Let $G\in\mathbb C^{n\times m}$ be the Moore-Penrose inverse of matrix $A\in\mathbb C^{m\times n}$, then we know $G$ satisfies the Penrose conditions $$ \begin{aligned} (1)\quad& AGA=A,\\ (2)\quad& GAG=G,\\ (3)\quad& (AG)^{\rm H} =AG,\\ (4)\quad& (GA)^{\rm H} =GA.\\ \end{aligned} $$ From which we can easily obtain $$ \begin{aligned} (i)\quad& AG=P_{\mathcal R(A),\mathcal N(A^{\rm H}\,)},\\ (ii)\quad& GA=P_{\mathcal R(A^{\rm H}\,),\mathcal N(A)}, \end{aligned} $$ where $P_{L,M}$ is the orthogonal projector on $L$ along $M=L^\perp$, $\mathcal R(A)$ and $\mathcal N(A)$ are range and null space of $A$, respectively. Someone says conditions (1)-(4) are equivalent to (i) and (ii). I have known how to obtain (1), (3), (4) by (i) and (ii). To obtain (1) I just need to use the fact that $\mathcal R(A)=\mathcal R(AG)$, then for any $x\in \mathbb C^n$ we have $Ax=AG(Ax)$ and thus $A=AGA$. But it seems that this method is not valid for proving $GAG=G$ since we cannot obtain $\mathcal R(G)=\mathcal R(GA)$ to provide $Gx=GAGx$ for all $x$ directly. Any other method to derive (2) from (i) and (ii)?
UPDATE: Maybe this claim is wrong, we should replace the condition (ii) by $GA=P_{\mathcal R(G),\mathcal N(G^{\rm H})}$. Then the method above is valid.
