 Solution

Fix a Fredholm operator $T$. Split its domain as $X\oplus N=(\ker T{)}^{\perp }\oplus \ker T$ and its target as $Y\oplus M=\mathrm{ran}T\oplus \ker T^{\ast }$. For a sufficiently small norm perturbation $R$, writewhere $A:X\to Y$ remains invertible by the Neumann series. Invertible triangular row and column operations reduce this matrix toThe second block maps the finite-dimensional space $N$ into $M$. Hence the perturbed operator has closed range, finite kernel and cokernel, andsince the rank of that finite block cancels from the difference. Thus the Fredholm set is open and its index is locally constant, hence norm continuous.

If $S,T$ are Fredholm, their product is Fredholm by multiplying their invertible images in the Calkin algebra. For index additivity use the following finite-dimensional exact sequence:Here $q$ is reduction modulo $\mathrm{ran}T$, the induced $S$-map sends $[x]$ to $[Sx]$, and the last map is the quotient by the larger range $\mathrm{ran}S$. These maps are well-defined. For example, $\ker q=\ker S\cap \mathrm{ran}T$ is the image of the preceding map. If $Sx\in \mathrm{ran}ST$, subtracting a suitable $Ty$ makes $x-Ty\in \ker S$, proving exactness at $\mathrm{coker}T$. The other positions follow directly from the definitions. Alternating dimensions in this exact sequence gives