Let $A$ be bounded a Fredholm operator on a separable Hilbert space, and let $B$ be another bounded operator on a separable Hilbert space.
Can anyone provide an example where $AB^\ast$ is Fredholm such that $\mathrm{index}(AB^\ast)=0$ yet $B$ is not Fredholm?
If $AB^\ast$ is Fredholm then it has a parametrix, say $C$, such that $CAB^\ast – 1,AB^\ast C-1$ are compact, so $CA$ could be a parametrix of $B^\ast$, but we don't know that $B^\ast CA-1$ is compact…
Best Answer
The answer is no, that is, $B$ is necessarily Fredholm.
The reason is that the image of $A$ in the Calkin algebra $B(H)/K(H)$ is invertible, so any right inverse of it must be itself invertible.
To follow through with the OP's attempt, let $L=AB^*C-1$, hence a compact operator, and let $D$ be a parametrix of $A$, so that $DA=1+K$, where $K$ is compact. Then $$ B^*CA = (DA-K)B^*CA = DAB^*CA - KB^*CA = $$$$ = D(L+1)A - KB^*CA = DLA + DA - KB^*CA = DLA + 1+K - KB^*CA\in 1+\mathscr K(H). $$