I have a question of an explicit example of a busby invariant of a extension, which can be found in Blackadars book "K-theory for Operator Algebras". 



> 
> Let $0\to B\to E\to A\to 0$ be a short exact sequence of $C^\*$-algebras, in particular $A=\mathbb{C}$, $B=C\_0((0,1))$. Now there are four possible choices of $E$, namely $C\_0((0,1))\oplus \mathbb{C}, C\_0((0,1]), C\_0([0,1)), C(S^1)$. 
> 
> 
> 


Why only these 4 $C^\*$-algebras, what is with $C^\*$-algebras which are isomorphic to one of the possibilities enumerated above?


A busby invariant of this extension is a $\ast$-homomorphism $\tau:E/B\cong A\to M(B)/B$, where $M(B)$ is the multiplier algebra of $B$ and this homomorphism comes from the map $E\to M(B)$ (which arises from the universal property of multiplier algebras) composed with the quotient map $M(B)\to M(B)/B$.


For the extension above it is $M(B)\cong C(\beta\mathbb{R})$, where $\beta\mathbb{R}$ is the Stone-Čech compactification of $\mathbb{R}$ (with euclidean topology) and $M(B)/B\cong C(\beta \mathbb{R}\setminus \mathbb{R})$.
I.e. $\tau$ must be a $\*$-homomorphism $\tau:\mathbb{C}\to C(\beta \mathbb{R}\setminus \mathbb{R}) $ and for every choice of $E$ we get a specific $\tau$. 


But (if we are in the situation that we do not know how many possible choices of $E$ we have), why are there only 4 possibilities of $\*$-homomorphism $\tau:\mathbb{C}\to C(\beta \mathbb{R}\setminus \mathbb{R}) $ ? We have to determine the images of $1\in \mathbb{C}$ in every case, for example one possible choice of $\tau$ is if we set $\tau (1)=0$, but in the book the author enumerates the other possibilities $1$ sending to the characteristic function of the component at $-\infty$, the characteristic function of the component at $\infty$ and to $1$. I don't see why this should be all possibilities. 


In addition to that, which of the choices of $\tau$ corresponds to $C(S^1)$ for example? 


Regards



busby invariant of extensions of $C^*$-algebras
The idempotents in the corona algebra are very restricted, because they lift to functions$\newcommand{\Real}{{\bf R}}\newcommand{\veps}{\varepsilon}$ $f\in C\_b(\Real)$ such that $f^2-f\in C\_0(\Real)$.


Fix such an $f$. Note that if $z$ is a complex number such that $|z^2-z|= |z(z-1)|$ is small (say less than $\veps$ where $\veps<1/4$) then $\min(|z|, |z-1|)$ is small. (I think you can get an upper bound of $4\veps$, but the upper bound $\veps^{1/2}$ is trivial and easy.) Using this, you see that outside some compact subset of $\Real$, $f$ is "close" to being $0-1$ valued. By connectedness, this means that $f(t)$ converges to $0$ or $1$ as $t\to+\infty$, and similarly as $t\to -\infty$. This gives you the four possibilities mentioned in your source; note that within each case you get an infinite number of $f$ satisfying those boundary conditions, but they all define the same Busby map, and hence the same extension.


(This is related to the notion of *ends* of topological spaces: $\Real$ has 2 ends, so you get $2^2$ possibilities; for $n\geq 2$, $\Real^n$ has only 1 end, so the analogous question with $\Real$ replaced by $\Real^n$ would have $2^1$ possibilities.)


As for your final question about $C(S^1)$, I think this is an instructive exercise to think about oneself. Given my previous remarks it shouldn't be too hard.


