Pseudo Inverse of product of Matrices Let $A$ and $B$ are two matrices where $A \in \mathbb{R}^{m\times p}$ and $B \in \mathbb{R}^{p\times n}$ and both $A$ and $B$ are full rank matrices  Now I really want to know in what cases
$(AB)^+ = B^+A^+$  ,where $A^+$ is Moore-Penrose Pseudo-inverse of $A$
Here $m,p$ and $n$ are in any order like $m<p<d$ or $m>p<d$ etc.
 A: Wikipedia

Example
Define matrices
$$
\mathbf{A} = 
\left[
\begin{array}{cc}
 0 & 1 \\
 3 & 2 \\
 0 & 2 \\
\end{array}
\right]
, \quad
%
\mathbf{B} =
\left[
\begin{array}{ccc}
 0 & 3 & 0 \\
 1 & 2 & 2 \\
\end{array}
\right]
, \quad
%
\mathbf{C} = \mathbf{A} \mathbf{B} =
\left[
\begin{array}{rrr}
 1 & 2 & 2 \\
 2 & 13 & 4 \\
 2 & 4 & 4 \\
\end{array}
\right]
$$
$$
\mathbf{A}^{\dagger} = \frac{1}{15}
\left[
\begin{array}{rrr}
 -2 & 5 & -4 \\
 3 & 0 & 6 \\
\end{array}
\right]
, \quad
%
\mathbf{B}^{\dagger} = \frac{1}{15}
\left[
\begin{array}{rr}
 -2 & 3 \\
 5 & 0 \\
 -4 & 6 \\
\end{array}
\right]
, \quad
%
\mathbf{C}^{\dagger} = \frac{1}{225}
\left[
\begin{array}{rrr}
 13 & -10 & 26 \\
 -10 & 25 & -20 \\
 26 & -20 & 52 \\
\end{array}
\right]
$$
Test premise
Does $\mathbf{C}^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$?
$$
\begin{align}
\mathbf{B}^{\dagger}\mathbf{A}^{\dagger} &= \frac{1}{15}
\left[
\begin{array}{rr}
 -2 & 3 \\
  5 & 0 \\
 -4 & 6 \\
\end{array}
\right]
\frac{1}{15}
\left[
\begin{array}{rrr}
 -2 & 5 & -4 \\
 3 & 0 & 6 \\
\end{array}
\right] \\[3pt]
%%
& =
%%
\frac{1}{225}
\left[
\begin{array}{rrr}
 13 & -10 & 26 \\
 -10 & 25 & -20 \\
 26 & -20 & 52 \\
\end{array}
\right] \\
&= \mathbf{C}^{\dagger}
\end{align}
$$
Conclusion
$$\therefore \qquad \left(\mathbf{A}\mathbf{B}\right)^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$$
