If $A,B\in\mathcal{B}(\mathcal{H}):AB=BA$ and $A$ does not have an inverse, then $AB$ does not have an inverse

Question

This question has already been asked at: Inverse of product of operators but the post does not help me: If we take $C=B(AB)^{-1}$, then we get that $C$ is a right inverse of $A$ as $AC = AB(AB)^{-1} = I$. But what about the left inverse? Why is $CA = B(AB)^{-1}A = I$? Is there a way to show that if $AB$ were to be invertible then $B$ commutes with $(AB)^{-1}$?

Answer 1

Assuming that $AB$ is invertible, it can be shown that $A$ is invertible.

$C=B(AB)^{-1}\implies AC= I\implies C$ is a right inverse of $A$.

So $A$ is onto.

For any $u,v\in H$, $Au=Av\implies BAu= BAv\implies ABu= ABv\implies (AB)^{-1}(AB)u= (AB)^{-1}(AB)v\implies u=v.$

It follows that $A$ is injective.