I am reading a book C*-algebra and finite-Dimensional Approximations. There are two conclusions (in the book) below.
Corollary 1.5.16. Let $E\subset A$ be an operator subsystem and $\phi: E \rightarrow M_{n}(C)$ be a completely positive map. THen $\phi$ extends to a completely positive map $A \rightarrow B(H)$. $~$($M_{n}(C)$ denotes the sets of all complex matrix.)
Theorem 1.6.1 Let $A$ be a unital C*-algebra and $E\subset A$ be an operator subsystem. Then, every contractive completely positive map $\phi:E\rightarrow B(H)$ extends to a contractive completely positive map $\bar{\phi}: A\rightarrow B(H)$.
Proof. Let $P_{i}\in B(H)$ be an increasing net of finite-rank projections which converge to the identity in the strong operator topology. For each $i$, we regard the contractive completely positive map $\phi_{i}: E \rightarrow P_{i}B(H)P_{i}$, $\phi_{i}(e)=P_{i}\phi(e)P_{ i}$ as taking values in a matrix algebra. Thus , by Corollary 1.5.16, we may assume that each $\phi_{i}$ is actually defined on all of $A$. Now we regard $\phi_{i}$ as taking values in $B(H)$ and apply compactness of the unit ball of $B(A,B(H))$ in the point-ultraweak topology to find a cluster point $\Phi:A:\rightarrow B(H)$. It is readily verified that $\Phi$ is completely positive and extends to $\phi$.
My question is how to explain take values of $\phi_{i}$ in a matrix algebra and, meanwhile, take values in $B(H)$ in the proof? ($B(H)$ denotes all the linear bounded operators on $H$)
The proof is making (liberal) use of the following identifications.
If $P$ is a finite-rank projection in $B(H) $, say of rank $n$, then $PH\simeq\mathbb C^n$ and $P\, B(H)\,P\simeq M_n(\mathbb C) $.
To see this last identification, construct a basis $\{e_i\}$ of $H$ such that $e_1,\ldots,e_n$ form a basis of $PH$. Construct the corresponding matrix units $E_{kj}\in B(H)$, i.e. $E_{kj}e_i=\delta_{ji}e_k$.
Note that $P=\sum_jE_{jj}$. Now consider a map $f:M_n(\mathbb C)\to P\,B(H)\,P$, by $$ f([c_{kj}])=\sum_{j,k=1}^nc_{kj}E_{kj}. $$ Exercise: show that $f$ is a $*$-homomorphism.
Exercise 2: show that $f$ is injective.
To see that $f$ is onto, show that for any $T\in B(H)$, $E_{kk}TE_{jj}=\langle Te_j,e_k\rangle\,E_{kj}$. So, if $T\in P\,B(H)\,P$, $$ T=PTP=\left(\sum_{k=1}^nE_{kk}\right)T\left(\sum_{j=1}^nE_{jj}\right)=\sum_{k,j=1}^nE_{kk}TE_{jj}=\sum_{k,j=1}^n\langle Te_j,e_k\rangle\,E_{kj}=f([\langle Te_j,e_k\rangle]). $$