Counterexample for Mazur–Ulam theorem 
We know,  Mazur–Ulam theorem states that if $V$ and $W$ are normed spaces over $\mathbb{R}$ and the mapping
  $f\colon V\to W$
  is a surjective isometry, then $f$ is affine.

Can somebody say counterexample when $f\colon V\to W$ is no surjective then f is not affine.

Define $f \colon \mathbb R \to \mathbb R^2$ by $f(x) = (x, \left|x\right|)$. If $\mathbb R$ is endowed with the usual norm given by the absolute value, and $\mathbb R^2$ with the maximum norm $\left|x\right|_\infty = \max\{|x_1|, |x_2|\}$, then $f$ is an isometry.
