Affine transformation, if $L_1, L_2 - $ skew lines, $f(L_1), \ f(L_2) $ are parallel, then $f$ is not injective

Question

Could you tell me how to prove that if $f$ is affine transformation, $L_1, L_2 $ are skew lines, $f(L_1), \ f(L_2) $ are parallel, then $f$ is not injective?

Answer 1

Suppose $L_i = a_i + \mathbb Rv_i$ are the two lines and $f$ is given by $x \mapsto b + Ax$ with linear $A$. Then we have $$ f(L_i) = b + Aa_i + \mathbb RAv_i, \quad i = 1,2 $$ Now, $L_1$ and $L_2$ being skew means $v_1$ and $v_2$ are linear independent, $f(L_1) \parallel f(L_2)$ gives that $Av_1$ and $Av_2$ are linear dependent, say $\lambda_1Av_1 + \lambda_2 Av_2 = 0$, for some $(\lambda_1, \lambda_2) \ne (0,0)$. As $\lambda_1v_1 + \lambda_2v_2 \ne 0$, $A$ is not injective, hence $f$ is not.