Let $T$ be a bounded compact operator on a Banach space; if $S$ is a bounded left inverse of $I-T$ , then $(I-T)S=I$

Question

Let $X$ be a Banach space . Let $S$ be a bounded operator on $X$ and $T$ be a compact operator on $X$.

How to prove that if $S(I-T)=I$, then $(I-T)S=I$ and then $I-(I-T)^{-1}=I-S$ is a compact operator ?

I can see that $S$ is an open map but I can't seem to be able to use the compactness of $T$.

Please help. Thanks in advance.

Answer 1

$I-T$ is injective since $S(I-T)=I$, since $T$ is compact we deduce that $I-T$ is invertible (every non zero element in the spectrum of a compact operator is an eigenvalue), thus its left inverse is equal to its right inverse.

https://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators#Statement