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 $A^T$.

翻译：标题陈述通常不成立，除非$C$和$R$具有满秩。假设$m\times r$矩阵$C$有$r$个独立列（秩为$r$），$r\times n$矩阵$R$有$r$个独立行（秩为$r$）。此时伪逆$C^+$是$C$的左逆，伪逆$R^+$是$R$的右逆。证明$A^+ = R^+C^+$的最简方法是通过验证确定任意矩阵$A$伪逆$A^+$的四条彭罗斯恒等式。我们的目标是基于基本原理给出$A^+ = R^+C^+$的另一种证明。我们从任意秩为$r$的$m\times n$矩阵$A$相关联的四个基本子空间入手，即$A$与$A^T$的列空间和零空间。随后对$A^+ = R^+C^+$的证明表明，该矩阵正确作用于$A$列空间中的每个向量以及$A^T$零空间中的每个向量。