Atkinson's theorem states:
$T ∈ L(H)$ is a Fredholm operator if and only if T is invertible modulo compact perturbation, i.e. $TS = I + C_{1}$ and $ST = I + C_{2}$ for some bounded operator S and compact operators $C_{1}$ and $C_{2}$.
$L(H)$= espace of linear and bounded operators.
Let $A$ and $B$ two fredholm operators. How to prove that $A\oplus B\ $ is a fredholm operator using the the Atkinson's theorem?
$\newcommand{\dim}{\text{dim}}\newcommand{\codim}{\text{codim}}\newcommand{\Ind}{\text{Ind}}\newcommand{\Ker}{\text{Ker}}\newcommand{\Ran}{\text{Ran}}$Notice that $\Ker(A\oplus B) = \Ker(A)\oplus \Ker(B)$, hence a finite dimensional space, and $\Ran(A\oplus B) = \Ran(A)\oplus \Ran(B)$, hence a finite co-dimensional space. Therefore $A\oplus B$ is Fredholm and $$ \Ind(A\oplus B) = \dim(\Ker(A\oplus B)) - \codim(\Ran(A\oplus B)) = $$ $$ = \dim(\Ker(A)) + \dim(\Ker(B)) - \codim(\Ran(A)) - \codim(\Ran(B)) = $$ $$ = \Ind(A) + \Ind(B). $$
Now, since $\Ker(A^*)^\perp = \Ran(A)$, we have $$ \Ind(A) = \dim(\Ker(A)) - \dim(\Ker(A^*)), $$ from where it follows that $\Ind(A^*) = -\Ind(A)$.