Let $X_1,X_2$ be Banach spaces and $P:X_1\rightarrow X_2$ be a Fredholm operator of index 0. Then I want to show that $P=E+F$ where $E$ is invertible and $F$ has finite rank. I wanted to use the description via Grushin problems but dont know how to use the index 0 here.
Any hints will be gratefully welcome, thanks in advance.
Since the index of $P$ is $0$, $dimKer P=codim ImP$. Let $(e_1,...,e_n)$ a basis of $KerP$. Let $p:X_2/Im(F)$ the quotient map and $(f_1,...,f_n)\in X_2$ such that $(p(f_1),...,p(f_n))$ is a basis of $X_2/Im(P)$. The linear form $a_i$ defined on $Vect(e_1,..,e_n)$ by $a_i(e_i)=1, a_i(e_j)=0, i\neq j$ can be extended to $X_2$ in $b_i$ by Hahn Banach.
Let $U=\cap_{i=1}^{i=n}Ker b_i$, $X_1=U\oplus Ker P$. To see this, write $x=(x-b_1(x)e_1+...+b_n(x)e_n)+b_1(x)e_1+..+b_n(x)e_n$, $b_i(x-b_1(x)e_1+..+b_n(x)e_n))=0$. Then show that $U\cap Ker P=0$.
Consider $E:X_1\rightarrow X_2$ defined by $E(x)=P(x)+b_1(x)f_1+..+b_n(x)f_n$. $E$ is injective: To see this, remark that $E(x)=0$ implies that $P(x)=-(b_1(x)f_1+...+b_n(x)f_n)\in ImP$, this implies that $p(b_1(x)f_1+...+b_n(x))f_n=b_1(x)p(f_1)+...+b_n(x)p(f_n)=0$, we deduce that $b_i(x)=0, i=1,..,n$ since $p(f_1),..,p(f_n)$ is a basis of $X_2/ImP$, this implies that $P(x)=0$ and $x=x_1e_1+..+x_ne_n$, we deduce that $E(x)=x_1f_1+...+x_nf_n=0$ and $x_1=..=x_n=0$.
$E$ is surjective. Let $y\in X_2$, $y=y_1+y_2, y_1\in ImP$, $y_2\in Vect(f_1,..,f_n)$. There exists $x_1\in X_1$ such that $P(x_1)=y_1$ since $ImP=ImP_{P\mid U}$, we can assume that $y_1\in U$, write $y_2=y_1f_1+...+y_nf_n, E(x_1+y_1e_1+...+y_ne_n)=y$. The Open mapping theorem implies that $E$ is invertible.
This implies that $P=E+F$ where $F(x)=-b_1(x)f_1-..-b_n(x)f_n$.