Atkinson's theorem

Remarks

If $T$ is Fredholm, choose closed complements $X = \ker T \oplus X_0$ and $Y = T(X) \oplus Y_0$, with $Y_0$ finite-dimensional. Then $T$ maps $X_0$ bijectively onto $T(X)$, and $S$ can be taken as its inverse on $T(X)$ and $0$ on $Y_0$. With this choice, $I_X - ST$ and $I_Y - TS$ are projections onto $\ker T$ and $Y_0$, so they are even finite-rank operators. Conversely, if $ST = I_X + K$ with $K$ compact, then $\ker T \subseteq \ker(I_X + K)$ is finite-dimensional, and $TS = I_Y + K'$ shows that $T(X) \supseteq (I_Y + K')(Y)$ has finite codimension. For a Hilbert space $H$, the theorem says that $T \in B(H)$ is Fredholm if and only if its image is invertible in the Calkin algebra $B(H)/K(H)$. Example: the right shift $S$ on $\ell^2$ and the left shift $S^*$ satisfy $S^* S = I$ and $S S^* = I - P$, where $P$ is the rank-one projection onto the first coordinate, so $S$ is Fredholm.

Going deeper

For a student

The theorem turns Fredholm theory into algebra: Fredholm operators are the preimages of the invertible elements of a quotient algebra. From this view, it follows at once that compact perturbations of Fredholm operators are Fredholm, and that products of Fredholm operators are Fredholm. It also leads to the essential spectrum of an operator $T$ on a Hilbert space, the spectrum of its image in the Calkin algebra.