Let $H$ be a separable infinite dimensional Hilbert space. Denote the Calkin algebra by $Q(H)=B(H)/K(H)$, and $U(Q(H))$ the group of unitaries in $Q(H)$.

I'm trying to show that the map $F: U(Q(H))/U(Q(H))_0 \to \mathbb{Z}$ (where $U(Q(H))_0$ denotes the identity component) defined by $F((A+K(H))+U(Q(H))_0)=index(A)$, where $A+K(H)$ is unitary in $Q(H)$, is well-defined, surjective and injective.

First, as $A+K(H)\in U(Q(H))$ in particular it is invertible in $Q(H)$ and by Atkinson Theorem $A$ is a Fredholm operator and $\forall k\in K(H), index(A)=index(A+k)$ is well-defined.

For well-definiteness of $F$ it is sufficient to check that if $A+K(H)$ is in the connected component of the identity, i.e. there is a path of unitaries in $Q(H)$ between $A+K(H)$ and $1_{B(H)}+K(H)$ then $index(A)=0$.
I couldn't show that. I know that if $F(H)$ denotes the Fredholm operators in $B(H)$ then $F_1,F_2 \in F(H)$ are in the same connected component iff they have the same index. But it's just in $B(H)$.

Also, I couldn't show $F$ is injective, i.e. if $index(A)=0$ then $A+K(H)\in U(Q(H))_0$.
I know that if $index(A)=0$ then $A$ is a perturbation of invertible operator in $B(H)$ by a compact operator. I also know that by the polar decomposition $|A|$ is unitary and that $GL(B(H))$ is path-connected. If I could take a unitary from the class of $A$ in $Q(H)$ then I could construct a path of invertible elements between $1$ and that unitary. By the polar decomposition we even can construct a path of unitaries between them, that could help.

Surjectivity is OK, just by taking the class of the unilateral shift in the calkin algebra. There it is unitary and its index equals to $-1$, so we can reach any element of $\mathbb{Z}$.

Any help will be appreciated.


These results follow from some quite general results about surjective homomorphisms between $C^{\ast}$-algebras which I will state without proof. In the following, $A$ and $B$ are unital $C^{\ast}$-algebras and $\varphi :A\to B$ is a unital surjective $\ast$-homomorphism.

[Proposition 4.3.14 in Higson & Roe's Analytic K-homology ] If $f: [0,1] \to U(B)$ is a continuous path of unitaries in $B$, then $\exists g:[0,1] \to U(A)$ continuous such that $\varphi\circ g = f$.

This answers your first question (well-definedness): if $A + K(H) \in U(Q(H))$ is connected to $1 +K(H)$ by a path of unitaries, then $A \sim_h 1_{B(H)}$ in $U(B(H))$, and so $ind(A) = 0$

For your second question (injectivity), you are on the right track: If $A + K(H) \in U(Q(H))$ satisfies $ind(A) = 0$, then $A\sim_h 1_{B(H)}$ in $F(H)$, so $A+K(H) \sim_h 1_{B(H)} + K(H)$ in $GL(Q(H))$. However, we have the following fact (which uses the polar decomposition to prove that $U(B)$ is a deformation retract of $GL(B)$):

[Proposition 2.1.8 in Rordam/Larsen/Laustsen's Introduction to K-theory] If $u,v\in U(B)$ such that $u\sim_h v$ in $GL(B)$, then $u\sim_h v$ in $U(B)$.


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