= Solution
For a <Fredholm operator> $T$, define $B$ to invert $T:(\ker T)^\perp\to\operatorname{ran}T$ and to vanish on the range complement. The <bounded inverse theorem> makes $B$ bounded, and
$$
BT=I-P_{\ker T},\qquad TB=I-P_{\ker T^*}.
$$
The 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 $Tx_j\to0$. Since $x_j=BTx_j+Kx_j$, 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 $Tx_j$ 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 $B^*T^*=I-L^*$. The same kernel argument makes $\ker T^*$ finite-dimensional. Therefore $T$ is <Fredholm>. \b[$\boxed{T\text{ is Fredholm}\iff[T]\text{ is invertible modulo compacts}}$], the <Atkinson theorem>. The finite-dimensional case also satisfies the parametrix formulation directly, though its quotient algebra is the zero algebra.
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