Is a map that preserves lines and fixes the origin necessarily linear? Let $V$ and $W$ be vector spaces over a field $\mathbb{F}$ with $\text{dim }V \ge 2$. A line is a set of the form $\{ \mathbf{u} + t\mathbf{v} : t \in \mathbb{F} \}$. A map $f: V \to W$ preserves lines if the image of every line in $V$ is a line in $W$. A map fixes the origin if $f(0) = 0$.
Is a function $f: V\to W$ that preserves lines and fixes the origin necessarily linear?
 A: Consider $V=W=\Bbb F_2^k$. Then, since a subset is a line if and only if it contains (at most) two points, any bijective map that sends $0$ to $0$ does the trick. However, there are $(2^k-1)!$ such maps, while the bijective linear maps are less than that for $k\ge 3$.
A: You may be interested in this paper
   Affinity of a Permutation of a Finite Vector Space
It discusses the problem of how many k-flats (cosets of a k-dimentional subspace) of an n-dimensional vector space over a finite field must be preserved by a permutation to force the permutation to preserve all k-flats.  See the references for the history of this problem for other fields. And, by the way, Vilmos Totik and Wen-Xiu Ma proved (personal communication) that if f is a transformation of Euclidean n-space, n  > 1, such that for all but countably many lines L the image f(L) is a line, then the image of any line is a line, hence f is an affine transformation.
A: Let me give an example in the Euclidean plane.
The function $f\colon\mathbb R^2\to\mathbb R$ given by $f(x,y)=x^3$ maps lines to lines.
A vertical line $x=a$ is mapped to the line $a^3$ — points are lines by the OP's definition.
Any other line is of the form $\{(t,a+bt);t\in\mathbb R\}$ for some $a,b\in\mathbb R$.
The image of any such line is $\mathbb R$.
Thus $f$ maps lines to lines, and it clearly fixes the origin.
Non-linearity is evident.
This $f$ can be promoted to a function $g\colon\mathbb R^2\to\mathbb R^2$ by letting $g(x,y)=(f(x,y),0)$.
This inherits the desired properties and is a function between two-dimensional spaces.

Below is a previous, erroneous answer. I left it here as a warning example. My actual answer is above.
I can delete this if it would be more appropriate.
Let me give an example with infinite fields.
The real line $\mathbb R$ is an infinite dimensional vector space over $\mathbb Q$.
Lines — other than the origin — are translations of the rationals ($r+\mathbb Q$ for some $r$).
Take the function $g\colon\mathbb R\to\mathbb R$,
$$
g(x)
=
\begin{cases}
x, & x\in\mathbb Q\\
0, & x\notin\mathbb Q.
\end{cases}
$$
The image of the line $r+\mathbb Q$ is the line $\mathbb Q$ if $r\in\mathbb Q$ and $\{0\}$ if $r\notin\mathbb Q$.
Therefore $g$ preserves lines and fixes the origin.
But it is not linear: $5=g(5)\neq g(5-\pi)+g(\pi)=0$.
This is not a valid example because I had misidentified lines.
For example, $2+\pi\mathbb Q$ is a line but its image $\{0,2\}$ is not.
A weaker statement is true: the image of every line is either a line or a set containing the origin and a non-zero rational number.
