Theorem: Let $B: \mathbb{R}^2 \to \mathbb{R}^2$, $B(0,0)=(0,0)$ a bijection that applies lines to lines (The image of a line through $B$ is contained in a line). If $k,l \subseteq \mathbb{R}^2$ are two lines such that $B(l) \subseteq \mathbb{R}^2$ then $B(l)=k$ and $l$ is the only line whose image is contained in $k$.
Proof: Suppose by contradiction $\exists q \in k, q \notin B(l)$. Since $B$ is a bijection, $\exists ! q_{0}$ such that $B(q_{0}) = q$. By supposition $ q_{0} \notin l$. Let $l_1$ line that contains $q_{0}$ and is perpendicular to $l$ in $q_{1} \in l$. Since B apply lines to lines and $B(q_0), B(q_1)\in k$, then $B(q_1) \subseteq k$. Then, let $p \in \mathbb{R}^2$. $p \in l_2$ line that intersepts $l_1 \cup l$ in at least two points, say $p_1,p_2$. Again, $B(p_1),B(p_2) \in k \Rightarrow B(l_2) \subseteq k$. This shows that $B(\mathbb{R}^2) \subseteq k$ , which is a contradiction, since $B$ is onto.
My problem is with the last part. Why is it a contradiction that the image of the plane is in a line? Since $\mathbb{R}^2$ and $\mathbb{R}$ have the same cardinality, there can be a bijection between them, then why is $B$ being onto a contradiction?
Already found the problem. $B: \mathbb{R}^2 \to \mathbb{R}^2$ is onto, then $\Im(B) = B(\mathbb{R}^2) = \mathbb{R}^2 $, which can't be contained in a line $k$.
There is no problem in having a surjection from $\mathbb{R}^2$ to $\mathbb{R}$.