A bijection of the plane that maps horizontal and vertical lines to horizontal and vertical lines respectively I want to understand why this "easy to see" implication is true.
If $C:\mathbb{R}^2 \to \mathbb{R}^2$ is a bijection that maps horizontal and vertical lines to horizontal and vertical lines, respectively, then $C(x,y)=(f(x),g(y))$, for $f,g: \mathbb{R} \to \mathbb{R}$ bijections.
Why can $C$ be written that way?
Any help would be appreciated.
 A: I guess that by "preserving horizontal lines" you mean that "$C$ maps any horizontal line to a horizontal line (not necessarily the same)", and similarly for the vertical lines.
In that case, one needs only to analyze this a bit:
"$C$ preserves vertical lines (in the above sense)" means that given any $\alpha \in \mathbb{R},$  there is $f(\alpha)$ such that for all $y\in \mathbb{R}$ we have $$C(\alpha, y)=(f(\alpha), g_{\alpha}(y)),$$
where the function $g_\alpha$ possibly depends on $\alpha$. Note that the function  $f$ is a bijection $f: \mathbb{R} \rightarrow \mathbb{R}$.
Similarly, by "preserving horizontal lines", given any $\beta \in \mathbb{R},$ there is $g(\beta) \in \mathbb{R}$ and a function $f_\beta$ such that 
$$\forall x \in \mathbb{R}: C(x, \beta)=(f_\beta(x), g(\beta)),$$
and as before, it is easy to see that $g$ is a bijection $\mathbb{R} \rightarrow \mathbb{R}$.
So if $(\alpha, \beta) \in \mathbb{R}^2$ is an arbitrary point, then by the above, the first coordinate of $C(\alpha, \beta)$ is $f(\alpha)$ and the second one is $g(\beta),$ hence
$$\forall (\alpha, \beta) \in \mathbb{R}^2: C(\alpha, \beta)=(f(\alpha), g(\beta)).$$
