$A$ and $B$ are bounded linear operators from the normed linear space $X$ to itself. If $AB$ is invertible are $A$ and $B$ invertible? I think I understand the proof for square matrices, such that $(AB)^{-1}=B^{-1}A^{-1}$, but I'm not sure if I can just say the same for the bounded linear operators A and B.
Does the existence of $(AB)^{-1}:X\rightarrow X$ imply the existence of $B^{-1}$ and $A^{-1}$?
 A: Not necessarily.  For example, define on $\ell^p$
$$
L(x_1,x_2,\dots) = (x_2,x_3,\dots)\\
R(x_1,x_2,\dots) = (0,x_1,x_2,\dots)
$$
Then $LR = \text{id}_{\ell^p}$ is invertible, but neither $L$ nor $R$ are bijective.
You might be able to say something in the case that both $AB$ and $BA$ are invertible.
A: The claim is false in the general infinite dimensional setting. For example, let $X$ be the Banach space $\ell^p$, and define $A,B$ by$$A(x_1,x_2,\ldots)=(x_2,x_3,\ldots),\qquad B(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots).$$Then $AB$ is the identity, but $A$ and $B$ are not invertible.
A: For the question in the title, the answer is No. Take $B : \ell^2 \to \ell^2$ be the right-shift operator
$$
(x_n) \mapsto (0,x_1,x_2,\ldots)
$$
and $A$ to be the left shift
$$
(x_n) \mapsto (x_2,x_3,\ldots)
$$
Then $AB$ is the identity operator, and so invertible, but neither $A$ nor $B$ are invertible.
But later on, you ask about $B^{-1} A^{-1}$ implying the existence of $B^{-1}$ or $A^{-1}$. Not sure what you are asking there.
