Most generic f in R4 who send straight lines in straight lines Suppose $f : \mathbb{R}^4\rightarrow\mathbb{R}^4$ is invertible and sends every straight lines onto straight lines; more precisely (this is a stronger request) every curve $t->At+B$ has to be sent in another $t->A't+B'$, ossia also the "constance" of the velocity has to be preserved. Is $f$ always of the form $f(x) = Ax + b$ for a vector $b$ and a matrix $A$?
If yes, where I can find some references?
MOTIVATION: Correct foundation of special relativity, avoiding every extra- not necessary-hypotesis, and the imprecisions found in a lot of Physics' textbooks.
 A: $\newcommand{\Reals}{\mathbf{R}}$tl; dr: Yes. A sketch of a proof follows.

Suppose $f$ is a bijection of Cartesian four-space and maps lines to lines, preserving constant-velocity paths (but not necessarily an isometry).
Lemma 1: The image of every plane is a plane.
Proof: Assume contrapositively that there exists a plane $P$ and points $q_{1}$ and $q_{2}$ of $f(P)$ such that the segment $\overline{q_{1}q_{2}}$ is not contained in $f(P)$. Because $f$ is a bijection of $\Reals^{4}$, it is a bijection from $P$ to $f(P)$. Consequently, there exist unique points $p_{1}$ and $p_{2}$ of $P$ with $q_{i} = f(p_{i})$. The segment $\overline{p_{1}p_{2}} \subset P$ does not map to $\overline{q_{1}q_{2}}$, so $f$ does not map lines to lines.
Lemma 2: $f$ maps parallel lines to parallel lines.
Proof: Two distinct parallel lines lie in a plane $P$. By Lemma 1, the image lines also lie in a plane, and are disjoint, hence parallel.
Lemma 3: The mapping $f_{0}(x) = f(x) - f(0)$ is linear.
Proof: The mapping $f_{0}$ preserves constant-speed lines (as a composition of transformations with this property), and by construction $f_{0}(0) = 0$. Particularly, if $y$ is an arbitrary vector, then
$$
f_{0}(ty) = tf_{0}(y)\quad\text{for all real $t$.}
$$
Now let $x$ and $y$ be arbitrary vectors. If $t$ is real, the vector $x + ty$ parametrizes a constant-speed line parallel to the line $ty$. By Lemma 2, $f_{0}(x + ty)$ parametrizes a constant-speed line, passing through $f_{0}(x)$ when $t = 0$ and parallel to the line $f_{0}(ty) = tf_{0}(y)$. That is,
$$
f_{0}(x + ty) - f_{0}(x) = tf_{0}(y),
$$
or $f_{0}(x + ty) = f_{0}(x) + tf_{0}(y)$ for all $x$ and $y$ in $\Reals^{4}$ and all real $t$.
Lemma 4: If $f(0) = 0$ and for each standard basis vector $e_{i}$ we have $f(e_{i}) = e_{i}$, then $f$ is the identity.
Proof: By Lemma 3, $f$ is linear. Since $f$ fixes the standard basis, $f$ is the identity.
Lemma 5: There exist an invertible matrix $A$ and a vector $b$ such that $f(x) = Ax + b$ for all $x$.
Proof: Let $b = f(0)$, $f_{0}(x) = f(x) - b$ as in Lemma 3, and let $A$ be the matrix with $i$th column $f_{0}(e_{i}) = f(e_{i}) - f(0)$. Because $f_{0}$ is linear and bijective, $A$ is invertible.
But the mapping $g(x) = A^{-1}f_{0}(x) = A^{-1}(f(x) - b)$ preserves constant-speed lines (again, as a composition of transformations with this property), and by construction $g(0) = 0$ and $g(e_{i}) = e_{i}$ for each $i$. Lemma 4 implies $g$ is the identity, i.e., that $f(x) = Ax + b$.
