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For $T$ compact, $I-T$ left or right invertible implies $I-T$ invertible
Let $S\in B(X)$ be a bounded linear operator from $X$ onto $X$ and let $T\in K(X)$ be a compact linear operator from $X$ onto $X$. Then
$$
S(I-T)=I \iff (I-T)S=I.
$$
I don't know if we need the fact ...
