I am working through Rordam's book on $K$-Theory for C$^{*}$-algebras and I am stuck on one detail in exercise 8.17.
Let $A$ be a unital C$^{*}$-algebra. In this book, the stabilization $\mathcal{K}A$ of $A$ is defined to be the limit of the following inductive sequence:
$$ A\overset{\varphi_{1}}{\longrightarrow}M_{2}(A)\overset{\varphi_{2}}{\longrightarrow}M_{3}(A)\overset{\varphi_{3}}{\longrightarrow}\cdots\longrightarrow (\mathcal{K}A,\{\kappa_{n}\}_{n=1}^{\infty}), $$
where $\phi_{n}\colon M_{n}(A)\to M_{n+1}(A)$ is given by $\varphi_{n}(a)=\pmatrix{a & 0\\ 0 & 0}$, and the $\kappa_{n}\colon M_{n}(A)\to\mathcal{K}A$ are injective.
Let $\mathcal{U}_{n}(A):=\{u\in M_{n}(A):u\text{ is unitary}\}$ and $e_{n}=\kappa_{n}(1_{n})$, and define $\rho_{n}\colon\mathcal{U}_{n}(A)\to\mathcal{U}(\widetilde{\mathcal{K}A})$ by $\rho_{n}(u)=\kappa_{n}(u)+1-e_{n}$.
For a unitary of the form $v=(b,1)\in\widetilde{\mathcal{K}A}$, I was able to prove that
$$ \lim_{n\to\infty}\|v-(e_{n}ve_{n}+(1-e_{n}))\|=0\qquad(*) $$
However, the exercise then states to use $(*)$ and exercise 2.8 to prove the following:
There is an $n\in\mathbb{N}$ and a $u\in \mathcal{U}_{n}(A)$ such that $v\sim_{h}\rho_{n}(u)$.$\qquad(**)$
This is the part I am unable to figure out. I know that two given two unitaries $w_{1},w_{2}$, then $w_{1}\sim_{h} w_{2}$ if $\|w_{1}-w_{2}\|<2$, but I am really unsure how to piece this, $(*)$, and exercise 2.8 together to prove $(**)$.
Below, I have copied exercise 2.8 for reference:
As a final note, I should point out that the ultimate purpose of this problem is to prove that $\mathcal{U}(\widetilde{\mathcal{K}A})/\mathcal{U}_{0}(\widetilde{\mathcal{K}A})\cong\mathcal{U}_{\infty}(A)/\sim_{1}\cong K_{1}(A)$, which, assuming $(**)$, I was able to prove.
Note that $e_n b e_n$ is in the image of $\varphi_n$ and that $e_nbe_n\to b$ (this is $(*)$ expressed in another way). Further since $1+b$ is unitary in $\widetilde{\mathcal {KA}}$ you have that $1+e_nbe_n$ converges to a unitary and as such $$(1+e_nb e_n)^*(1+e_nbe_n) = 1 + e_n (b+b^*+ b^* e_nb )e_n$$ converges to $1$, ie $e_n(b+b^*+b^*e_nb)e_n\to 0$ and $\|e_n-(e_n+e_nbe_n)^*(e_n+e_nbe_n)\|\to 0$, the same with $e_n-(e_n+e_nbe_n)(e_n+e_nb e_n)^*$. Replace every $e_n$ by $1_n$ and apply the exercise in the picture to find an $n$ and a unitary $u$ in $M_n(\mathcal A)$ for which $\|u-(e_n+e_nbe_n)\|$ is arbitrarily small.
Then $\rho_n(u) = u+(1-e_n)$ is unitary in $\widetilde{\mathcal{KA}}$ and $$\|\rho_n(u) - (1+b)\| ≤ \|u-(e_n+e_nbe_n)\|+\|e_nbe_n-b\|.$$ Now both terms can be made arbitrarily small in particular $<2$, which by the criterium in the question implies $\rho_n(u) \sim_h (1,b)$.