Prove: For T Compact, left or right invertible implies invertible

[CornMuffin]

Homework Statement

$X$ is a Banach space
$S\in B(X)$ (Bounded linear transformation from X to X)
$T\in K(X)$ (Compact bounded linear transformation from X to X)

$S(I-T)=I$ if and only if $(I-T)S=I$

The question also asks to show that either of these equalities implies that $I-(I-T)^{-1}$ is compact.

Homework Equations

The Attempt at a Solution

I have tried using the adjoint, cause S is invertible if and only if S* is invertible. but that didn't get me anywhere.

If there happens to be a theorem that says ST = TS, then it would be easy, but i couldn't find anything like that. For the second part:
$S(I-T)=I\Rightarrow S-ST=I \Rightarrow S=I+ST$
$I-(I-T)^{-1} = I-S = I-(I+ST) = ST$
And ST is compact since T is compact

 

[morphism]

Hint: Fredholm alternative.