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Mar 24, 2017 at 8:27 comment added Ali Taghavi @lvan thank you. What about the same question for $\mathbb{R}^3 \to \mathbb{R}^2$.?
Mar 22, 2017 at 14:29 comment added Ivan Izmestiev 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).
Mar 22, 2017 at 13:40 comment added Ali Taghavi 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?
Mar 21, 2017 at 20:14 history edited Michael Albanese CC BY-SA 3.0
Inserted letters with accents.
Mar 21, 2017 at 17:45 history edited Ivan Izmestiev CC BY-SA 3.0
added 7 characters in body
Mar 21, 2017 at 15:33 history edited Ivan Izmestiev CC BY-SA 3.0
added example of non-orientable manifold (first paragraph)
Mar 21, 2017 at 14:14 vote accept Ali Taghavi
Mar 21, 2017 at 13:54 history answered Ivan Izmestiev CC BY-SA 3.0