Brezis shows that a compact operator $T \colon E \to E$ ($E$ a normed vector space) satisfies $0 \in \sigma(T)$ by showing that if $0 \notin \sigma(T)$, then $T$ is bijective and hence $I = T \circ T^{-1}$ is compact, a contradiction to infinite dimensionality. How do I see that 'hence $T \circ T^{-1}$ is compact'?
EDIT: are there any related results when $T$ is only surjecive or injective?
This follows from the more general result
Indeed, since $S$ is bounded it maps bounded sets in $E$ to bounded sets in $E$. Since $T$ is compact, it maps bounded sets in $E$ to precompact sets in $E$. Thus $TS$ maps bounded sets in $E$ to precompact sets in $E$, and therefore $TS$ is compact.
Now assuming $T$ is compact and $0\notin\sigma(T)$, it has a bounded inverse $T^{-1}$ and applying the above result with $S=T^{-1}$, we see that $I=TT^{-1}$ is compact.