In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following:
Let $X$ be a Banach space, $T:X \to X$ be a compact operator and $A:=T-I$. We proved that $\mbox{ker}(A)< \infty$ and that there is a closed subspace $V\subset X$ such that $X=V\oplus \mbox{ker}(A)$. Why does it follow that
$X/V \cong \mbox{ker}(A)?$
Thanks in advance!
Best Answer
Consider the linear operators $$i: \ker A \to X: x \mapsto x$$ and $$\pi: X \to X/V: x \mapsto [x].$$ $i$ is the inclusion map and $\pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y \in \ker \pi$ iff $y \in V$ and equivalent $y \notin \ker A$ as by assumption $X = \ker A \oplus V$.
The composition $\Phi := \pi \circ i$ is the isomorphism you are looking for: Let $x \in \ker \Phi$. Then $\pi (ix) = [0]$ and therefore $ix \in V$. Since $X = \ker A \oplus V$ this means $x = 0$ and hence $\Phi$ is injective.
It is also surjective. Let $[x] \in X/V$. If $[x] = [0]$ then $\Phi(0) = [x]$. If $[x] \neq [0]$ then $x \notin V$ and therefore $x \in \ker A$ and $\Phi(x) = [x]$.