Let $X$ be Banach space, $N$ a (closed) subspace of $X$ and $\pi\in X\to X/N$ the natural quotient map.
Call $Y\subseteq\mathbb{Z}\to X\sqcup X/N$ the set of functions on $\mathbb{Z}$ such that $f\in Y$ iff
- $f$ is bounded,
- $f(n)$ is in $X$ for $n\leq 0$, and
- $f(n)$ is in $X/N$ for $n>0$.
Consider the (densely defined) operators $A$ and $B$ on $Y$ defined by: \begin{gather*} (Af)(n)=\begin{cases}f(n) & n=1 \\ 0 & n\neq1\end{cases} \\ (Bf)(n)=\begin{cases}\pi(f(0)) & n=1 \\ f(n-1) & n\neq1\end{cases} \end{gather*} where $\pi$ is the natural map from $X$ to $X/N$.
I am confused about two things. First, why is $R(A)\subset R(B)$?
Second: Assume that there exists an operator $C$ on $Y$ such that $A=BC$. My book says:
Let $D_1$ be the map from $X/N$ to $Y$ and $D_2$ the map from $Y$ to $X$ defined by \begin{gather*} (D_1x)(n)=\begin{cases}x&n=1 \\ 0&n\neq1\end{cases} \\ D_2f=f(0) \end{gather*} Then $E=D_2CD_1$ is a map from $X/N$ to $X$ such that $I-E\pi$ is a bounded projection of $X$ onto $N$. Thus if we choose $N$ to be a subspace for which no bounded projection exists, then we arrive at a contradiction and see that there exists no operator $C$ on $F$ for which $A=BC$.
I cannot make heads nor tails of this quote. What does it mean?