Question about Fredholm operator $X,Y$ are Banach spaces and $A\in B(X,Y)$ is a Fredholm operator (that is, the dimensions of ker($A$) and coker($A$) are both finite), then are closed linear subspaces ker($A$) and Im($A$) complemented? (A closed linear subspace $H$ in a Banach space $Z$ is called complemented iff there is a closed linear subspace $G$ such that $H+G=Z$ and $H \cap G=0$)
 A: Yes.


*

*A finite-dimensional subspace $E$ of a Banach space $X$ is closed. Choose a basis $e_{1},\ldots,e_{n}$ of $E$, use Hahn–Banach to extend the dual functionals $\varphi_{i} :E \to \mathbb{R}$ determined by $\varphi_i(e_j) = \delta_{ij}$ to continuous linear functionals $e_{i}^\ast: X \to \mathbb{R}$ and check that $P : X \to X$ given by $Px = \sum_{i=1}^n e_{i}^\ast(x)\cdot e_i$ is a projection of norm $\|P\| \leq n$ and range $E$. 
Note that every $x \in X$ can be uniquely written as $x = Px + (1-P)x$ and that $F = \ker{P} = \operatorname{im}(1-P)$ so that $E \oplus F \to X$ given by $(e,f) \mapsto e+f$ is a continuous linear map with inverse $(Px,(1-P)x)$. In other words, every finite-dimensional subspace of a Banach space is complemented.

*Every closed subspace $E$ of finite co-dimension in a Banach space $X$ is complemented.
To see this, choose a basis $e_{1},\ldots,e_{n}$ of the Banach space $X/E$, choose pre-images $f_{1},\ldots,f_n$ of $e_1,\ldots,e_n$ under the canonical projection $X \to X/E$ and note that the linear span $F$ of $f_1,\ldots,f_n$ is a complement of $E$.

*If the range $R = T(E)$ of a bounded linear operator $T: E \to F$ between Banach space has finite codimension in $F$ then $R$ is closed.
To see this, consider the Banach space $\bar{E} = E/\ker{T} = \operatorname{coim}{T}$ and factor $T$ over the injective bounded operator $\bar{T}: \bar{E} \to F$. Note that $R = T(E) = \bar{T}(\bar{E})$. Choose a basis $f_{1},\ldots,f_{n}$ of an algebraic complement of $R$ and consider the operator
$$
S: \bar{E} \oplus \mathbb{R}^n \to F,\quad S(\bar{e},x) \mapsto \bar{T}(\bar{e}) + \sum_{i=1}^n x_i f_i.
$$
Now observe that $S$ is a continuous bijection, hence a homeomorphism by the open mapping theorem and conclude by observing that $R = S(\bar{E} \oplus 0)$ is the image of a closed subspace of $\bar{E} \oplus \mathbb{R}^n$, hence $R=T(E)$ is closed (of course, this argument also shows that $R = T(E)$ is complemented in $F$). 
Now let $T:E \to F$ be a bounded operator with finite-dimensional kernel and whose range has finite co-dimension in $F$. By point 1. $K = \ker{T}$ is closed and complemented, by point 3. $T(E)$ is closed and therefore by 2. or otherwise $T(E)$ is complemented in $F$.
This implies that every Fredholm operator $T: E \to F$ is isomorphic to the operator
$$
K \oplus \bar{E} \longrightarrow T(E) \oplus C, \quad (k,\bar{e}) \longmapsto (\bar{T}\bar{e},0)
$$
where $\bar{E} = E/\ker{T}$, $C = F/T(E)$ and $\bar{T}:\bar{E} \to T(E)$ is an isomorphism of Banach spaces. This is often called the canonical factorization of a Fredholm operator.
In more algebraic terms, a Fredholm operator induces two split short exact sequences
$$
\ker{T} \rightarrowtail E \twoheadrightarrow \operatorname{coim}T\quad\text{ and }
\quad\operatorname{im}T \rightarrowtail F \twoheadrightarrow \operatorname{coker}{T},
$$
and, also contrary to general operators, the induced map $\bar{T}:\operatorname{coim}T \to \operatorname{im}T$ is an isomorphism of Banach spaces. These split exact sequences and the isomorphism $\bar{T}$ are ultimately the reason for index theory to work the way it works and why “Fredholm operators behave very much like matrices”.
One of the best sources on basic Fredholm theory I know is in one of the first chapters of:


*

*Richard S. Palais: Seminar on the Atiyah–Singer Index Theorem, Annals of Mathematics Studies 57, Princeton University Press, Princeton NJ 1965,


a slightly expanded form of which can be found in in Hirzebruch–Scharlau's Einführung in die Funktionalanalysis if you read German.
