Universal $C^\ast$-algebra generated by the unilateral shift. Consider the Hilbert space $L^2(\mathbb{S}^1)$ and the Hardy subspace $\mathcal{H}(\mathbb{S}^1)$, along with the projection $\pi:L^2(\mathbb{S}^1)\to\mathcal{H}(\mathbb{S}^1)$. For each $f\in C(\mathbb{S}^1)$, let $T_f\in B(\mathcal{H}(\mathbb{S}^1))$ be the associated Toeplitz operator.
Let now $\mathcal{T}$ be the $C^\ast$-algebra generated by $\{T_f:f\in C(\mathbb{S}^1)\}$ and the compact operators $K$ on $\mathcal{H}(\mathbb{S}^1)$, called the Toeplitz algebra, which fits into the short exact sequence  $0\to K\to\mathcal{T}\overset{\sigma}{\to} C(\mathbb{S}^1)\to0$.
There is a couple of things I don't understand here (this is from the notes from my teacher):
How is this map $\sigma$ defined, and why is this sequence in fact short exact?
More importantly, the teacher mentions that $\mathcal{T}$ is in fact the universal $C^\ast$-algebra generated by $s$ and $s^\ast$ satisfying $s^\ast s=1$, and he hints at $S$ being the unilateral shift on $\ell^2(\mathbb{Z}_{\ge0})\cong\mathcal{H}(\mathbb{S}^1)$. How can one see/prove this?
 A: The map $\sigma$ is the quotient map by the compacts. It is not hard to see that 
$$
\mathcal T=\{T_f+K:\ f\in C(\mathbb S^1),\ K\ \text{ compact }\}=C^*(\{T_z\}).
$$
For this, one can show that $T_f$ is never compact, and that $T_fT_g-T_{fg}$ is always compact. Thus $\sigma $ is defined by $\sigma(T_f+K)=f$. From $T_f-T_g=T_{f-g}$, one can see that $\sigma$ is well-defined, and the other aforementioned properties, that it is a $*$-homomorphism. 
Here is a sketch of the proof that $\mathcal T=C^*(\{T_z\})$. Write $\{e_n\}$ for the canonical orthonormal basis of $L^2(\mathbb S^1)$. 


*

*Show that $C^*(T_z)$ is irreducible. If a projection $P$ commutes with $T_z$, it commutes with $I-T_zT_z^*$, which is the projection onto the span of the constant function $e_0=1$. So $Pe_0=\lambda\,e_0$, with $\lambda=0$ or $\lambda=1$. In any case, $$Pe_n=PT_z^ne_0=T_z^nPe_0=\lambda T_z^ne_0=\lambda e_n.$$ Thus either $P=0$ of $P=I$. 

*Show that $C^*(T_z)$ contains the compact operators. Since $C^*(T_z)$ contains a compact operator ($I-T_zT_z^*$, for instance), and it is irreducible, we get that $C^*(T_z) $ contains the compact operators (this is definitely not a trivial result, it requires some work). 

*Show that $C^*(T_z)$ contains $\{T_f:\ f\in C(\mathbb S^1)\}$. These are limits of linear combinations of powers of $T_z$. 

*The steps above show that $\mathcal T\subset C^*(T_z)$. The reverse inclusion follows from the fact that $T_fT_g-T_{fg}$ is compact. 
As for your second question, the Wold Decomposition lets you write any isometry as $$U\oplus \bigoplus_j S,$$ where $U$ is a unitary and $S$ the unilateral shift on $\ell^2(\mathbb N)$. We can then replace $S$ with $T_z$ (they are unitarily equivalent). As the universal C$^*$-algebra generated by a unitary is also $C(\mathbb S^1)$, we get that the C$^*$-algebra generated by any isometry is a quotient of $\mathcal T$. 
