 Solution

For a Fredholm operator $T$, define $B$ to invert $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ and to vanish on the range complement. The bounded inverse theorem makes $B$ bounded, andThe errors are finite rank, so $T$ is invertible in the Calkin algebra.

Conversely suppose $BT=I-K$ and $TB=I-L$ for bounded $B$ and compact $K,L$. On $\ker T$, $K$ is the identity. An infinite-dimensional closed kernel would contain an orthonormal sequence, whose images under $K$ have no convergent subsequence, contradicting compactness. Thus the kernel is finite-dimensional.

There is a positive lower bound for $T$ on $(\ker T{)}^{\perp }$. Otherwise choose unit vectors $x_j$ there with $T{x}_j\to 0$. Since $x_j=BT{x}_j+K{x}_j$, compactness gives a norm-convergent subsequence with unit limit $x$. The limit lies in the kernel complement and has $Tx=0$, a contradiction. This lower bound proves closed range: if $T{x}_j$ converges, remove each kernel component; the remaining $x_j$ form a Cauchy sequence and converge to a preimage of the limit.

Finally adjointing $TB=I-L$ gives $B^{\ast }{T}^{\ast }=I-L^{\ast }$. The same kernel argument makes $\ker T^{\ast }$ finite-dimensional. Therefore $T$ is Fredholm. $\boxed{T\text{ is Fredholm}⟺[T]\text{ is invertible modulo compacts}}$, the Atkinson theorem. The finite-dimensional case also satisfies the parametrix formulation directly, though its quotient algebra is the zero algebra.