A parametric version of the Borsuk Ulam theorem
Is there a topological space $X$, which is not a singleton, and satisfies the following property?
For every continuous function $f: Xtimes S^2tomathbb{R}^2$ there exist a point $xin S^2$ such that $f(t, x)=f(t,-x),;forall t in X$?
Is there a classification of all spaces $X$ with such property?
at.algebraic-topology gn.general-topology
asked yesterday
Ali Taghavi
7652083
add a comment |
Is there a topological space $X$, which is not a singleton, and satisfies the following property?
For every continuous function $f: Xtimes S^2tomathbb{R}^2$ there exist a point $xin S^2$ such that $f(t, x)=f(t,-x),;forall t in X$?
Is there a classification of all spaces $X$ with such property?
at.algebraic-topology gn.general-topology
asked yesterday
Ali Taghavi
7652083
3
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
3
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday
add a comment |
Is there a topological space $X$, which is not a singleton, and satisfies the following property?
For every continuous function $f: Xtimes S^2tomathbb{R}^2$ there exist a point $xin S^2$ such that $f(t, x)=f(t,-x),;forall t in X$?
Is there a classification of all spaces $X$ with such property?
at.algebraic-topology gn.general-topology
asked yesterday
Ali Taghavi
7652083
Is there a topological space $X$, which is not a singleton, and satisfies the following property?
For every continuous function $f: Xtimes S^2tomathbb{R}^2$ there exist a point $xin S^2$ such that $f(t, x)=f(t,-x),;forall t in X$?
Is there a classification of all spaces $X$ with such property?
at.algebraic-topology gn.general-topology
at.algebraic-topology gn.general-topology
asked yesterday
Ali Taghavi
7652083
asked yesterday
Ali Taghavi
7652083
edited yesterday
asked yesterday
Ali Taghavi
7652083
asked yesterday
Ali Taghavi
7652083
asked yesterday
Ali Taghavi
7652083
7652083
3
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
3
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday
add a comment |
3
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
3
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday
3
3
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
3
3
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday
add a comment |
1 Answer
1
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Theorem. For a topological space $X$ the following conditions are equivalent:
1) for any continuous map $f:Xtimes S^2tomathbb R^2$ there exists a point $sin S^2$ such that $f(x,s)=f(x,-s)$ for any $xin X$;
2) any continuous map $f:Xto mathbb R$ is constant.
Proof. (1) $Rightarrow$ (2) Assume that $X$ admits a non-constant map $hbar :Xto mathbb R$. We lose no generality assuming that ${0,1}subset hbar(X)subset[0,1]$.
Then there are points $x_0,x_1in X$ such that $hbar(x_i)=i$ for $iin{0,1}$.
Let $p:S^2tomathbb R^2$, $p:(x,y,z)mapsto (x,y)$, be the projection of the sphere $S^2={(x,y,z)inmathbb R^3:x^2+y^2+z^2=1}$ onto the plane.
Let $varphi:S^2to S^2$ be any homeomorphism of the sphere $S^2$ such that $pcirc varphi(0,0,1)ne pcirc varphi(0,0,-1)$.
Using the Tietze-Urysohn Theorem, find a continuous map $psi:[0,1]times S^2tomathbb R^2$ such that $psi(0,s)=p(s)$ and $psi(1,x)=pcircvarphi(s)$ for all $sin S$. Then the continuous map $$f:Xtimes S^2tomathbb R^2,;;f:(x,s)mapsto psi(hbar(x),s),$$ has the following property:
if $f(x_0,s)=f(x_0,-s)$ for some $sin S^2$, then $sin{(0,0,1),(0,0,-1)}$ and $f(x_1,s)ne f(x_1,-s)$.
(2) $Rightarrow$ (1) Assume that each continuous map $Xtomathbb R$ is constant and take any continuous function $f:Xtimes S^2tomathbb R^2$. Fix any point $x_0in X$ and using the Borsuk-Ulam Theorem, find a point $sin S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $sin S^2$ the function $f{restriction}Xtimes{s}$ is constant. So, for every $xin X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$
Remark. For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".
answered yesterday
Taras Banakh
15.9k13191
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
add a comment |
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Theorem. For a topological space $X$ the following conditions are equivalent:
1) for any continuous map $f:Xtimes S^2tomathbb R^2$ there exists a point $sin S^2$ such that $f(x,s)=f(x,-s)$ for any $xin X$;
2) any continuous map $f:Xto mathbb R$ is constant.
