Is there a compact $n$-dimensional manifold $M$ or, more generaly, a compact $n$-dimensional topological space $M$ with the following property?

"For every continuous map $f:M \to \mathbb{R}^{n}$ there are points $a,b,c \in M $ with $f(a)=f(b)=f(c)$."

This is motivated by the following obvious consequence of the Borsuk-Ulam theorem:

"For every continuous map $f:S^n \to \mathbb{R}^n$ there are points $a, b \in S^n$ with $f(a)=f(b)$."


If M is allowed to be a simplicial complex, five $n$-simplices with a common $(n-1)$-dimensional face should do the job. If $M$ should be a manifold, take any closed non-orientable one.

The multiplicity of maps between manifolds can be studied with the help of characteristic classes, see this preprint of Roman Karasev and this preprint of Roman Karasev and Pavle Blagojevic. For example, any continuous map $\mathbb{R}\mathrm{P}^4 \to \mathbb{R}^4$ sends some four points in $\mathbb{R}\mathrm{P}^4$ to the same point in $\mathbb{R}^4$.

Another generalization of the Borsuk-Ulam theorem (more close to it in the spirit) deals with spaces with a group action (like $\mathbb{Z}_2$-action on the sphere) such that for any map to $\mathbb{R}^n$ some orbit is sent to a point. This is discussed in detail in Section 6 of

Matoušek, Jiří, Using the Borsuk-Ulam theorem. Lectures on topological methods in combinatorics and geometry. Written in cooperation with Anders Björner and Günter M. Ziegler, Universitext. Berlin: Springer (ISBN 978-3-540-00362-5/pbk; 978-3-540-76649-0/ebook). xii, 214~p. (2008). ZBL1234.05002.

  • $\begingroup$ Thank you again for your very interesting answer. It seems that the powerfull method of characteristic classes is more applicable in compact cases for example $\mathbb{R}P^4$ rather than non compact bases. As an example of non compact case, I am wondering if this question is an elementary question: "Is there a continuous map $f:\mathbb{R}^2 \to \mathbb{R}$ such that every level set is a finite set? $\endgroup$ – Ali Taghavi Mar 22 '17 at 13:40
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    $\begingroup$ There is no such a map. Let $p$ be an interior point of $f(\mathbb{R}^2)$ (if there is no interior point, then $f$ is constant). The restriction of $f$ to $\mathbb{R}^2 \setminus f^{-1}(p)$ is a continuous map onto a disconnected set. Hence $\mathbb{R}^2 \setminus f^{-1}(p)$ is disconnected, in particular $f^{-1}(p)$ is infinite (even uncountable). $\endgroup$ – Ivan Izmestiev Mar 22 '17 at 14:29
  • $\begingroup$ @lvan thank you. What about the same question for $\mathbb{R}^3 \to \mathbb{R}^2$.? $\endgroup$ – Ali Taghavi Mar 24 '17 at 8:27

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