# 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 $$f: X\times S^2\to\mathbb{R}^2$$ 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?

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

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