Proof. (1) $Rightarrow$ (2) Assume that $X$ admits a non-constant map $hbar :Xto mathbb R$. We lose no generality assuming that ${0,1}subset hbar(X)subset[0,1]$.
Then there are points $x_0,x_1in X$ such that $hbar(x_i)=i$ for $iin{0,1}$.
Let $p:S^2tomathbb R^2$, $p:(x,y,z)mapsto (x,y)$, be the projection of the sphere $S^2={(x,y,z)inmathbb R^3:x^2+y^2+z^2=1}$ onto the plane.
Let $varphi:S^2to S^2$ be any homeomorphism of the sphere $S^2$ such that $pcirc varphi(0,0,1)ne pcirc varphi(0,0,-1)$.
Using the Tietze-Urysohn Theorem, find a continuous map $psi:[0,1]times S^2tomathbb R^2$ such that $psi(0,s)=p(s)$ and $psi(1,x)=pcircvarphi(s)$ for all $sin S$. Then the continuous map $$f:Xtimes S^2tomathbb R^2,;;f:(x,s)mapsto psi(hbar(x),s),$$ has the following property:
if $f(x_0,s)=f(x_0,-s)$ for some $sin S^2$, then $sin{(0,0,1),(0,0,-1)}$ and $f(x_1,s)ne f(x_1,-s)$.
(2) $Rightarrow$ (1) Assume that each continuous map $Xtomathbb R$ is constant and take any continuous function $f:Xtimes S^2tomathbb R^2$. Fix any point $x_0in X$ and using the Borsuk-Ulam Theorem, find a point $sin S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $sin S^2$ the function $f{restriction}Xtimes{s}$ is constant. So, for every $xin X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$
Remark. For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".
answered yesterday
Taras Banakh
15.9k13191
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
add a comment |
Theorem. For a topological space $X$ the following conditions are equivalent:
1) for any continuous map $f:Xtimes S^2tomathbb R^2$ there exists a point $sin S^2$ such that $f(x,s)=f(x,-s)$ for any $xin X$;
2) any continuous map $f:Xto mathbb R$ is constant.
Proof. (1) $Rightarrow$ (2) Assume that $X$ admits a non-constant map $hbar :Xto mathbb R$. We lose no generality assuming that ${0,1}subset hbar(X)subset[0,1]$.
Then there are points $x_0,x_1in X$ such that $hbar(x_i)=i$ for $iin{0,1}$.
Let $p:S^2tomathbb R^2$, $p:(x,y,z)mapsto (x,y)$, be the projection of the sphere $S^2={(x,y,z)inmathbb R^3:x^2+y^2+z^2=1}$ onto the plane.
Let $varphi:S^2to S^2$ be any homeomorphism of the sphere $S^2$ such that $pcirc varphi(0,0,1)ne pcirc varphi(0,0,-1)$.
Using the Tietze-Urysohn Theorem, find a continuous map $psi:[0,1]times S^2tomathbb R^2$ such that $psi(0,s)=p(s)$ and $psi(1,x)=pcircvarphi(s)$ for all $sin S$. Then the continuous map $$f:Xtimes S^2tomathbb R^2,;;f:(x,s)mapsto psi(hbar(x),s),$$ has the following property:
if $f(x_0,s)=f(x_0,-s)$ for some $sin S^2$, then $sin{(0,0,1),(0,0,-1)}$ and $f(x_1,s)ne f(x_1,-s)$.
(2) $Rightarrow$ (1) Assume that each continuous map $Xtomathbb R$ is constant and take any continuous function $f:Xtimes S^2tomathbb R^2$. Fix any point $x_0in X$ and using the Borsuk-Ulam Theorem, find a point $sin S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $sin S^2$ the function $f{restriction}Xtimes{s}$ is constant. So, for every $xin X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$
Remark. For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".
answered yesterday
Taras Banakh
15.9k13191
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
add a comment |
Theorem. For a topological space $X$ the following conditions are equivalent:
1) for any continuous map $f:Xtimes S^2tomathbb R^2$ there exists a point $sin S^2$ such that $f(x,s)=f(x,-s)$ for any $xin X$;
2) any continuous map $f:Xto mathbb R$ is constant.
Proof. (1) $Rightarrow$ (2) Assume that $X$ admits a non-constant map $hbar :Xto mathbb R$. We lose no generality assuming that ${0,1}subset hbar(X)subset[0,1]$.
