Given an injective linear map $A:\mathbb R^n\to\mathbb R^m$, there exists a linear map $\hat{A}$ which is a left-inverse of A. What can be said about $\hat{A}(\mathbb R^m \backslash (Image(A))$?
In other words, given $ x \in \mathbb R^m$ with $x \notin Image(A)$, can it be said that $\hat{A}(x)$ is equal to $\hat{A}(y)$ where y is the orthogonal projection of x onto the Image of the linear map $A$?
On
Suppose that $\{e_1, \dots, e_n\}$ is a basis of $\mathbb R^n$. By hypothesis, $\{Ae_1, \dots, Ae_n\}$ is a linearly independent family of vectors of $\mathbb R^m$. You can complement this family with $\{f_{n+1}, \dots, f_m\}$ in order to get a basis $\{Ae_1, \dots, Ae_n,f_{n+1}, \dots, f_m\}$.
Name $F$ the linear space spanned by $\{Ae_1, \dots, Ae_n\}$ and $G$ the one spanned by $\{f_{n+1}, \dots, f_m\}$.
$\hat{A}$ can be any linear map providing that $\hat{A}(Ae_i) = e_i$ for $1 \le i \le n$.
Finally, for $y \in \mathbb R^m$, $\hat{A}(y) = \hat{A}(x)$ where $x$ is the oblique projection of $y$ onto $F$ using the direction $G$.
There are many possible $\hat A$. They differ exactly in how $A\hat A$ projects everything outside the image of $A$ onto the image of $A$.
More specifically, the kernel of $A\hat A$ (and thus of $\hat A$) can be any subspace of $\Bbb R^m$ fulfilling the two requirements
Given any such kernel, the rest of $A\hat A$ (and thus of $\hat A$) is uniquely determined. This is because for any $v\in\Bbb R^m$, there is a unique $a$ in the image of $A$ and $k$ in the kernel of $\hat A$ such that $v=k+a$. This gives $A\hat A v=a$, and thus $\hat Av=\hat AA\hat Av=\hat Aa$, which is uniquely determined by the injectivity of $A$.
Letting $\hat A$ be the Moore-Penrose inverse indeed corresponds to the option where the kernel is orthogonal to the image. This is because $A\hat A$ is required to be symmetric, and thus the eigenspace with eigenvalue 1 (i.e. the image) must be orthogonal to the eigenspace with eigenvalue 0 (i.e. the kernel).