This is from Higson's Analytic K-Homology.
Let $X$ be a non-empty and compact subset of $\mathbb{C}$. An index function for $X$ is an integer-valued function on the set of bounded components of the complement of $X$ in $\mathbb{C}$.
Brown-Douglas-Fillmore Theorem: If $X$ is any non-empty and compact subset of $\mathbb{C}$ then the semigroup Ext $(X)$ is a group. The map which associates to $[T] \in \operatorname{Ext}(X)$ the index function of $T$ is an isomorphism from Ext $(X)$ to the group of all index functions for $X$.
Using BDF theorem, I wish to prove $\text { Index : } \operatorname{Ext}\left(\mathrm{S}^1\right) \rightarrow \mathbb{Z} $ is an isomorphism.
In the above definition of index function, putting $X=S^1$. So the interior ${S^1}^\circ$ is the only bounded component. How do I prove the group of integer valued functions on ${S^1}^\circ$ is $\mathbb{Z}$?
Any help is appreciated.
The definition of index function says that a bounded component is mapped to a single integer. Since the set of bounded components is $\{{S^1}^\circ\}$, we have $\mathbb{Z}$-many choices to map ${S^1}^\circ$ to.