Let $A$ be separable $C^*$-algebra and $B$ a arbitrary $C^*$-algebra and $\operatorname{CPC(A,B)}:=\{f:A\to B | f\;\text{ is completely positive contractive}\}$. Let $\{a_1,a_2,,\}$ a countable dense subset in $A^1=\{a\in A:\|a\|\le 1\}$. For $f,g\in \operatorname{CPC(A,B)}$ define a metric $$d_B(f,g):=\sum\limits_{k=1}^\infty 2^{-k}\|f(a_k)-g(a_k)\|.$$ ( see my question metric for completely positive contractive maps ). I have several questions concerning the proof of theorem 2.1 in Arveson's paper notes "notes on the lifting theorem". The theorem states (I'm interested in a slightly different version for CPC-maps instead of UCP-maps, but there should be no difference in the proof):
Theorem: Let $A$ be a separable $C^*$-algebra, $B$ a arbitrary $C^*$-algebra and let $J$ be an ideal in $B$. Then the set of all liftable CPC-maps (with CPC-lifts) from $A$ to $B/J$ is $d_{B/J}-$closed in $\operatorname{CPC}(A,B/J)$.
Proof: Let $(\phi_n)_{n\in\mathbb{N}}$ be a sequence of liftable maps in $\operatorname{CPC}(A,B/J)$ ( i.e. for all $n\in\mathbb{N}$ there exists a CPC-map $\hat{\phi_n}:A\to B$ such that $\pi\circ \hat{\phi_n}=\phi_n$, where $\pi:B\to B/J$ is the canonical quotient map) and such that $(\phi_n)_{n\in\mathbb{N}}$ converges to a map $\phi\in \operatorname{CPC}(A,B/J)$ relative to $d_{B/J}$. We have to check that $\phi$ is liftable (and the lift is cpc). Since $(\phi_n)_{n\in\mathbb{N}}$ converges to a map $\phi\in \operatorname{CPC}(A,B/J)$ relative to $d_{B/J}$, we can arrange that $d_{B/J}(\phi_n, \phi)<\frac{1}{2^{n+1}}$ for $n\in\mathbb{N}$.
We claim that there is a sequence $(\psi_n)_{n\in\mathbb{N}} \subseteq \operatorname{CPC}(A,B)$ satisfying $\pi\circ \psi_n =\phi_n$ and $d_{B}(\psi_n, \psi_{n+1})<\frac{1}{2^{n}}$ for all $n\in\mathbb{N}$.
Indeed, let $\psi_1$ be any CPC-lifting of $\phi_1$. Assuming that $\psi_1,...,\psi_n$ have been defined and satisfy the stated conditions, choose any lifting $\lambda$ of $\phi_{n+1}$. We then have $d_{B/J}(\pi\circ\psi_n,\pi\circ \lambda )=d_{B/J}(\phi_{n},\phi_{n+1} )<\frac{1}{2^{n}}$. It's now possible to conclude that there is a CPC-map $\psi_{n+1}\in \operatorname{CPC}(A,B)$ satisfying $\pi\circ\psi_{n+1} = \pi\circ \lambda =\phi_{n+1}$ and $d(\psi_n, \psi_{n+1})<\frac{1}{2^{n}}$. Since $(\psi_n)$ is a cauchy-sequence in $\operatorname{CPC}(A,B)$ relative to $d_B$ and $\operatorname{CPC}(A,B)$ is $d_B$-complete, we can define a CPC-limit $\psi$ as the $d_B$-limit of $(\psi_n)$. Since $\pi\circ \psi_n=\phi_n$ and $(\psi_n)_{n\in\mathbb{N}}$ converges to $\phi$, while $\pi\circ \psi_n$ converges to $\pi\circ \psi$, $\psi$ is a lifting of $\phi$.
My questions:
In the induction basis, if we choose an arbitrary lifting $\psi_1$, why we can arrange that $d_{B}(\psi_n, \psi_{n+1})<\frac{1}{2^{n}}$ for $n\in\mathbb{N}$, or repsectively how to arrange that $d_{B}(\psi_n, \psi_{n+1})<\frac{1}{2^{n}}$ for $n\in\mathbb{N}$?
Why is $\operatorname{CPC}(A,B)$ is $d_B$-complete?
My thoughts to my second question (can you check if everything is correct?): As stated here metric for completely positive contractive maps for nets $(f_{\lambda})_\lambda\subseteq \operatorname{CPC(A,B)}$, it is
$$d_B(f_{\lambda},f)\to 0 \iff \|f_\lambda (a_k)-f(a_k)\|\to 0\ \text{ for all k}\in \mathbb{N}.$$ It follows that a sequence $(\epsilon_n)\subseteq \operatorname{CPC(A,B)}$ is $d_B$-cauchy iff $(\epsilon_n(a))\subseteq B$ is a Cauchy sequence for fixed, but arbitrary $a\in A$. Since $B$ is complete, for all $a\in A$ the limit $$\|\enspace \|_B-\lim\limits_{n\to\infty} \epsilon_n(a)=\epsilon(a)$$ in $B$ exists. Thus we can define a map $\epsilon :A\to B$ which is positive. That $\epsilon$ is completely positive I would argue similar. And $\epsilon$ is contractive, since for all $a\in A$ it is $$\|\epsilon (a)\|=\lim\limits_{n\to\infty}\|\epsilon_n(a)\|\le \limsup\|\epsilon_n\|\|a\|\le \|a\|.$$ Is it correct?
A appreciate your help.
In your copy of Arveson's text, you wrote "It's now possible to conclude that..." instead of the original "Lemma 1.2 implies that..." Maybe less obvious is that there is a typo in Arveson's text and it should have said "Lemma 2.2".
Lemma 2.2 is the exact statement that you are skipping in your text. Given $\psi_1,\lambda:A\to B$, and $\varepsilon>0$, there exists ucp $\psi_2:A\to B$ such that $$d(\psi_1,\psi_2)\leq d(\pi\circ\psi_1,\pi\circ\lambda)+\varepsilon.$$
Arveson claims that the details for the estimate in the proof of this lemma are in his 1977 Duke paper. I may have a paper copy of that paper somewhere, but don't have electronic access to it.
Your argument for completeness is correct. It is the standard one every time a topology on functions is defined by pointwise convergence in some other topology.