Let $X$ be a Banach Space and $Y$ a dense subset of $X$.
An operator $T:X \to X$ is said to be Fredholm if it has closed range, $\dim \ker(T)<\infty$ and $\dim coker(T) < \infty$.
Here is my question: If $S:X \to X$ is an injective operator and its restriction to $Y$, $S:Y \to Y$ is Fredholm of index $0$ (In fact I know that it is invertible), does this imply that $S:X \to X$ is Fredholm of index $0$? (and hance invertible, as it is injective).
Thanks in advance.