= Solution
A bounded operator $T:H\to H$ is a <Fredholm operator> when its range is closed and both $\ker T$ and $\operatorname{ran}(T)^\perp=\ker T^*$ have finite dimension. Its <Fredholm index> is
$$
\operatorname{ind}T=\dim\ker T-\dim\ker T^*.
$$
For index zero, choose an isomorphism $J:\ker T\to\ker T^*$ and extend it by zero on $(\ker T)^\perp$. The restricted map $T:(\ker T)^\perp\to\operatorname{ran}T$ is a bounded isomorphism, by the <bounded inverse theorem>. The <finite-rank operator> $J$ fills exactly the missing kernel-to-cokernel block. Thus $T+J$ is invertible and $T=(T+J)-J$ is an invertible operator plus a <compact operator>.
Conversely suppose $T=A+K$ with $A$ invertible and $K$ compact. The <Atkinson theorem> proved in part (b) makes every $A+tK$, $0\leq t\leq1$, <Fredholm>: its quotient image is always $[A]$. The block argument in part (c) proves <local constancy of the Fredholm index>, so its index is constant along this interval and equals $\operatorname{ind}A=0$. This invokes the actual parametrix and block proofs below, not an unproved perturbation assertion. \b[The claimed equivalence holds.]
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