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A bounded operator $T:H\to H$ is a Fredholm operator when its range is closed and both $\ker T$ and $\mathrm{ran}(T{)}^{\perp }=\ker T^{\ast }$ have finite dimension. Its Fredholm index isFor index zero, choose an isomorphism $J:\ker T\to \ker T^{\ast }$ and extend it by zero on $(\ker T{)}^{\perp }$. The restricted map $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ 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 $A+tK$, $0\le t\le 1$, 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 $\mathrm{ind}A=0$. This invokes the actual parametrix and block proofs below, not an unproved perturbation assertion. The claimed equivalence holds.

On the Hardy space of the circle, a continuous symbol gives a Fredholm Toeplitz operator exactly when it never vanishes. Nonvanishing gives a parametrix $T_{1/f}$ modulo compact semicommutators. Write $f(z)=z^n{e}^{h(z)}$, where $n$ is its counterclockwise winding number and $h$ is continuous. Homotopy through nonvanishing symbols and local constancy of the Fredholm index reduce the index to that of the shift $T_{z^n}$, namely $-n$. If $f$ vanishes at a point, normalized Hardy reproducing kernels concentrating there are weakly null unit vectors on which $T_f$ tends to zero, contradicting a Fredholm parametrix.