# Isomorphisms in the proof of the Fredholm alternative/Theorem of Riesz-Schauder (for compact operators)

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

$$X/V \cong \mbox{ker}(A)?$$

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 $$\Phi(x) = [x]$$.