injective open map between two euclidean spaces 
Does there exists an injective function from $\mathbb R^2$ to $\mathbb R$ such that image of every open set is open ?  Thank you.

 A: There does not exist such a function even if we do not assume continuity.
$f(\mathbb R^2)\subset \mathbb R$ must be an open set, and we have a bijective continuous function $f^{-1}:f(\mathbb R^2)\to \mathbb R^2$. This map is a homeomorphism onto its image when restricted to each closed subinterval of $f(\mathbb R^2)$, and a subspace of $\mathbb R^2$ homeomorphic to a closed interval in $\mathbb R$ has empty interior. Since $f(\mathbb R^2)$ is a countable union of such closed intervals, so is $\mathbb R^2$. But this contradicts the Baire category theorem.
A: I assume by map you mean 'continuous function.'  If so, there does not exist such a function.
Assume for contradiction such a map $f:\mathbb{R}^2 \to \mathbb{R}$ exists.  Then  its image must be a connected set in $\mathbb{R}$.  Now, because $f$ is injective by hypothesis, the pre-image of any point in the image of $f$ is a unique point in the domain. But if I remove a point in the image, the image ceases to be connected; however $\mathbb{R}^2$ less one point remains connected.
In that spirit, let $a\in f(\mathbb{R}^2)$.  Then using $f$ I can define the continuous injective map $g: \mathbb{R}^2 \setminus f^{-1}(a) \to \mathbb{R}$ via $(x,y) \mapsto f(x,y)$.  By construction, $g$ is continuous and injective since $f$ is.  But $g$ cannot exist: it continuously maps a connected set into a disconnected one.  Contradiction.
