Throughout this question, $H$ is an infinite-dimensional, separable Hilbert space. You may use, without proof, any theorems about compact operators that you require.

Define a Fredholm operator $T$, on a Hilbert space $H$, and define the index of $T$.

(i) Prove that if $T$ is Fredholm then $\operatorname{im} T$ is closed.

(ii) Let $F \in \mathcal{B}(H)$ and let $F$ have finite rank. Prove that $F^{*}$ also has finite rank.

(iii) Let $T=I+F$, where $I$ is the identity operator on $H$ and $F$ has finite rank; let $E=\operatorname{im} F+\operatorname{im} F^{*}$. By considering $T \mid E$ and $T \mid E^{\perp}$ (or otherwise) prove that $T$ is Fredholm with ind $T=0$.

(iv) Let $T \in \mathcal{B}(H)$ be Fredholm with ind $T=0$. Prove that $T=A+F$, where $A$ is invertible and $F$ has finite rank.

[You may wish to note that $T$ effects an isomorphism from $(\operatorname{ker} T)^{\perp}$ onto $\operatorname{im} T$; also ker $T$ and $(\operatorname{im} T)^{\perp}$ have the same finite dimension.]

(v) Deduce from (iii) and (iv) that $T \in \mathcal{B}(H)$ is Fredholm with ind $T=0$ if and only if $T=A+K$ with $A$ invertible and $K$ compact.

(vi) Explain briefly, by considering suitable shift operators on $\ell^{2}$ (i.e. not using any theorems about Fredholm operators) that, for each $k \in \mathbb{Z}$, there is a Fredholm operator $S$ on $H$ with ind $S=k$.