When I read the article "Index Theory" in Handbook of global analysis, I meet a result as below(Corollary 2.13):

If every $F_0\in \mathcal {F}(H_1,H_2)$, there is an open neighborhood $U_0\subseteq \mathcal {B}(H_1,H_2)$, such that $F\in U_0$ implies $F((KerF_0)^\perp)\oplus F_0(H_1)^\perp =H_2$

I didn't find this result in other books. I can't understand the proof about it. $Fv+w=F(v-f_0)+w$? Why?

Edit: $H_1$ and $H_2$ are separable Hilbert spaces.

$\mathcal {F}(H_1,H_2)$ is the spaces of Fredholm operators.

$\mathcal {B}(H_1,H_2)$ is the spaces of bounded operators.

In the proof, construct a $\overline{F}:H_1\oplus F_0(H_1)^\perp \to H_2\oplus kerF_0$ by $\overline{F}(v,w)=(Fv-w,\pi_{KerF_0}v)$, this is a isomorphism. Since $\overline{F}$ is onto, for any $(u,\ f_0)\in H_2\oplus kerF_0$, there is $(v,w)\in H_1\oplus F_0(H_1)^\perp$, with $u=Fv-w$ and $\pi_{KerF_0}v=f_0$.

$\pi_{KerF_0}: H_1\to KerF_0$