Is there a topological space $X$, which is not a singleton, and satisfies the following property?

For every continuous function $f: X\times S^2\to\mathbb{R}^2$ there exist a point $x\in S^2$ such that $f(t, x)=f(t,-x),\;\forall t \in X$?

Is there a classification of all spaces $X$ with such property?

  • 3
    $\begingroup$ Who downvoted this? It's a totally reasonable question! $\endgroup$ – Dylan Wilson Jan 4 at 18:08
  • $\begingroup$ At least we may consider any set $X$ with the trivial topology:) $\endgroup$ – Aleksei Kulikov Jan 4 at 18:17
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    $\begingroup$ A less stringent condition would be that there is a continuous (instead of constant) map $t\to x(t)$ with $f(t,x(t))=f(t,-x(t))$. $\endgroup$ – BS. Jan 5 at 11:07

Theorem. For a topological space $X$ the following conditions are equivalent:

1) for any continuous map $f:X\times S^2\to\mathbb R^2$ there exists a point $s\in S^2$ such that $f(x,s)=f(x,-s)$ for any $x\in X$;

2) any continuous map $f:X\to \mathbb R$ is constant.

Proof. (1) $\Rightarrow$ (2) Assume that $X$ admits a non-constant map $\hbar :X\to \mathbb R$. We lose no generality assuming that $\{0,1\}\subset \hbar(X)\subset[0,1]$. Then there are points $x_0,x_1\in X$ such that $\hbar(x_i)=i$ for $i\in\{0,1\}$.

Let $p:S^2\to\mathbb R^2$, $p:(x,y,z)\mapsto (x,y)$, be the projection of the sphere $S^2=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\}$ onto the plane.

Let $\varphi:S^2\to S^2$ be any homeomorphism of the sphere $S^2$ such that $p\circ \varphi(0,0,1)\ne p\circ \varphi(0,0,-1)$.

Using the Tietze-Urysohn Theorem, find a continuous map $\psi:[0,1]\times S^2\to\mathbb R^2$ such that $\psi(0,s)=p(s)$ and $\psi(1,x)=p\circ\varphi(s)$ for all $s\in S$. Then the continuous map $$f:X\times S^2\to\mathbb R^2,\;\;f:(x,s)\mapsto \psi(\hbar(x),s),$$ has the following property:

if $f(x_0,s)=f(x_0,-s)$ for some $s\in S^2$, then $s\in\{(0,0,1),(0,0,-1)\}$ and $f(x_1,s)\ne f(x_1,-s)$.

(2) $\Rightarrow$ (1) Assume that each continuous map $X\to\mathbb R$ is constant and take any continuous function $f:X\times S^2\to\mathbb R^2$. Fix any point $x_0\in X$ and using the Borsuk-Ulam Theorem, find a point $s\in S^2$ such that $f(x_0,s)=f(x_0,-s)$. By our assumption, for every $s\in S^2$ the function $f{\restriction}X\times\{s\}$ is constant. So, for every $x\in X$ 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".

  • $\begingroup$ Thank you very much for your attention to my question and your very interesting answer. $\endgroup$ – Ali Taghavi Jan 4 at 20:53

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