Then there are points $x_0,x_1in X$ such that $hbar(x_i)=i$ for $iin{0,1}$.
Let $p:S^2tomathbb R^2$, $p:(x,y,z)mapsto (x,y)$, be the projection of the sphere $S^2={(x,y,z)inmathbb R^3:x^2+y^2+z^2=1}$ onto the plane.
Let $varphi:S^2to S^2$ be any homeomorphism of the sphere $S^2$ such that $pcirc varphi(0,0,1)ne pcirc varphi(0,0,-1)$.
Using the Tietze-Urysohn Theorem, find a continuous map $psi:[0,1]times S^2tomathbb R^2$ such that $psi(0,s)=p(s)$ and $psi(1,x)=pcircvarphi(s)$ for all $sin S$. Then the continuous map $$f:Xtimes S^2tomathbb R^2,;;f:(x,s)mapsto psi(hbar(x),s),$$ has the following property:
if $f(x_0,s)=f(x_0,-s)$ for some $sin S^2$, then $sin{(0,0,1),(0,0,-1)}$ and $f(x_1,s)ne f(x_1,-s)$.
(2) $Rightarrow$ (1) Assume that each continuous map $Xtomathbb R$ is constant and take any continuous function $f:Xtimes S^2tomathbb R^2$. Fix any point $x_0in X$ and using the Borsuk-Ulam Theorem, find a point $sin S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $sin S^2$ the function $f{restriction}Xtimes{s}$ is constant. So, for every $xin X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$
Remark. For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".
answered yesterday
Taras Banakh
15.9k13191
Theorem. For a topological space $X$ the following conditions are equivalent:
1) for any continuous map $f:Xtimes S^2tomathbb R^2$ there exists a point $sin S^2$ such that $f(x,s)=f(x,-s)$ for any $xin X$;
2) any continuous map $f:Xto mathbb R$ is constant.
Proof. (1) $Rightarrow$ (2) Assume that $X$ admits a non-constant map $hbar :Xto mathbb R$. We lose no generality assuming that ${0,1}subset hbar(X)subset[0,1]$.
Then there are points $x_0,x_1in X$ such that $hbar(x_i)=i$ for $iin{0,1}$.
Let $p:S^2tomathbb R^2$, $p:(x,y,z)mapsto (x,y)$, be the projection of the sphere $S^2={(x,y,z)inmathbb R^3:x^2+y^2+z^2=1}$ onto the plane.
Let $varphi:S^2to S^2$ be any homeomorphism of the sphere $S^2$ such that $pcirc varphi(0,0,1)ne pcirc varphi(0,0,-1)$.
Using the Tietze-Urysohn Theorem, find a continuous map $psi:[0,1]times S^2tomathbb R^2$ such that $psi(0,s)=p(s)$ and $psi(1,x)=pcircvarphi(s)$ for all $sin S$. Then the continuous map $$f:Xtimes S^2tomathbb R^2,;;f:(x,s)mapsto psi(hbar(x),s),$$ has the following property:
if $f(x_0,s)=f(x_0,-s)$ for some $sin S^2$, then $sin{(0,0,1),(0,0,-1)}$ and $f(x_1,s)ne f(x_1,-s)$.
(2) $Rightarrow$ (1) Assume that each continuous map $Xtomathbb R$ is constant and take any continuous function $f:Xtimes S^2tomathbb R^2$. Fix any point $x_0in X$ and using the Borsuk-Ulam Theorem, find a point $sin S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $sin S^2$ the function $f{restriction}Xtimes{s}$ is constant. So, for every $xin X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$
Remark. For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".
answered yesterday
Taras Banakh
15.9k13191
edited yesterday
answered yesterday
Taras Banakh
15.9k13191
answered yesterday
Taras Banakh
15.9k13191
answered yesterday
Taras Banakh
15.9k13191
15.9k13191
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
add a comment |
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
Thank you very much for your attention to my question and your very interesting answer.
– Ali Taghavi
yesterday
add a comment |
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3
Who downvoted this? It's a totally reasonable question!
– Dylan Wilson
yesterday
At least we may consider any set $X$ with the trivial topology:)
– Aleksei Kulikov
yesterday
3
A less stringent condition would be that there is a continuous (instead of constant) map $tto x(t)$ with $f(t,x(t))=f(t,-x(t))$.
– BS.
yesterday