural map in other direction is - I think - not possible. You can at most define a natural map $Btomathbb R^{Gamma}$ where $B$ denotes the subset containing functions $f:mathbb R^2tomathbb R$ that respect $sim$ in the sense that $xsim yimplies f(x)=f(y)$. For every such $f$ a unique $g$ exists with $f=gcircpi$.
$endgroup$
– drhab
Jan 18 at 9:23
add a comment |
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