Remark

Let E, F and G be locally convex spaces, $u: E\rightarrow F$ and $v: F\rightarrow G$ continuous linear mappings.

1) Suppose that $u$ is a Fredholm mapping and let $u_1$ be a quasi-inverse of $u$. Since $u$ is a quasi-inverse of $u_1$, the mapping $u_1$ is a Fredholm mapping.

2) Suppose that $u$ and $v$ are mappings of Fredholm, and let $u_1$ (resp. $v_1$) be a quasi-inverse of $u$ (resp. $v$). Then $v\circ u$ is a mapping of Fredholm from E into G and $u_1\circ v_1$ is a quasi-inverse of $v\circ u$. Indeed, one calculates

3) Suppose that $u$ and $v\circ u$ are mappings of Fredholm, and let $w_1$ be a quasi-inverse of $v\circ u$. Then $v$ is a mapping of Fredholm and $u\circ w_1$ is a quasi-inverse of $v$.

Indeed, $w_1$ is a mapping of Fredholm according to the first remark. Let $u_1$ be a quasi-inverse of $u$; according to the second remark, $u\circ w_1$ is a mapping of Fredholm and $(v\circ u)\circ u_1$ is a quasi-inverse of it. We have $u\circ u_1\equiv 1_F$, whence $v\circ u\circ u_1\equiv v$; this proves the assertion.

4) Suppose that $v$ and $v\circ u$ are mappings of Fredholm, and let $w_1$ be a quasi-inverse of $v\circ u$. Then $u$ is a mapping of Fredholm and $w_1\circ v$ is a quasi-inverse of $u$.

The proof is analogous to that of the preceding remark.