It is well known that every operator $A$ on a finite-dimensional space has a polar decomposition $A = UP$ where $U$ is a unitary operator and $P = \sqrt{A^*A}$. Further, when $A$ is not invertible the choice of $U$ is not unique. However, in the below wikipedia article it states that even if $A$ is not invertible, there exists a "natural" choice for $U$. It says the following:
$AP^+$ is a unitary operator from the range of $A$ to itself, which can be extended by the identity on the kernel of $A^∗$. The resulting unitary operator $U$ then yields the polar decomposition of $A$
where $P^+$ is the Moore-Penrose inverse of $P$. However, I don't see how $AP^+$ defines a unitary operator from the range of $A$ to itself. It seems to me that $AP^+$ actually defines a unitary operator from $\textrm{Ker}(P)^\perp = \textrm{Ran}(P) = \textrm{Ran}(A^*)$ to $\textrm{Ran}(A)$. Hence, if $AP^+$ is extended to be the identity on $\textrm{Ker}(A^*)$ then the extension will not necessarily be unitary.
Is what I say correct? If so, it seems like it is not possible to have a "natural" choice for $U$?
https://en.wikipedia.org/wiki/Square_root_of_a_matrix#Polar_decomposition