The Pseudoinverse of $A=CR$ is $A^+=R^+C^+$ (?)

The statement in the title is not generally true, unless $C$ and $R$ have full rank. Then the $m$ by $r$ matrix $C$ is assumed to have $r$ independent columns (rank $r$). The $r$ by $n$ matrix $R$ is assumed to have $r$ independent rows (rank $r$). In this case the pseudoinverse $C^+$ is the left inverse of $C$, and the pseudoinverse $R^+$ is the right inverse of $R$. The simplest proof of $A^+ = R^+C^+$ verifies the four Penrose identities that determine the pseudoinverse $A^+$ of any matrix $A$. Our goal is a different proof of $A^+ = R^+C^+$, starting from first principles. We begin with the four fundamental subspaces associated with any $m$ by $n$ matrix $A$ of rank $r$. Those are the column space and nullspace of $A$ and $A^T$. The proof of $A^+ = R^+C^+$ then shows that this matrix acts correctly on every vector in the column space of $A$ and on every vector in the nullspace of $A^T$.