= Solution
We prove <compact perturbation invariance of Fredholm operators> using a two-sided inverse modulo compact operators. If $L$ is <Fredholm>, decompose
$$
H=\ker L\oplus(\ker L)^\perp,\qquad
H=\operatorname{ran}L\oplus(\operatorname{ran}L)^\perp.
$$
The restriction of $L$ from $(\ker L)^\perp$ to its closed range is a bounded bijection, so the <bounded inverse theorem> supplies a bounded inverse there. Extend that inverse by zero on $(\operatorname{ran}L)^\perp$ to obtain a bounded operator $B$. If $P,Q$ are the <orthogonal projections> onto $\ker L$ and $(\operatorname{ran}L)^\perp$, respectively, then
$$
BL=I-P,\qquad LB=I-Q.
$$
Both $P$ and $Q$ have finite rank.
Set $T=L+K$. Products of a <compact operator> and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hence
$$
BT=I+C,\qquad TB=I+D,\qquad
C=-P+BK,\quad D=-Q+KB,
$$
where $C,D$ are compact.
Here is why these identities force $T$ to be <Fredholm>. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ with
$$
\|Th\|\geq c\|h\|\qquad(h\perp\ker T).
$$
Otherwise unit vectors $h_n\perp\ker T$ could satisfy $Th_n\to0$. From $h_n=-Ch_n+BTh_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp\ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $Th_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.
For finite codimension, take the <adjoint operator> of $TB=I+D$, giving $B^*T^*=I+D^*$. The operator $D^*$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in <operator norm> to $D^*$, and norm limits of compact operators are compact. On $\ker T^*$, $D^*=-I$, so this kernel is finite-dimensional. Finally,
$$
(\operatorname{ran}T)^\perp=\ker T^*.
$$
Because the range is closed, its <cokernel> is isomorphic to this finite-dimensional orthogonal complement. This verifies all three <Fredholm operator> conditions for $L+K$.
For the converse, start with $L+K$ and perturb by the compact operator $-K$. \b[Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm], with no self-adjointness assumption.
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