Why does a unitary in the Calkin algebra always lift to an (co-)isometry? Cf. the the title, consider a separable infinite-dimensional Hilbert space H, and the short exact sequence $$0 \to \mathcal{K}(H) \to \mathcal{B}(H) \to \mathcal{Q}(H) \to 0,$$where $\mathcal{K}(H)$ is the compact operators and $\mathcal{Q}(H)$ is the quotient, also known as the Calkin algebra.
Let $u$ be a unitary in $\mathcal{Q}(H)$. This means that $u =  T + \mathcal{K}$, for a $T \in \mathcal{B}(H)$ where the differences $TT^* - I$ and $T^*T - I$ are compact. How can one show that $u$ lifts to an isometry, or a co-isometry? That is, why is there an isometry (or co-isometry) $S \in \mathcal{B}(H)$ such that $\pi (S) = u$, where $\pi$ denotes the quotient mapping? (Equivalently $S - T$ is compact)
This is an exercise in Rørdam's book on K-theory. In the previous part of the exercise one shows that whenever $E$ and $F$ are projections in $\mathcal{B}(H)$ with $Rank(E) \leq Rank(F)$ there exists a partial isometry $V$ with $V^*V = E$ and $VV^* \leq F$. I am assuming that one should use this somehow, possibly together with some results about the index map $\delta_1$, but I am really stuck on this one. Any help is appreciated!

If $u \in Q(H)$ is unitary, let $T_1 \in B(H)$ such that $\pi(T_1) = u$, then consider
$$
T = T_1R^{-n}
$$
where $R$ denotes the left or right shift (which exists on your separable infinite dimensional Hilbert space) so that $T$ has index zero. Now $\exists S_1 \in GL(B(H))$ such that
$$
T-S_1 \in K(H)
$$
Hence, $S:=S_1R^n$ satisfies
$$
T_1 -S\in K(H)
$$
and is an isometry or co-isometry.
