Do isometries send lines to lines?

My question is if any isometry $f:V\to W$ between real normed spaces sends lines to lines. I've seen several questions/answers about this but only in euclidean spaces.

So I thought it was false on general (real) normed spaces. However I found this theorem of Mazur-Ulam: any surjective isometry $f:V\to W$ is an affine map, hence it maps lines to lines.

But if my isometry is not surjective, would this still apply? I think that considering the image space $f(V)$ it would be the same, because $f:V\to f(V)$ is affine and any line $L$ would be sent to a line $f(L)$ in $f(V)$ which is also a line in $W$.

Is this correct?

Thank you.

No, this is not correct, because $f(V)$ doesn't have to be a vector space.
Consider, for instance, the norm $\bigl\|(x,y)\bigr\|=\max\{|x|,|y|\}$ in $\mathbb{R}^2$ and the usual norm in $\mathbb R$. Now, consider the map$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb{R}^2\\&x&\mapsto&\bigl(x,\sin(x)\bigr).\end{array}$$Then $f$ is an isometry, but it is not affine and it does not send lines into lines.
Of course, the only important property of the sine function here is that $(\forall x\in\mathbb{R}):|\sin x|\leqslant|x|$. I could have even used a function which is discontinuous everywhere, such has $\chi_{\mathbb Q}$.
• The codomain is $\mathbb{R}^2$. How do you want to find a surface there? Besides, the only line in $\mathbb R$ is $\mathbb R$ itself and its image is the graph of the sine function. – José Carlos Santos Jul 23 '17 at 7:26