Continuous mapping between $mathbb{S}^2$, $mathbb{R}P^2$ and $mathbb{R}^4$
$begingroup$
Let $mathbb{S}^2$ be the unit sphere in $mathbb{R}^3$. Let $f:mathbb{S}^2rightarrowmathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$
prove that $f$ determines a continuous map $tilde{f} : mathbb{R}P^2 → mathbb{R}^4$ where $mathbb{R}P^2$ is the real projective plane and that $tilde{f}$ defines a homeomorphism onto a subset of $mathbb{R}^4$
$mathbb{R}P^2$ is homeomorphic to the upper hemisphere of the sphere so is the first proof to show that $f$ is continuous on the upper hemisphere and that we can map the sphere continuously onto the upper hemisphere?
[as an aside question if $f = gcdot h$ and $f,g$ are continuous, is $h$ continuous?]
Fir the second part, is the subset it is homeomorphic to just the image of the upper hemisphere?
general-topology continuity
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
$endgroup$
add a comment |
$begingroup$
Let $mathbb{S}^2$ be the unit sphere in $mathbb{R}^3$. Let $f:mathbb{S}^2rightarrowmathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$
prove that $f$ determines a continuous map $tilde{f} : mathbb{R}P^2 → mathbb{R}^4$ where $mathbb{R}P^2$ is the real projective plane and that $tilde{f}$ defines a homeomorphism onto a subset of $mathbb{R}^4$
$mathbb{R}P^2$ is homeomorphic to the upper hemisphere of the sphere so is the first proof to show that $f$ is continuous on the upper hemisphere and that we can map the sphere continuously onto the upper hemisphere?
[as an aside question if $f = gcdot h$ and $f,g$ are continuous, is $h$ continuous?]
Fir the second part, is the subset it is homeomorphic to just the image of the upper hemisphere?
general-topology continuity
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
$endgroup$
1
$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52
add a comment |
$begingroup$
Let $mathbb{S}^2$ be the unit sphere in $mathbb{R}^3$. Let $f:mathbb{S}^2rightarrowmathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$
prove that $f$ determines a continuous map $tilde{f} : mathbb{R}P^2 → mathbb{R}^4$ where $mathbb{R}P^2$ is the real projective plane and that $tilde{f}$ defines a homeomorphism onto a subset of $mathbb{R}^4$
$mathbb{R}P^2$ is homeomorphic to the upper hemisphere of the sphere so is the first proof to show that $f$ is continuous on the upper hemisphere and that we can map the sphere continuously onto the upper hemisphere?
[as an aside question if $f = gcdot h$ and $f,g$ are continuous, is $h$ continuous?]
Fir the second part, is the subset it is homeomorphic to just the image of the upper hemisphere?
general-topology continuity
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
$endgroup$
Let $mathbb{S}^2$ be the unit sphere in $mathbb{R}^3$. Let $f:mathbb{S}^2rightarrowmathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$
prove that $f$ determines a continuous map $tilde{f} : mathbb{R}P^2 → mathbb{R}^4$ where $mathbb{R}P^2$ is the real projective plane and that $tilde{f}$ defines a homeomorphism onto a subset of $mathbb{R}^4$
$mathbb{R}P^2$ is homeomorphic to the upper hemisphere of the sphere so is the first proof to show that $f$ is continuous on the upper hemisphere and that we can map the sphere continuously onto the upper hemisphere?
[as an aside question if $f = gcdot h$ and $f,g$ are continuous, is $h$ continuous?]
Fir the second part, is the subset it is homeomorphic to just the image of the upper hemisphere?
general-topology continuity
general-topology continuity
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
edited Dec 22 '18 at 19:43
Clayton
19.3k33287
19.3k33287
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
asked Dec 22 '18 at 19:41
Toby PeterkenToby Peterken
1496
1496
1
$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52
add a comment |
1
$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52
1
1
$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52
$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52
add a comment |
1 Answer
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$begingroup$
$mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = pm (x,y,z)$. So let
$$(*) phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy in mathbb{C}$, $c' = x' + iy' in mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= epsilon c$ with $epsilon = pm 1$. In other words, $(x',y') = epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' in { -1, 1}$, hence $(x',y',z') = pm (x,y,z)$.
Case 2. $(x,y) ne (0,0)$. W.lo.g. let $x ne 0$. Then $epsilon x z' = x'z' = xz$, hence $z' = epsilon z$. This shows $(x',y',z') = epsilon (x,y,z)$.
Concerning your final question: The image $R = tilde{f}(mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f mid_{S^2_+}$ is no embedding because $f mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
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add a comment |
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$begingroup$
$mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = pm (x,y,z)$. So let
$$(*) phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy in mathbb{C}$, $c' = x' + iy' in mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= epsilon c$ with $epsilon = pm 1$. In other words, $(x',y') = epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' in { -1, 1}$, hence $(x',y',z') = pm (x,y,z)$.
Case 2. $(x,y) ne (0,0)$. W.lo.g. let $x ne 0$. Then $epsilon x z' = x'z' = xz$, hence $z' = epsilon z$. This shows $(x',y',z') = epsilon (x,y,z)$.
Concerning your final question: The image $R = tilde{f}(mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f mid_{S^2_+}$ is no embedding because $f mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
$endgroup$
add a comment |
$begingroup$
$mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = pm (x,y,z)$. So let
$$(*) phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy in mathbb{C}$, $c' = x' + iy' in mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= epsilon c$ with $epsilon = pm 1$. In other words, $(x',y') = epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' in { -1, 1}$, hence $(x',y',z') = pm (x,y,z)$.
Case 2. $(x,y) ne (0,0)$. W.lo.g. let $x ne 0$. Then $epsilon x z' = x'z' = xz$, hence $z' = epsilon z$. This shows $(x',y',z') = epsilon (x,y,z)$.
Concerning your final question: The image $R = tilde{f}(mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f mid_{S^2_+}$ is no embedding because $f mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
$endgroup$
add a comment |
$begingroup$
$mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = pm (x,y,z)$. So let
$$(*) phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy in mathbb{C}$, $c' = x' + iy' in mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= epsilon c$ with $epsilon = pm 1$. In other words, $(x',y') = epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' in { -1, 1}$, hence $(x',y',z') = pm (x,y,z)$.
Case 2. $(x,y) ne (0,0)$. W.lo.g. let $x ne 0$. Then $epsilon x z' = x'z' = xz$, hence $z' = epsilon z$. This shows $(x',y',z') = epsilon (x,y,z)$.
Concerning your final question: The image $R = tilde{f}(mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f mid_{S^2_+}$ is no embedding because $f mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
$endgroup$
$mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = pm (x,y,z)$. So let
$$(*) phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy in mathbb{C}$, $c' = x' + iy' in mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= epsilon c$ with $epsilon = pm 1$. In other words, $(x',y') = epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' in { -1, 1}$, hence $(x',y',z') = pm (x,y,z)$.
Case 2. $(x,y) ne (0,0)$. W.lo.g. let $x ne 0$. Then $epsilon x z' = x'z' = xz$, hence $z' = epsilon z$. This shows $(x',y',z') = epsilon (x,y,z)$.
Concerning your final question: The image $R = tilde{f}(mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f mid_{S^2_+}$ is no embedding because $f mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
edited Dec 23 '18 at 15:25
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
answered Dec 22 '18 at 23:58
Paul FrostPaul Frost
11.4k3934
11.4k3934
add a comment |
add a comment |
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$begingroup$
To answer your aside question, $f,g$ continuous and $f=gcirc h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 19:52