Show that an operator is well-defined

Question

Let $v\in B(H)$, Define $u:|v|H\to H$ such that $u(|v|\xi) = v\xi$ . To show the map $u$ is well-defined, the author writes $$\||v|\xi\|^2=\langle v^*v\xi,\xi\rangle = \|v\xi\|^2$$ But I do not know why the above equality shows that the map is well-defined. Please help me. Thanks.

Answer 1

The equality implies that if $\lvert v\rvert\xi_1 = \lvert v\rvert\xi_2$, then also $v\xi_1 = v\xi_2$, so it is immaterial which element $\xi\in \lvert v\rvert^{-1}(\eta)$ is chosen to define $u(\eta) = v(\xi)$. All choices yield the same result. Thus $u = v\circ \lvert v\rvert^{-1}$ is well-defined although in general $\lvert v\rvert^{-1}$ is not a map $\lvert v\rvert H\to H$.