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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 $\pi_{KerF_0}v=f_0$.

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

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  • $\begingroup$ While it is probably obvious for people in the field, you should really define all your notations why asking a question on MO. Just so that readers don't have to guess that you meant $\mathcal{F}$ to be the Fredholm operators and $\mathcal{B}$ to be the bounded operators. $\endgroup$
    – Willie Wong
    Commented Aug 26, 2010 at 12:59
  • $\begingroup$ ... I meant "when" instead of "why" above, of course. $\endgroup$
    – Willie Wong
    Commented Aug 26, 2010 at 13:00
  • $\begingroup$ I agree with Willie. The question should be reformulated so as to clarify all the notations used. $\endgroup$
    – André Henriques
    Commented Aug 26, 2010 at 13:14
  • $\begingroup$ Thanks, Willie Wong and André Henriques, I edit my question. $\endgroup$
    – Chen
    Commented Aug 26, 2010 at 13:39
  • $\begingroup$ And what is it that you don't understand? The only slightly non-trivial step I see in your sketch of the proof is the fact that $\bar{F}$ is an isomorphism. Once you have that, picking $f_0 = 0$ you see that every element of $H_2$ can be written as $Fv + w$ where $v\in H_1$ and $w\in F_0(H_1)^\perp$. On the other hand, by definition $\bar{F} |_{(ker F_0)^\perp\oplus F(H_1)^\perp}$ clearly has range contained in $H_2$. Hence your corollary. $\endgroup$
    – Willie Wong
    Commented Aug 26, 2010 at 14:46

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