oa.operator algebras - A result about Fredholm operator

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

If every $F_0in mathcal {F}(H_1,H_2)$, there is an open neighborhood $U_0subseteq mathcal {B}(H_1,H_2)$, such that $Fin U_0$ 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 $overline{F}:H_1oplus F_0(H_1)^perp to H_2oplus kerF_0$ by
$overline{F}(v,w)=(Fv-w,pi_{KerF_0}v)$, this is a isomorphism. Since $overline{F}$ is onto, for any $(u, f_0)in H_2oplus kerF_0$, there is $(v,w)in H_1oplus F_0(H_1)^perp$, with $u=Fv-w$ and $pi_{KerF_0}v=f_0$.

$pi_{KerF_0}: H_1to KerF_0$

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