Let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff $Ax=0$ implies $x=0$?

Question

Let $X$ be a Banach space and let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff ($Ax=0$ implies $x=0$)?

I can't see how this is implied by the common definitions of Fredholm operators.

Answer 1

It seems you are thinking of the Fredholm alternative, which ensures that the index of $I+K$ is zero whenever $K$ is compact.