A parametric version of the Borsuk Ulam theorem












10














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







share|cite|improve this question




















  • 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.
    21 hours ago
















10














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







share|cite|improve this question




















  • 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.
    21 hours ago














10












10








10


2





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Ali Taghavi

7152083




7152083








  • 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.
    21 hours ago














  • 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.
    21 hours ago








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.
21 hours ago




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.
21 hours ago










1 Answer
1






active

oldest

votes


















16















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".







share|cite|improve this answer























  • Thank you very much for your attention to my question and your very interesting answer.
    – Ali Taghavi
    yesterday











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1 Answer
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16















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".







share|cite|improve this answer























  • Thank you very much for your attention to my question and your very interesting answer.
    – Ali Taghavi
    yesterday
















16















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".







share|cite|improve this answer























  • Thank you very much for your attention to my question and your very interesting answer.
    – Ali Taghavi
    yesterday














16












16








16







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".







share|cite|improve this answer















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".








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























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


















  • 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


















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