Infinite dimensional operator inverse A is a linear operator on V and there exist a single operator B on V such that AB = I or BA = I. Prove that then A is monomorfic and epimorfic.
On infinite dimensions, left and right inverses need not be the same. But how to show that given at least one exists and is unique, the operator is regular?
 A: A linear operator is injective if and only if it has a left-inverse, and it is surjective if and only if it has a right-inverse.
To show that the uniqueness of the one-sided inverse implies the invertibility of $A$, show that if $A$ is not invertible, but has either a left- or a right-inverse, the one-sided inverse is not unique.
Say $A$ has a left-inverse, but is not invertible. Then $A$ is injective, and not surjective, so $A(V)$ is a proper subspace of $V$. Let $C$ be a complementary subspace of $A(V)$, i.e. $V = A(V)\oplus C$. A left-inverse of $A$ is then only determined on the subspace $A(V)$, but on $C$, it can be arbitrary. Then there are two different left-inverses given by $B_1(c) = 0$ and $B_2(c) = c$ for all $c\in C$. Since by assumption $C\neq\{0\}$, $B_1\neq B_2$, and $A$ has more than one left-inverse.
If we assume that $A$ has a right-inverse, but is not invertible, then $A$ is surjective but not injective, i.e. $\ker A \neq \{0\}$. If $B_1$ is a right-inverse of $A$, what modifications of $B_1$ lead to other right-inverses of $A$?
