My textbook says that

For n>=2, consider the set

$\displaystyle A_{n-1} = \{ x = (x_{1},....,x_{n}) \in R^{n} : x_{n}=0\}$ with metric induced by d (usual metric). Then $\displaystyle A_{n-1}$ is a subset of $\displaystyle R_{n}$ and is isometric to $\displaystyle R_{n-1}$ under correspondence

(a) $\displaystyle (x_{1},...,x_{n-1}, 0) \leftrightarrow (x_{1},...,x_{n-1})$

My question is

Isometry should be a bijective function between $\displaystyle A_{n-1}$ and $\displaystyle R_{n-1}$. (a) does not look bijective for me if each $\displaystyle x_{k}$ in $\displaystyle A_{n-1}$ maps to $\displaystyle x_{k}$ in $\displaystyle R_{n-1}$. ( how do we map 0? it should be bijective by definition of isometry)