Proposition 12

Let E and F be Banach spaces and $u$ an element of $\mathscr{Q}\mathscr{M}(E; F)$. Every $v\in \mathscr{L}(E; F)$ such that $\|v-u\|<((u))$ belong to $\mathscr{Q}\mathscr{M}(E; F)$ and satisfy the relations

dim Ker($v$)$\leqslant$ dim Ker($u$),

dim Coker($v$)$\leqslant$ dim Coker($u$),

ind($v$) $=$ ind($u$).

Since $u$ is strict, one has $((u))>0$. Let B denote the open ball with center $v$ and radius $((u))$ in $\mathscr{L}(E; F)$. For every $v\in$ B, one has dim Ker($v$)$\leqslant$ dim Ker($u$)(III, p. 65, prop. 9) and $v\in \mathscr{Q}\mathscr{M}(E; F)$ (prop. 11).

For $r\in \mathbf{Z}\cup \{+\infty \}$, let us denote by $B_r$ the set of elements $v\in B$ such that ind($v$) $=r$. If $r\in \mathbf{Z}$, the set $B_r$ is the set of Fredholm mappings from E into F of index $r$ which belong to B (III, p. 52, Prop. 11); it is open in B by Theorem 1 of III, p. 58.

Let us prove that the sets $B_r$ are closed in B. Let $v\in B$ be an adherent point of $B_r$. Since the sets $B_s$, for $s\in \mathbf{Z}$, are open in B and pairwise disjoint, we have $v\in B_r$ or $v\in B_{+\infty}$.

Suppose $v\in B_{+\infty}$. Let $n$ be the dimension of Ker($u$). Choose a vector subspace T of F of dimension $r+ 2n+ 1$ whose intersection with Im($v$) is reduced to 0, and a topological complement S of Ker($v$) in E (III, p. 55, Prop. 1). Consider the continuous linear mapping $f: (s, t)\mapsto v(s) +t$ of $S\times T$ into F. It is injective. The linear mapping $v$ is a strict morphism since $v$ belongs to $B\subset \mathscr{Q}\mathscr{M}(E; F)$. The image of $v$ is therefore closed (EVT, IV, p. 28, Theorem 1). By Prop. 1 of III, p. 39, the restriction of $v$ to S is a strict morphism with closed image from S into F, and $f$ is a strict morphism with closed image from $S\times T$ into F.

By Prop. 10, there exists a neighbourhood U of $v$ in B such that, for every $w\in U$, the linear mapping $(s, t)\mapsto w(s) +t$ of $S\times T$ into F is injective. Let $w$ be an element of U. Then Ker($w$)$\cap S =\{0\}$, so that $w$ defines by restriction and passing to the quotient an injective linear mapping of Ker($w$) into $E/S$, which implies that Ker($w$) is of dimension at most $n$. We also have $w(S)\cap T =\{0\}$, hence

codim$_F$(Im($w$))$\geqslant$ dim(T) $-$ codim$_E(S) =r+n+ 1$,

this yields ind($w$)$\geqslant r+ 1$ and contradicts the hypothesis that $v$ is adherent to $B_r$.

If $r$ is an element of $\mathbf{Z}$ such that $B_r$ is nonempty, one has $B_r= B$ since $B_r$ is open and closed in B and B is connected. If $B_r$ is empty for every $r\in \mathbf{Z}$, one has ind($v$) $= +\infty$ for every $v\in B$. The mapping $v\mapsto$ ind($v$) is therefore constant on B. Lastly, for every $v\in B$, one has

dim Coker($v$) $=$ ind($v$) $+$ dim Ker($v$)

$\leqslant$ ind($u$) $+$ dim Ker($u$) $=$ dim Coker($u$).

This concludes the proof.