Fredholm operator

Remarks

If $T$ is Fredholm, then its image $T(X)$ is automatically closed in $Y$. Example: let $S$ be the right shift on $\ell^2$, $S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$. Its kernel is $\set{0}$, and its image is $\set{x \in \ell^2 : x_1 = 0}$, of codimension $1$, so $S$ is Fredholm. Non-example: the diagonal operator $D(x_n) = (x_n / n)$ on $\ell^2$ has kernel $\set{0}$. Its image contains every finitely supported sequence, so it is dense, but it misses $(1/n)$, so it is not closed. By the previous remark, the image of a bounded operator with finite-dimensional cokernel is closed, so clause 2 fails. If $K \colon X \to X$ is a compact operator, then $I - K$ is Fredholm. Every linear map between finite-dimensional spaces is Fredholm.