Given the Calkin Algebra $\pi: B(E) \rightarrow B(E)/K(E)$ we know that $T \in B(E)$ is Fredholm iff $\pi(T)$ is invertible.
I am missing some intuition here. Is there a way to say how such an equivalence class (coset) $\pi(T)$ looks like explicitly (or some example)?
Also we know that $T \in B(E)$ is Fredholm iff there is an operator S such that $ST=I+C_{1}$ and $TS=I+C_2$ with $C_{1},C_2\in K(E)$. My question is - are the compact operator $C_{1}$ and $C_2$ telling us something about in which equivalence class the Fredholm operator lies?
Thanks to everyone that could help me to shed some light on this.
EDIT: Also how does the identity look like in the Calkin Algebra?
There is no uniqueness in what $C_1$ and $C_2$ are, because there is no uniqueness for $S$. Given $S$ with $ST=I+C_1$ and $TS=I+C_2$, take $K$ any compact operator and form $S'=S+K$. Then $$ S'T=ST+KT=I+(C_1+K),\qquad\qquad TS'=TS+TK=I+(C_2+K). $$ In other words, you are completely free to prescribe at least one of $C_1,C_2$.
I don't think you can expect to get much intuition about what $\pi(T)$ is. After all, the Calkin algebra is what you get when you forget everything finite-dimensional about $B(H)$. What I mean is this. Let $f$ be any linear functional on $B(H)/K(H)$. Then you can define a linear functional $f'$ on $B(H)$ by $f'(T)=f(\pi(T))$. This linear functional $f'$ is zero on every compact operator. In particular it cannot depend in any reasonable way on the entries of $T$ with respect to a basis.